Answer:
Explanation:
a ) Height to be cleared = 5 - 1.6 = 3.4 m
Horizontal distance to be cleared = 5 m .
angle of throw = 56°
here y = 3.4 , x = 5 , θ = 56
equation of trajectory
y = x tanθ - 1/2 g ( x/ucosθ)²
3.4 = 5 tan56 - 1/2 g ( 5/ucos56)²
3.4 = 7.4 - 122.5 / .3125u²
122.5 / .3125u² = 4
u² = 98
u = 9.9 m /s
Range = u² sin 2 x 56 / g
= 9.9 x 9.9 x .927 / 9.8
= 9.27 m
horizontal distance be-yond the fence will the rock land on the ground
= 9.27 - 5
= 4.27 m
Answer:
a) [tex]u=13.3032\ m.s^{-1}[/tex]
b) [tex]s=3.7597\ m[/tex]
Explanation:
Given:
horizontal distance between the fence and the point of throwing, [tex]r=14\ m[/tex]
height of the fence from the ground, [tex]h_f=5\ m[/tex]
height of projecting the throw above the ground, [tex]h'=1.6\ m[/tex]
angle of projection of throw from the horizontal, [tex]\theta=56^{\circ}[/tex]
Let the minimum initial speed of projection of the throw be u meters per second so that it clears the top of the fence.Now the effective target height, [tex]h=h_f-h'=5-1.6=3.4\ m[/tex]The horizontal component of the velocity that remains constant throughout the motion:
[tex]u_x=u\cos\theta[/tex]
Now the time taken to reach the distance of the fence:
use equation of motion,
[tex]t_f=\frac{r}{u_x}[/tex]
[tex]t_f=\frac{14}{u.\cos56}[/tex] .................................(1)
Now the time taken to reach the fence height (this height must be attained on the event of descending motion of the rock for the velocity to be minimum).
Maximum Height of the projectile:
[tex]v_y^2=u_y^2-2\times g.h_m[/tex]
[tex]h_m=\frac{u_y^2}{19.6}[/tex] ........................(4)
Now the height descended form the maximum height to reach the top of the fence:
[tex]\Delta h=h_m-h'[/tex]
[tex]\Delta h=(\frac{u_y^2}{19.6} -3.4)\ m[/tex]
time taken to descent this height from the top height:
[tex]\Delta h=u_{yt}.t_d+\frac{1}{2} \times g.t_d^2[/tex]
where:
[tex]u_{yt}=[/tex] initial vertical velocity at the top point
[tex]t_d=[/tex] time of descend
[tex](\frac{u_y^2}{19.6} -3.4)=0+0.5\times 9.8\times t_d^2[/tex]
[tex]t_d=\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}[/tex]..............................(2)
So we find the time taken by the rock to reach the top of projectile where the vertical velocity is zero:
[tex]v_y=u_y-g.t_t[/tex]
where:
[tex]u_y=[/tex] initial vertical velocity
[tex]v_y=[/tex] final vertical velocity
[tex]t_t=[/tex] time taken to reach the top height of the projectile
[tex]0=u_y-g.t_t[/tex]
[tex]t_t=\frac{u_y}{9.8}\ seconds[/tex] .................................(3)
Now the combined events of vertical and horizontal direction must take at the same time as the projectile is thrown:
So,
[tex]t_f=t_t+t_d[/tex]
[tex]\frac{14}{u.\cos56}=\frac{u_y}{9.8} +\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}[/tex]
[tex]\frac{14}{u.\cos56}=\frac{u\sin56}{9.8} +\sqrt{(\frac{(u.\sin56)^2}{96.04} -\frac{3.4}{4.9} )}[/tex]
[tex]\frac{196}{u^2.\cos^2 56} +\frac{u^2\sin^2 56}{96.04} -2.857\times \tan56=\frac{u^2\sin^2 56}{96.04} -0.694[/tex]
[tex]u=13.3032\ m.s^{-1}[/tex]
Max height:
[tex]h_m=\frac{(u.\sin 56)^2}{19.6}[/tex]
[tex]h_m=\frac{(13.3032\times \sin56)^2}{19.6}[/tex]
[tex]h_m=6.2059\ m[/tex]
Now the rock hits down the ground 1.6 meters below the level of throw.
Time taken by the rock to fall the gross height [tex]h_g=h_m+h'[/tex]:
[tex]h_g=u_{yt}.t_g+\frac{1}{2} g.t_g^2[/tex]
[tex]7.8059=0+0.5\times 9.8\times t_g^2[/tex]
[tex]t_g=1.2621\ s[/tex]
Time taken to reach the the top of the fence from the top, using eq. (2):
[tex]t_d=\sqrt{(\frac{u_y^2}{96.04} -\frac{3.4}{4.9} )}[/tex]
[tex]t_d=\sqrt{(\frac{(u.\sin56)^2}{96.04} -\frac{3.4}{4.9} )}[/tex]
[tex]t_d=0.7567\ s[/tex]
Time difference between falling from top height and the time taken to reach the top of fence:
[tex]\Delta t=t_g-t_d[/tex]
[tex]\Delta t=1.2621-0.7567[/tex]
[tex]\Delta t=0.5054\ s[/tex]
b)
Now the horizontal distance covered in this time:
[tex]s=u.\cos56\times\Delta t[/tex]
[tex]s=13.3032\times \cos56\times 0.5054[/tex]
[tex]s=3.7597\ m[/tex] is the horizontal distance covered after crossing the fence.
A juggler throws a bowling pin straight up with an initial speed of 8.20 m/s. How much time elapses until the bowling pin returns to the juggler’s hand?
The time it takes for the bowling pin to return to the juggler's hand is approximately [tex]\( 1.13 \, \text{s} \)[/tex].
To find the time it takes for the bowling pin to return to the juggler's hand, you can use the kinematic equation for vertical motion under constant acceleration. The equation is:
[tex]\[ h = v_0 t - \frac{1}{2}gt^2 \][/tex]
Where:
- [tex]\( h \)[/tex] is the displacement (in this case, the height the bowling pin reaches, which is zero when it returns to the hand),
- [tex]\( v_0 \)[/tex] is the initial velocity,
- [tex]\( t \)[/tex] is the time,
- [tex]\( g \)[/tex] is the acceleration due to gravity (approximately [tex]\( 9.8 \, \text{m/s}^2 \))[/tex].
In this case, the final height [tex](\( h \))[/tex] is zero because the bowling pin returns to the juggler's hand. The initial velocity [tex](\( v_0 \))[/tex] is given as [tex]\( 8.20 \, \text{m/s} \)[/tex], and [tex]\( g \) is \( 9.8 \, \text{m/s}^2 \)[/tex].
Plugging in these values, the equation becomes:
[tex]\[ 0 = (8.20 \, \text{m/s}) \cdot t - \frac{1}{2}(9.8 \, \text{m/s}^2) \cdot t^2 \][/tex]
Now, you can solve this quadratic equation for [tex]\( t \)[/tex]. The general form of a quadratic equation is [tex]\( at^2 + bt + c = 0 \)[/tex], where [tex]\( a = -\frac{1}{2}(9.8 \, \text{m/s}^2) \), \( b = 8.20 \, \text{m/s} \), and \( c = 0 \)[/tex]. The solutions to this equation give you the times when the bowling pin is at the initial and final heights.
You can use the quadratic formula to solve for [tex]\( t \)[/tex]:
[tex]\[ t = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \][/tex]
where [tex]\( a = -\frac{1}{2}(9.8 \, \text{m/s}^2) \), \( b = 8.20 \, \text{m/s} \), and \( c = 0 \)[/tex].
[tex]\[ t = \frac{-8.20 \, \text{m/s} \pm \sqrt{(8.20 \, \text{m/s})^2 - 4 \cdot \left(-\frac{1}{2}(9.8 \, \text{m/s}^2)\right) \cdot 0}}{2 \cdot \left(-\frac{1}{2}(9.8 \, \text{m/s}^2)\right)} \][/tex]
Simplifying further:
[tex]\[ t = \frac{-8.20 \, \text{m/s} \pm \sqrt{67.24}}{-9.8} \][/tex]
Now, calculate the two possible values for [tex]\( t \)[/tex] using both the plus and minus signs:
[tex]\[ t_1 = \frac{-8.20 + \sqrt{67.24}}{-9.8} \][/tex]
[tex]\[ t_2 = \frac{-8.20 - \sqrt{67.24}}{-9.8} \][/tex]
Calculating these values:
[tex]\[ t_1 \approx 1.13 \, \text{s} \][/tex]
[tex]\[ t_2 \approx -0.58 \, \text{s} \][/tex]
Since time cannot be negative in this context, we discard the negative solution. Therefore, the time it takes for the bowling pin to return to the juggler's hand is approximately [tex]\( 1.13 \, \text{s} \)[/tex].
Some region of space contains uniform electric field directed towards south with magnitude 100 V/m. At point A electric potential is 400 V. What is electric potential at point B which is 2 meters directly towards the north from A
The value of electric potential at point B which is 2 meters directly towards the north from point A is 600 V.
What is electric field?The electric field is the field, which is surrounded by the electric charged. The electric field is the electric force per unit charge.
In terms of potential difference, the electric field can be given as,
[tex]E=\dfrac{\Delta V}{d}[/tex]
Here, (ΔV) is the potential difference and (d) is the distance.
Some region of space contains uniform electric field directed towards south with magnitude 100 V/m.
At point A electric potential is 400 V. The distance of point B directly towards the north from point A is 2 meters. Thus, by the above formula,
[tex]100=\dfrac{ V_b-400}{2}\\V_b=600\rm\; V[/tex]
Hence, the value of electric potential at point B which is 2 meters directly towards the north from point A is 600 V.
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Final answer:
The electric potential at point B, which is 2 meters directly north of point A within a uniform electric field pointing south with a magnitude of 100 V/m and an initial potential at A of 400 V, is 600 V.
Explanation:
The student asks about the electric potential at a point located 2 meters to the north of another point A within a uniform electric field pointing south, with an initial potential at point A of 400 V. Since the direction towards point B is opposite the direction of the electric field, the potential will increase as we move from A to B. The electric field has a magnitude of 100 V/m, which means that for each meter we move against the electric field, the potential increases by 100 V. As point B is 2 meters north of point A, we simply calculate the potential at B as follows:
VB = VA + E × d
VB = 400 V + (100 V/m) × 2 m = 600 V
So the electric potential at point B is 600 V.
Oscilloscopes are found in most science laboratories. Inside, they contain deflecting plates consisting of more-or-less square parallel metal sheets, typically about 2.50 cm on each side and 2.00 mm apart. In many experiments, the maximum potential across these plates is about 30.0 V.
1.For this maximum potential, what is the strength of the electric field between the plates? [V/m]
2.For this maximum potential, what magnitude of acceleration would this field produce on an electron midway between the plates? [m/s^2]
Answer:
15000 V/m
[tex]2.634467618\times 10^{15}\ m/s^2[/tex]
Explanation:
V = Voltage = 30 V
d = Separation = 2 mm
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
Electric field is given by
[tex]E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{30}{2\times 10^{-3}}\\\Rightarrow E=15000\ V/m[/tex]
The electric field between the plates is 15000 V/m
Acceleration is given by
[tex]a=\dfrac{qE}{m}\\\Rightarrow a=\dfrac{1.6\times 10^{-19}\times 15000}{9.11\times 10^{-31}}\\\Rightarrow a=2.634467618\times 10^{15}\ m/s^2[/tex]
The acceleration is [tex]2.634467618\times 10^{15}\ m/s^2[/tex]
A bus leaves New York City, takes a non-direct route and arrives in St. Louis, Missouri 23 hours, 16 minutes later. If the distance between the two cities is 1250 km, what is the magnitude of the bus' average velocity?
Answer:
vavg = 53.7 km/h
Explanation:
In order to find the magnitude of the bus'average velocity, we need just to apply the definition of average velocity, as follows:
[tex]vavg =\frac{xf-xo}{t-to}[/tex]
where xf - xo = total displacement = 1250 Km
If we choose t₀ = 0, ⇒ t = 23h 16'= 23h + 0.27 h = 23.27 h
⇒ [tex]vavg =\frac{1250 km}{23.27h} = 53.7 Km/h[/tex]
The average velocity of the bus, calculated by dividing the total displacement (1250 km) by the total time (23.27 hours), is approximately 53.69 km/h. This refers to the magnitude of the average velocity, not the direction.
Explanation:The subject of the question is average velocity, which is defined in physics as the total displacement divided by the total time taken. The displacement in this scenario is the straight-line distance between New York City and St. Louis, Missouri, which is 1250 km. The time taken for the bus to travel this route is 23 hours and 16 minutes, or approximatively 23.27 hours when converted into decimal hours to facilitate calculation. Using the formula for average velocity (v = d / t), we substitute the given values to find the average velocity of the bus:
v = 1250 km / 23.27 hours
Upon calculating this, we find that the average velocity of the bus is approximately 53.69 km/h. please note that this is the magnitude of the average velocity, indicating the speed rather than the direction of the bus's travel.
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A circular rod with a gage length of 3.2 mm and a diameter of 2 cmcm is subjected to an axial load of 57 kNkN . If the modulus of elasticity is 200 GPaGPa , what is the change in length?
To solve this problem we will apply the concepts related to the change in length given by the following relation,
[tex]\delta_l = \frac{Pl}{AE}[/tex]
Here the variables mean the following,
P = Load
l = Length
A = Area
E = Modulus of elasticity
Our values are,
[tex]l = 3.2 m[/tex]
[tex]\phi = 2cm = 0.02m[/tex]
[tex]P = 57kN = 57*10^3N[/tex]
[tex]E = 200Gpa[/tex]
We can obtain the value of the Area through the geometrical relation:
[tex]A = \frac{\pi}{4} \phi^2[/tex]
Replacing,
[tex]A = \frac{\pi}{4} (0.02)^2[/tex]
[tex]A = 3.14*10^{-4}m^2[/tex]
Using our first equation,
[tex]\delta_l = \frac{Pl}{AE}[/tex]
[tex]\delta_l = \frac{(57*10^3)(3.2)}{(3.14*10^{-2})(200*10^9)}[/tex]
[tex]\delta_l = 0.000029044m[/tex]
[tex]\delta_l = 0.029044mm[/tex]
Therefore the change in length is 0.029mm
Two balls, made of different materials, are rubbed against each other, resulting in 0.30 nC of charge moving from one ball to the other. The balls are then held 0.90 m apart. What is the magnitude of the dipole moment of the two balls?
Answer:
[tex]5.4\times 10^{-10}C-m[/tex]
Explanation:
We are given that
Charge=[tex]q=0.30 nC=0.3\times 10^{-9} C[/tex]
[tex]1 nC=10^{-9}C[/tex]
Distance between two balls=l=0.90 m
We have to find the magnitude of dipole moment of the two balls.
We know that
Dipole moment=[tex]\mid p\mid=2lq[/tex]
Where q= Charge
l=Distance between two bodies
Using the formula
Magnitude of dipole moment=[tex]\mid P\mid=2\times 0.3\times 10^{-9}\times 0.9=5.4\times 10^{-10}C-m[/tex]
Hence, the magnitude of the dipole moment of the two balls=[tex]5.4\times 10^{-10}C-m[/tex]
In a rainstorm with a strong wind, what determines the best position in which to hold an umbrella?
Explanation:
The best way to hold a strong umbrella in rainstorm with a strong wind will be against the direction of the wind. This can provide with the maximum protection in rain. Moreover, it should be placed slightly upward also, at an angle. This will again call for maximum protection.
The best position to hold an umbrella in a rainstorm with wind is determined by the direction of the wind. You should hold your umbrella facing towards the wind's direction, including both horizontal and vertical direction, to best protect from the rain.
Explanation:In a rainstorm with a strong wind, the best position in which to hold an umbrella is largely determined by the direction of the wind. Since rain in a storm tends to fall diagonally due to wind rather than vertically, you should position your umbrella in such a way that it faces the direction from which the wind and rain are coming.
This includes both the horizontal direction (north, south, east, or west) and the vertical direction (upwards or downwards), as wind and rain can also come from above or below. For example, if the wind is blowing from the north, you should hold your umbrella to the north.
If it's also blowing downwards, you should tilt your umbrella accordingly. By doing so, you can protect yourself from the rain to the greatest extent possible.
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3. In a physics lab, 0.500-kg cart (Cart A) moving rightward with a speed of 100 m/s collides with a 1.50-kg cart (Cart B) moving leftward with a speed of 20 m/s. The two carts stick together and move as a single object after the collision. Determine the post-collision speed of the two carts.
Answer:
The speed of the two carts after the collision is 10 m/s.
Explanation:
Hi there!
The momentum of the system Cart A - Cart B is conserved because there is no external force acting on the system at the instant of the collision. Then, the momentum of the system before the collision will be equal to the momentum of the system after the collision. The momentum of the system is calculated as the sum of momenta of cart A and cart B:
initial momentum = mA · vA1 + mB · vB1
final momentum = (mA + mB) · vAB2
Where:
mA = mass of cart A = 0.500 kg
vA1 = velocity of cart A before the collision = 100 m/s
mB = mass of cart B = 1.50 kg.
vB1 = velocity of cart B before the collision = - 20 m/s
vAB2 = velocity of the carts that move as a single object = unknown.
(notice that we have considered leftward as negative direction)
Since the momentum of system remains constant:
initial momentum = final momentum
mA · vA1 + mB · vB1 = (mA + mB) · vAB2
Solving for vAB2:
(mA · vA1 + mB · vB1) / (mA + mB) = vAB2
(0.500 kg · 100 m/s - 1.50 kg · 20 m/s) / (0.500 kg + 1.50 kg) = vAB2
vAB2 = 10 m/s
The speed of the two carts after the collision is 10 m/s.
The post-collision speed of the two carts is 10 m/s moving in the positive x-direction.
Explanation:In order to determine the post-collision speed of the two carts, we can use the principle of conservation of momentum. The total momentum before the collision is equal to the total momentum after the collision.
Before the collision, Cart A has a mass of 0.500 kg and a velocity of 100 m/s, while Cart B has a mass of 1.50 kg and a velocity of -20 m/s (negative because it is moving leftward). After the collision, the two carts stick together, so their masses add up to 2 kg.
Using the conservation of momentum equation: (Momentum before collision) = (Momentum after collision)
(0.500 kg × 100 m/s) + (1.50 kg × -20 m/s) = 2 kg × v
By solving this equation, we find that the post-collision speed of the two carts is 10 m/s moving in the positive x-direction.
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A light rope is attached to a block with mass 4.10 kg that rests on a frictionless, horizontal surface. The horizontal rope passes over a frictionless, massless pulley, and a block with mass m is suspended from the other end. When the blocks are released, the tension in the rope is 14.7 N.
a. Draw two free-body diagrams: one for each block.
b. What is the acceleration of either block?
c. Find m.
d. How does the tension compare to the weight of the hanging block?
Answer:
a) please find the attachment
(b) 3.65 m/s^2
c) 2.5 kg
d) 0.617 W
T<weight of the hanging block
Explanation:
a) please find the attachment
(b) Let +x be to the right and +y be upward.
The magnitude of acceleration is the same for the two blocks.
In order to calculate the acceleration for the block that is resting on the horizontal surface, we will use Newton's second law:
∑Fx=ma_x
T=m1a_x
14.7=4.10a_x
a_x= 3.65 m/s^2
c) in order to calculate m we will apply newton second law on the hanging
block
∑F=ma_y
T-W= -ma_y
T-mg= -ma_y
T=mg-ma_y
T=m(g-a_y)
a_x=a_y
14.7=m(9.8-3.65)
m = 2.5 kg
the sign of ay is -ve cause ay is in the -ve y direction and it has the same magnitude of ax
d) calculate the weight of the hanging block :
W=mg
W=2.5*9.8
=25 N
T=14.7/25
=0.617 W
T<weight of the hanging block
A runner wants to run 11.8 km. Her running pace is 7.4 mi/hr. How many minutes must she run? Express your answer using two significant figures.
Answer:
She must run 59 min to run 11.8 km.
Explanation:
Hi there!
First let's convert mi/h into km/min:
7.4 mi/h · (1.61 km /1 mi) · (1 h / 60 min) = 0.20 km/min (notice how the units mi and h cancel).
The runner runs at 0.20 km/ min, i.e., every minute she travels 0.20 km.
If 0.20 km are traveled in 1 min, then 11.8 km will be traveled in:
11.8 km / 0.20 km/min = 59 min
She must run 59 min to run 11.8 km.
CP Bang! A student sits atop a platform a distance h above the ground. He throws a large firecracker horizontally with a speed. However, a wind blowing parallel to the ground gives the firecracker a constant horizontal acceleration with magnitude a. As a result, the firecracker reaches the ground directly below the student. Determine the height h in terms of v, a, and g. Ignore the effect of air resistance on the vertical motion.
Answer:
h = v₀ g / a
Explanation:
We can solve this problem using the kinematic equations. As they indicate that the air does not influence the vertical movement, we can find the time it takes for the body to reach the floor
y = [tex]v_{oy}[/tex] t - ½ g t²
The vertical start speed is zero
t² = 2t / g
The horizontal document has an acceleration, with direction opposite to the speed therefore it is negative, the expression is
x = v₀ₓ t - ½ a t²
Indicates that it reaches the same exit point x = 0
v₀ₓ t = ½ a t2
v₀ₓ = ½ a (2h / g)
v₀ₓ = v₀
h = v₀ g / a
Your car's blinker has a period of 0.85 s and at the moment is in phase with a faster blinker on the car in front of you. They drift out of phase but then get back in phase after 16 s. What is the period of the other car's blinker in s?
Answer:
The time period of the other car's blinker is 0.807
Solution:
As per the question:
Time period of the car blinker, T = 0.85 s
Time taken by the blinkers to get back in phase, t = 16 s
Now,
To find the time period of the other car's blinker:
No. of oscillations, [tex]n = \frac{t}{T}[/tex]
Thus for the blinker:
[tex]n = \frac{16}{0.85}[/tex]
Now,
For the other car's blinker with time period, T':
[tex]n' = \frac{16}{T'}[/tex]
Time taken to get back in phase is t = 16 s:
n' - n = 1
[tex]\frac{16}{T'} - \frac{16}{0.85} = 1[/tex]
[tex]\frac{1}{T'} = \frac{1}{16} + \frac{1}{0.85}[/tex]
[tex]\frac{1}{T'} = 1.2389[/tex]
[tex]T' = \frac{1}{1.2389} = 0.807[/tex]
The period of the other car's blinker is approximately 1.06 s.
Explanation:To solve this problem, we need to understand the concept of phase and period. The period is the time it takes for a complete cycle of a periodic motion. In this case, the period of your car's blinker is given as 0.85 s. The phase refers to the position within a cycle at a given time. If your blinker is in phase with the other car's blinker initially, it means they are both starting their cycles at the same time.
However, they drift out of phase and then get back in phase after 16 s. This means that the other car's blinker completes a whole number of cycles in 16 s. Let's call the period of the other car's blinker T. So, in 16 s, the other car's blinker completes 16/T cycles. We know that the two cars get back in phase after 16 s, which means they complete the same number of cycles in that time.
Therefore, we can set up the following equation: 16/T = 16/0.85. Solving for T, we find that the period of the other car's blinker is approximately 1.06 s.
(Schaum’s 18.25) A 55 g copper calorimeter (c=377 J/kg-K) contains 250 g of water (c=4190 J/kg-K) at 18o When a 75 g metal alloy at 100o C is dropped into the calorimeter, the final equilibrium temperature is 20.4o C. What is the specific heat of the alloy?
Answer:
1205.77 J/kg.K
Explanation:
Heat lost by alloy = heat gained by water + heat gained by the calorimeter
c₁m₁(t₂-t₃) = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)................. Equation 1
Where c₁ = specific heat capacity of the alloy, m₁ = mass of the alloy, t₂ = initial temperature of the alloy, t₃ = equilibrium temperature, c₂ = specific heat capacity of water, m₂ = mass of water, t₁ = initial temperature of water and calorimter, c₃ = specific heat capacity of calorimter, m₃ = mass of calorimter.
Making c₁ the subject of the equation,
c₁ = c₂m₂(t₃-t₁) + c₃m₃(t₃-t₁)/m₁(t₂-t₃)........................ Equation 2
Given: c₂ = 4190 J/kgK, m₂ = 250 g = 0.25 kg, m₁ = 75 g = 0.075 kg, m₃ = 55 g = 0.055 kg, c₃ = 377 J/kg.K, t₁ = 18 °C, t₂ = 100 °C, t₃ = 24.4 °C.
Substitute into equation 2
c₁ = [0.25×4190×(24.4-18) + 0.055×377×(24.4-18)]/[0.075(100-24.4)]
c₁ = (6704+132.704)/5.67
c₁ = 6836.704/5.67
c₁ = 1205.77 J/kg.K
Thus the specific heat capacity of the alloy = 1205.77 J/kg.K
Final answer:
The student's question is about calculating the specific heat capacity of a metal alloy using the principles of calorimetry and the conservation of energy in a heat exchange process.
The student is asking about finding the specific heat capacity of a metal alloy using calorimetry. We know that when objects at different temperatures are combined, they will exchange heat energy until they reach thermal equilibrium. We can use the equation Q = mc ext{ extdegree}T (where Q is heat energy, m is mass, c is specific heat capacity, and ext{ extdegree}T is the change in temperature) to find the specific heat capacity. In this scenario, the heat lost by the metal alloy will equal the heat gained by the copper calorimeter and the water contained within it. By setting these two equations equal to each other and solving for the specific heat capacity of the alloy, we can find that value.
A small, solid sphere of mass 0.9 kg and radius 47 cm rolls without slipping along the track consisting of slope and loop-the-loop with radius 4.75 m at the end of the slope. It starts from rest near the top of the track at a height h, where h is large compared to 47 cm. If the g = 9.8 m/s^2 and I(solid sphere) = 2/5 mr^2, what is the minimum value of h such that the sphere completes the loop?
The minimum height that the sphere should start from to complete a loop-the-loop is 23.75 meters, as calculated through the conservation of energy and dynamics principles.
Explanation:In this physics problem, the minimum height (h) that the solid sphere needs to start from to ensure it completes the loop-the-loop involves applying principles of conservation of energy and dynamics. Initially, the sphere has potential energy equal to mgh, and no kinetic energy as it starts from rest. As it descends, it gains kinetic energy and loses potential energy.
For the ball to successfully complete the loop, the force at the top must be equivalent to the weight of the sphere plus the force necessary to maintain circular motion. This can be written as: mg + mv²/r = 5mg. From here, we can derive the equation for v²: v² = 4gr.
Since the kinetic energy at the top of the loop is (1/2)mv² and the potential energy is 2mgr, by equating total energy at the top of the loop (potential plus kinetic) to the initial potential energy (mgh), we obtain: mgh = (1/2)m(4gr) + 2mgr.
From this equation, we can solve for h and find that h = 5r = 5*4.75m = 23.75m. This is the minimum height the sphere must start from to complete the loop.
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Calculate the osmotic pressure at 36.6 degrees C of a solution made by dissolving 9.18 g of glucose in 34.2 mL of solution. Enter your answer using 2 decimal places!!!!
Answer:
38.35 bar
Explanation:
We are given that
Temperature=T=36.6 degree Celsius=36.6+273=309.6 K
Given mass of glucose=9.18 g
Molar mass of glucose([tex]C_6H_{12}O_6=6(12)+12(1)+6(16)[/tex]=180 g
Mass of c=12 g,mass of hydrogen=1 g, mass of O=16 g
Volume of solution=34.2 mL
Molarity of solution=[tex]\frac{given\;mass}{molar\;mass\times volume}\times 1000[/tex]
Where volume (in mL)
Molarity of solution=[tex]\frac{9.18}{180\times 34.2}\times 1000=1.49 M[/tex]
We know that
Osmotic pressure=[tex]\pi=MRT[/tex]
Where M=Molarity of solution
R=Constant=0.08314 Lbar/mol k
T=Temperature in kelvin
Using the formula
[tex]\pi=1.49\times 0.08314\times 309.6=38.35 bar[/tex]
Hence, the osmotic pressure=38.35 bar
Two point charges are placed on the x axis. The first charge, q1 = 8.00 nC, is placed a distance 16.0 m from the origin along the positive x axis; the second charge, q2 = 6.00 nC, is placed a distance 9.00 m from the origin along the negative x axis.Calculate the electric field at point A, located at coordinates (0 mm, 12.0 mm ). Give the x and y components of the electric field as an ordered pair. Express your answer in newtons per coulomb to three significant figures. EAx, EAy =
The x and y components of the electric field at point A are [tex]EAx = 3.11 * 10^4 N/C~ and~ EAy = 0 N/C.[/tex]
The net electric field at point A due to two point charges can be found by calculating the electric field contributed by each charge independently and then summing the components to find the net electric field as an ordered pair.
Finally, we sum the x-components and sum the y-components of the electric fields from both charges to get the net electric field at point A as an ordered pair (EAx, EAy).
For q1:
r1x = 16.0 m
r1y = 0.012 m (converting 12.0 mm to meters)
[tex]r1 = sqrt(r1x^2 + r1y^2)\\r_1 = sqrt(16.0^2 + 0.012^2) \\r_1 = 16.0001 m[/tex]
For q2:
[tex]r2x = -9.0 m\\r2y = 0.012 m (same as for q1)\\r2 = sqrt(r2x^2 + r2y^2) \\r2 = sqrt((-9.0)^2 + 0.012^2)\\ r2 = 9.0001 m[/tex]
Now, we can calculate the electric field contributions from each charge:
[tex]E1x = (8.99 * 10^9) * (8 * 10^-9) / (16.0001)^2 \\E1x = 1.94 * 10^4 N/C\\E1y = 0 E2y = (8.99 * 10^9) * (6 * 10^-9) / (9.0001)^2 \\E2y= 1.17 * 10^4 N/C\\E2y = 0 E1x[/tex]
Finally, we add the x-components of the electric fields vectorially:
[tex]EAx = E1x + E2x \\= 1.94 * 10^4 + 1.17 * 10^4 \\= 3.11 * 10^4 N/C[/tex]
The y-component of the electric field, EAy, is 0 since both charges are on the x-axis.
So, the x and y components of the electric field at point A are [tex]EAx = 3.11 * 10^4 N/C~ and~ EAy = 0 N/C.[/tex]
An electric heater has a Nichrome heating element with a resistance of 9 Ω at 20oC. When 112 V are applied, the electric current heats the Nichrome wire to 1090oC. What is the operating wattage of this heater? (The temperature coefficient of resistivity of Nichrome is α = 0.0004 Co-1)
Answer:
975.28 W.
Explanation:
Using,
R' = R(1+αΔt)....................... Equation 1
Where R' = Resistance at the final temperature, R = Resistance at the initial temperature, α = temperature coefficient of resistivity of Nichorome, Δt = Temperature rise.
Given: R = 9 Ω, α = 0.0004/°C, Δt = 1090-20 = 1070 °C
Substitute into equation 1
R' = 9(1+0.0004×1070)
R' = 9(1.428)
R' = 12.862 Ω.
Note: Operating wattage of the heater means the operating power of the heater
The power of the heater is given as,
P = V²/R'...................... Equation 2
Where P = Operating wattage of the heater, V = Voltage, R' = Operating resistance.
Given: V = 112 V, R' = 12.862 Ω
Substitute into equation 2
P = 112²/12.862
P = 975.28 W.
The operating wattage for this heater can be calculated using Ohm's Law, resulting in approximately 1405.33 Watts assuming constant resistance. However, in reality, resistance alters with temperature, reflecting the importance of considering temperature effects in physics.
Explanation:The operating wattage for this electric heater, or the power (P), can be calculated using Ohm's Law where power equals voltage (V) times current (I), or P=IV. Because I = V/R, where R is resistance, the formula can also be written as P = V2/R. With the provided values, we have P = (112V)2 / 9Ω, which gives approximately 1405.33 Watts, assuming that the resistance remains constant over the temperature change.
However, in reality, the resistance changes with temperature according to the equation R = R0[1 + α(T - T0)] where R0 is the original resistance, α is the temperature coefficient of resistivity, T is the final temperature, and T0 is the initial temperature. Considering the provided values and the significant temperature increase, we would need to adjust the resistance for the increased temperature before calculating the power, underlining the importance of temperature effects in practical physics.
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A car travels in the + x-direction on a straight and level road. For the first 4.00 s of its motion, the average velocity of the car is vav-x = 6.25 m/s. How far does the car travel in 4.00 s?
Answer:
25 m
Explanation:
The relationship between Velocity, distance and time is given as
S = v/t........................... Equation 1
Where S = average velocity of the car, d = distance covered by the car, t = time
Making d the subject of the equation,
d = vt.................... Equation 2
Given: v = 6.25 m/s, t = 4.00 s.
Substitute into equation 2,
d = 6.25(4)
d = 25 m.
Hence, the distance traveled by the car = 25 m
The distance traveled by the car will be "25 m".
The given values are:
Speed,
v = 6.25 m/s,Time,
t = 4.00 s
As we know the formula,
→ [tex]Distance = Speed\times Time[/tex]
or,
→ [tex]d = v\times t[/tex]
By substituting the values, we get
[tex]= 6.25\times 4[/tex]
[tex]=25 \ m[/tex]
Thus the above answer is right.
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A sample of nitrogen gas exerts a pressure of 9.80 atm at 32 C. What would its temperature be (in C) when its pressure is increased to 11.2 atm?
Answer:
T₂ = 111.57 °C
Explanation:
Given that
Initial pressure P₁ = 9.8 atm
T₁ = 32°C = 273 + 32 =305 K
The final pressure P₂ = 11.2 atm
Lets take the final temperature = T₂
We know that ,the ideal gas equation
If the volume of the gas is constant ,then we can say that
[tex]\dfrac{P_2}{P_1}=\dfrac{T_2}{T_1}[/tex]
[tex]T_2=\dfrac{P_2}{P_1}\times T_1[/tex]
Now by putting the values in the above equation ,we get
[tex]T_2=\dfrac{11.2}{9.8}\times 305\ K[/tex]
[tex]T_2=348.57\ K[/tex]
T₂ = 384.57 - 273 °C
T₂ = 111.57 °C
An unknown gas effuses 2.3 times faster than N2O4 at the same temperature. What is the identity of the unknown gas?
Answer:
The molar mass of the unknown gas is 17.3 g/mol. The molar mass matches that of ammonia (NH₃) the most (17 g/mol)
Explanation:
Let the unknown gas be gas 1
Let N₂O₄ gas be gas 2
Rate of effusion ∝ [1/√(Molar Mass)]
R ∝ [1/√(M)]
R = k/√(M) (where k is the constant of proportionality)₁₂
R₁ = k/√(M₁)
k = R₁√(M₁)
R₂ = k/√(M₂)
k = R₂√(M₂)
k = k
R₁√(M₁) = R₂√(M₂)
(R₁/R₂) = [√(M₂)/√(M₁)]
(R₁/R₂) = √(M₂/M₁)
R₁ = 2.3 R₂
M₁ = Molar Mass of unknown gas
M₂ = Molar Mass of N₂O₄ = 92.01 g/mol
(2.3R₂/R₂) = √(92.01/M₁)
2.3 = √(92.01/M₁)
92.01/M₁ = 2.3²
M₁ = 92.01/5.29
M₁ = 17.3 g/mol
The molar mass matches that of ammonia the most (17 g/mol)
The unknown gas in the system has been ammonia.
The rate of diffusion of the two gases has been proportional to the molar mass of the gases.
The ratio of the rate of two gases can be given as:
[tex]\rm \dfrac{RateA}{RateB}\;=\;\sqrt{\dfrac{Molar\;mass\[A}{Molar\;mass\;B} }[/tex]
The two gases can be given as:
Gas A = Nitrogen tetraoxide = 2.3x
Gas B = x
[tex]\rm \dfrac{2.3x}{x}\;=\;\sqrt{\dfrac{92.011}{m} }[/tex]
Mass of the unknown gas = 17.39 grams.
The mass has been equivalent to the mass of the Ammonia. Thus, the unknown gas in the system has been ammonia.
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An airplane is dropping bales of hay to cattle stranded in a blizzard on the Great Plains. The pilot releases the bales at 150 m above the level ground when the plane is flying at 75 m/s in a direction 55° above the horizontal. How far in front of the cattle should the pilot release the hay so that the bales land at the point where the cattle are stranded?
Answer:
Δx=629.35 m
The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.
Explanation:
Step 1:
Finding initial velocity components:
Initial velocity=v=75 m/s
α=55
[tex]v_{ox}=vcos\alpha\\v_{ox}=75cos55^o\\v_{ox}=43.018 m/s\\v_{oy}=vsin\alpha\\v_{oy}=75sin55^o\\v_{oy}=61.436 m/s[/tex]
Step 2:
[tex]y_o=150\ m[/tex]
Newton Second Equation:
[tex]y-y_o=v_{oy}t+\frac{1}{2}g t^2[/tex]
g=-9.8 m/s^2 (Downward direction)
[tex]v_{oy}=61.436\ m/s[/tex]
y=0 m
Above equation will become:
-150=(61.436)t-(4.90)t^2
Solving the above quadratic equation we will get:
t=-2.09 sec , t=14.63 sec
t= 14.63 sec
Step 3:
Finding the distance:
Using Again Newton equation of motion in x-direction:
[tex]x-x_o=v_{ox}t+\frac{1}{2}a_{x} t^2[/tex]
Since velocity is constant in x- direction, [tex]a_x[/tex] will be zero.
Above equation will be:
[tex]\Delta x=v_{ox}t[/tex]
Δx=(43.018)(14.63)
Δx=629.35 m
The pilot release the hay 629.35 m in front of the cattle so that the bales land at the point where the cattle are stranded.
The pilot should release the hay at a height of 629.35 m.
Given information,
Initial velocity = 75 m/s
Velocity For x-component,
[tex]\bold {V_0x = Vcos \alpha}\\\\\bold {V_0x = 75 cos 55^o}\\\\\bold {V_0x = 43. 018m/s}[/tex]
Velocity for Y-component
[tex]\bold {V_0y = Vsin \alpha}\\\\\bold {V_0y = 75 sin 55^o}\\\\\bold {V_0y = 61. 43m/s}[/tex]
Using Newton's second equation for y-axis,
[tex]\bold {y-y_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]
Where,
g - gravitational acceleration
put the values in the equation,
[tex]\bold {-150=(61.436)t-(4.90)t^2}[/tex]
Solving this quadratic equation, we get 2 values
t = 14.29 s
To find the distance, use Newton's second equation,
[tex]\bold {x-x_0 = V_0t + \dfrac {1}{2} gt^2}[/tex]
Since acceleration is zero because the velocity is constant in x-axis hence .
So,
[tex]\bold {x-x_0 = V_0_xt }[/tex]
[tex]\bold {x- x_0=(43.018)(14.63)}\\\\\bold {x - x_0=629.35 m}[/tex]
Therefore, the pilot should release the hay at 629.35 m.
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Cathode ray tubes (CRTs) used in old-style televisions have been replaced by modern LCD and LED screens. Part of the CRT included a set of accelerating plates separated by a distance of about 1.54 cm. If the potential difference across the plates was 27.0 kV, find the magnitude of the electric field (in V/m) in the region between the plates. HINT
Answer:
1753246.75325 V/m
Explanation:
d = Distance of separation = 1.54 cm
V = Potential difference = 27 kV
When the voltage is divided by the distance between the plates we get the electric field.
Electric field is given by
[tex]E=\dfrac{V}{d}\\\Rightarrow E=\dfrac{27\times 10^3}{1.54\times 10^{-2}}\\\Rightarrow E=1753246.75325\ V/m[/tex]
The magnitude of the electric field in the region between the plates is 1753246.75325 V/m
A gas had an initial pressure of 4.80atm in a 5.50L container. After transfering it to a 9.60L container, the gas was found to have a pressure of 2.10atm and a temperature of 25.00∘C. What was the initial temperature in degrees Celsius?
To solve this problem we will apply the concepts related to the ideal gas equations. Which defines us that the relationship between pressure, temperature and volume in the first state must be equivalent in the second state of matter. In mathematical terms this is
[tex]\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}[/tex]
If we rearrange the equation to find the Temperature at state 1 we have that
[tex]T_1 = \frac{P_1V_1T_2}{P_2V_2}[/tex]
Replacing our values we have that
[tex]T_1 = \frac{(4.8*5.5*298.15)}{(2.1*9.6)}[/tex]
[tex]T_1 = 390.435K[/tex]
Therefore the temperature is 390.435K
Answer:
117 ∘C
Explanation:
Use the combined gas law.
P1V1/T1 = P2V2/T2
Let the subscript 2 represent the 9.60L of gas at 25.0∘C and the subscript 1 represent the gas at the initial volume of 5.50L.
Remember to covert the temperature from degrees Celsius to Kelvin by adding 273.15.
Therefore, we have that T2=298.15K, P2=2.10atm, V2=9.60L, P1=4.80atm, V1=5.50L, and T1 is unknown.
Rearrange the equation for T1 and substitute in the known values to solve for the initial temperature.
T1T1T1=P1V1T2P2V2=(4.80atm)(5.50L)(298.15K)(2.10atm)(9.60L)=390.434K
Now, convert this temperature from Kelvin to degrees Celsius.
T1=390.434K−273.15 = 117.28∘C
Therefore, after rounding this value to three significant figures, we find that the initial temperature is 117∘C.
The force, F, of the wind blowing against a building is given by where V is the wind speed, rho the density of the air, A the cross-sectional area of the building, and CD is a constant termed the drag coefficient. Determine the dimensions of the drag coefficient.
Answer:
dimensions of the drag coefficient is [tex][M^0 L^0 T^0][/tex]
Drag coefficient is a dimensionless quantity
Explanation:
force is given by[tex]F=\frac{C_{D} \rho V^2 A}{2}[/tex]
we get expression for drag coefficient [tex]C_{D} =\frac{2F}{\rho V^2 A}[/tex]
By substituting the dimensions of the F,V,A and density , we get
[tex]C_{D} =\frac{[F]}{[\rho ][V]^2[A]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L T^{-1}]^2[L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[ML^{-3} ][L^2 T^{-2}][L^2]} \\C_{D} =\frac{[MLT^{-2}]}{[MLT^{-2}]}\\C_{D}=[M^0 L^0 T^0][/tex]
Drag coefficient is a dimensionless
The dimensions of the drag coefficient, CD, are kg/m.
Explanation:The dimensions of the drag coefficient, CD, can be determined by examining the equation for force, F, of the wind blowing against a building. In this equation, the dimensions for force are mass x acceleration, which are kg x m/s^2. On the other side of the equation, the wind speed, V, has dimensions of m/s, the density, rho, has dimensions of kg/m^3, and the cross-sectional area, A, has dimensions of m^2. Therefore, in order for the equation to be balanced, the dimensions of the drag coefficient must be kg/m.
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During the battle of Bunker Hill, Colonel William Prescott ordered the American Army to bombard the British Army camped near Boston. The projectiles had an initial velocity of 41 m/s at 38° above the horizon and an initial position that was 35 m higher than where they hit the ground. How far did the projectiles move horizontally before they hit the ground? Ignore air resistance.
Answer:
[tex]s=203.149\ m[/tex]
Explanation:
Given:
initial velocity of projectile, [tex]u=41\ m.s^{-1}[/tex]angle of projection above horizontal, [tex]\theta=38^{\circ}[/tex]height of the initial projection point above the ground, [tex]y=35\ m[/tex]
Vertical component of the velocity:
[tex]u_y=u.\sin\theta[/tex]
[tex]u_y=41\times \sin38[/tex]
[tex]u_y=25.242\m.s^{-1}[/tex]
The time taken in course of going up:
(at top the final velocity will be zero)
[tex]v_y=u_y-g.t[/tex]
[tex]0=25.242-9.8\times t[/tex]
[tex]t=2.576\ s[/tex]
In course of going up the maximum height reached form the initial point:
(at top height the final velocity is zero. )
using eq. of motion,
[tex]v_y^2=u_y^2-2\times g.h[/tex]
where:
[tex]v_y=[/tex] final vertical velocity while going up.=0
[tex]h=[/tex] maximum height
[tex]0^2=25.242^2-2\times 9.8\times h[/tex]
[tex]h=32.5081\ m[/tex]
Now the total height to be descended:
[tex]h'=h+y[/tex]
[tex]h'=32.5081+35[/tex]
[tex]h'=67.5081\ m[/tex]
Now the time taken to fall the gross height in course of falling from the top:
[tex]h'=v_y.t'+\frac{1}{2} g.t'^2[/tex]
[tex]67.5081=0+4.9\times t'^2[/tex]
[tex]t'=3.7118\ s[/tex]
Now the total time the projectile spends in the air:
[tex]t_t=t+t'[/tex]
[tex]t_t=2.576+3.7118[/tex]
[tex]t_t=6.2878\ s[/tex]
Now the horizontal component of the initial velocity:
(it remains constant throughout the motion)
[tex]u_x=u.\cos\theta[/tex]
[tex]u_x=41\times \cos38[/tex]
[tex]u_x=32.3084\ m.s^{-1}[/tex]
Therefore the horizontal distance covered in the total time;
[tex]s=u_x\times t_t[/tex]
[tex]s=32.3084\times 6.2878[/tex]
[tex]s=203.149\ m[/tex]
Answer:
Explanation:
initial velocity, u = 41 m/s
angle, θ = 38 °
height, h = 35 m
Let the time is t.
Use second equation of motion in vertical direction
h = ut + 1/2 gt²
- 35 = 41 Sin 38 t - 0.5 x 9.8 x t²
4.9t² - 25.2 t - 35 = 0
[tex]t = \frac{25.2 \pm \sqrt{25.2^{2}+4\times 4.9\times 35}}{2\times 4.9}[/tex]
t = 6.3 second
Horizontal distance traveled in time t is
d = uCos 38 x t
d = 41 x Cos 38 x 6.3
d = 203.54 m
flat sheet is in the shape of a rectangle with sides of lengths 0.400 mm and 0.600 mm. The sheet is immersed in a uniform electric field of magnitude 76.7 N/CN/C that is directed at 20 ∘∘ from the plane of the sheeta- Find the magnitude of the electric flux through the sheet?
Answer:
[tex]6.29591\times 10^{-6}\ N/C^2[/tex]
Explanation:
Flux is given by
[tex]\phi=EAcos\theta[/tex]
A = Area
[tex]A=0.4\times 10^{-3}\times 0.6\times 10^{-3}[/tex]
E = Electric field = 76.7 N/C
Angle is given by
[tex]\theta=90-20\\\Rightarrow \theta=70^{\circ}[/tex]
[tex]\phi=76.7\times 0.4\times 10^{-3}\times 0.6\times 10^{-3}\times cos70\\\Rightarrow \phi=6.29591\times 10^{-6}\ N/C^2[/tex]
The flux through the sheet is [tex]6.29591\times 10^{-6}\ N/C^2[/tex]
An unknown sample has a volume of 3.61 cm3 and a mass of 9.93 g. What is the density (g/cm3) of the unknown?
Answer:
2.75 g/cm³
Explanation:
given,
Volume of unknown sample, V = 3.61 cm³
mass of the sample, m = 9.93 g
density = ?
We know,
[tex]density = \dfrac{mass}{volume}[/tex]
[tex]\rho= \dfrac{9.93}{3.61}[/tex]
[tex]\rho = 2.75\ g/cm^3[/tex]
Hence, density of the unknown sample is equal to 2.75 g/cm³
Honeybees acquire a charge while flying due to friction with the air. A 120 mg bee with a charge of + 23 pC experiences an electric force in the earth's electric field, which is typically 100 N/C, directed downward.What is the ratio of the electric force on the bee to the bee's weight?F/W = ______What electric field strength would allow the bee to hang suspended in the air?E= _______What electric field direction would allow the bee to hang suspended in the air?Upward? or downward?
Answer:
[tex] \frac{F}{W}=1.95\times10^{-6}[/tex][tex] 51304447 \frac{N}{C} [/tex]UpwardExplanation:
The weight of the bee is:
[tex] W=mg=(120\times10^{-6}kg)(9.81\frac{m}{s^{2}})=1.18\times10^{-3}N[/tex]
with m the mass and g the gravity acceleration.
Electric force of the bee is related with the electric field of earth by:
[tex]F_{e}=qE=(23\times10^{-12}C)(100\frac{N}{C})=2.3\times10^{-9} [/tex]
with q the charge, E the electric field and Fe the electric force.
So:
[tex] \frac{F}{W}=\frac{2.3\times10^{-9}}{1.18\times10^{-3}}=1.95\times10^{-6}[/tex]
Because Newton's first law we should make the net force on it equals cero:
[tex] F+F_e+W=0[/tex]
[tex]F=-(F_e+W)=-(2.3\times10^{-9} +1.18\times10^{-3})=-1.1800023\times10^{-3} [/tex]
with W the weight, Fe the electric force on the bee due earth's electric field and F the force.
So, the applied electric field should be:
[tex]E_a=\frac{-1.1800023\times10^{-3}}{23\times10^{-12}}=-51304447 \frac{N}{C} [/tex]
The negative sign indicates that the electric field should be opposite to earth's electric field, so it should be upward.
Final answer:
The ratio of the electric force on the bee to the bee's weight is approximately 0.002. An electric field strength of about 51.1 N/C directed upward would allow the bee to hang suspended in the air.
Explanation:
To find the ratio of the electric force on the bee to the bee's weight (F/W), we first need to calculate both forces. The electric force (F) can be calculated using the equation F = qE, where q is the charge in coulombs (C) and E is the electric field in newtons per coulomb (N/C). The weight (W) of the bee can be calculated using W = mg, where m is the mass in kilograms (kg) and g is the acceleration due to gravity, approximately 9.8 m/s².
Given, q = 23 pC (23 × 10⁻¹² C) and E = 100 N/C. Thus, F = 23 × 10⁻¹² C × 100 N/C = 2.3 × 10⁻¹⁴ N. The mass of the bee is 120 mg (0.120 g or 0.000120 kg), so the weight W = 0.000120 kg × 9.8 m/s² = 1.176 × 10⁻³ N. Hence, the ratio F/W = (2.3 × 10⁻¹⁴ N) / (1.176 × 10⁻³ N) ≈ 0.002.
To allow the bee to hang suspended in the air, the upward electric force must equal the downward force of gravity. Therefore, the required electric field strength (E) can be found by rearranging F = qE to E = W/q. Substituting the values gives E = (1.176 × 10⁻³ N) / (23 × 10⁻¹² C) ≈ 51.1 N/C directed upward.
A 3-kW resistance heater in a water heater runs for 3 hours to raise the water temperature to the desired level. Determine the amount of electric energy used in both kWh and kJ. The amount of electricity used, in kWh, is kWh. The amount of electricity used, in kJ, is
Answer:
Energy, 9 kWh or 32400 kJ
Explanation:
Given that,
The power of heater, P = 3 kW
It runs for 3 hours to raise the water temperature to the desired level. We need to find the amount of electric energy used. We know that the electrical power of an object is given by total energy delivered per unit time. It is given by :
[tex]P=\dfrac{E}{t}[/tex]
[tex]E=P\times t[/tex]
[tex]E=3\ kW\times 3\ h[/tex]
E = 9 kWh
Since, 1 kWh = 3600 kJ
E = 32400 kJ
So, the amount of electric energy used is 9 kWh or 32400 kJ. Hence, this is the required solution.
The amount of electric energy used by a 3-kW resistance heater running for 3 hours would be 9kWh, which is equivalent to 32,400 kJ.
Explanation:To determine the amount of electric energy used, we use the formula E = Pt, where E represents energy, P is power, and t is time. Here, the power used is 3kW (or kilowatts) and time is 3 hours. So, E = 3kW * 3 hours = 9 kWh (kilowatt-hours).
Moving forward, 1 kWh is equal to 3600 kilojoules (kJ). Therefore, to convert the energy we obtained in kilowatt-hours to kilojoules, we multiply it by 3600. This amounts to: 9 kWh * 3600 kJ/kWh = 32,400 kJ.
Therefore, the amount of electric energy used is 9 kWh or 32,400 kJ.
Learn more about Electric Energy Consumption here:https://brainly.com/question/34856812
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How do astronomers use the Doppler effect to determine the velocities of astronomical objects?
Astronomers apply the Doppler effect because from there it is possible to obtain information about the change of light, which in turn affects the light spectrum and determines the movement of a body moving away or approaching us. The extent of the shift is directly proportional to the source's radial velocity relative to the observer.
The phenomenon that occurs to determine this process is linked to the wavelength. When the wave source moves towards you, the wavelength tends to decrease. This leads to a change in the color of the light moving towards the end of the spectrum, that is, towards the color blue. (It is really violet, but by convention the color blue was chosen as it is a more common color) When the source moves away from you and the wavelength lengthens, we call the color change a shift to red. Because the Doppler effect was first used with visible light in astronomy, the terms "blue shift" and "red shift" were well established.
Final answer:
Astronomers use the Doppler effect to calculate the velocities of stars and galaxies by observing changes in light wavelengths due to motion towards or away from the observer. It also helps in exoplanet detection and measuring a star's rotation speed by analyzing the broadened spectral lines.
Explanation:
Astronomers utilize the Doppler effect to determine the velocities of astronomical objects such as stars and galaxies. To calculate the radial velocity of an object, they require the speed of light, the original wavelength of the light emitted by the object, and the observed change in this wavelength due to the Doppler shift. This shift occurs because the object is moving relative to Earth—approaching objects cause a blue shift, where the wavelength shortens, while receding objects cause a red shift, where the wavelength lengthens.
The Doppler effect is also instrumental in exoplanet detection through stellar radial velocity measurements. When a planet orbits a star, it imparts a gravitational tug that causes the star to wobble slightly. This wobble changes the star's radial velocity, which can be detected as small shifts in the star's spectral lines, irrespective of the star's distance, as long as it can be observed with a high-resolution spectrograph.
Additionally, the Doppler effect helps measure the rotation speed of distant stars. By analyzing broadened spectral lines, which result from the spread of Doppler shifts due to the rotating star's edges moving towards and away from us, astronomers can infer how fast a star is spinning.