3 trams must be added
Explanation:
In this problem, there are 12 trams along the ring road, spaced at regular intervals.
Calling L the length of the ring road, this means that the space between two consecutive trams is
[tex]d=\frac{L}{12}[/tex] (1)
In this problem, we want to add n trams such that the interval between the trams will decrease by 1/5; therefore the distance will become
[tex]d'=(1-\frac{1}{5})d=\frac{4}{5}d[/tex]
And the number of trams will become
[tex]12+n[/tex]
So eq.(1) will become
[tex]\frac{4}{5}d=\frac{L}{n+12}[/tex] (2)
And substituting eq.(1) into eq.(2), we find:
[tex]\frac{4}{5}(\frac{L}{12})=\frac{L}{n+12}\\\rightarrow n+12=15\\\rightarrow n = 3[/tex]
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Answer: 3
Explanation:
We need to do (1/(5-1))x12
How many times does a typical person blink her eyes in a lifetime?
Answer:
415,224,000
Explanation:
a person blinks 10 times per minute ,60 minutes in a hour so 600 per hour,24 hours per day so 14,400 blinks per day and there are 365 days in a year so 5,256,000 blinks per year and an average person lives to 79 years so 415224000 in an average lifetime
The rate constant of a reaction is 7.8 × 10−3 s−1 at 25°C, and the activation energy is 33.6 kJ/mol. What is k at 75°C? Enter your answer in scientific notation.
The rate constant at 75°C is calculated using the two-point form of the Arrhenius equation. The original conditions, the new temperature, and the activation energy are substituted into the equation and solved for the new rate constant, k2. The result is k2 = 0.048, or 4.8 x 10^-2 s^-1.
Explanation:For calculating the rate constant at a different temperature, we can use the Arrhenius equation: k = Ae^(-Ea/RT) where k is the rate constant, A is the pre-exponential factor, Ea is the activation energy, R is the gas constant and T is the temperature in Kelvin.
To find the new temperature constant, we can transform the Arrhenius equation into the two-point form: ln(k2/k1) = (-Ea/R)(1/T2 - 1/T1).
Given:
k1 = 7.8 × 10−3 s−1, T1=25°C = 25 + 273 = 298K
Ea = 33.6 kJ/mol = 33,600 J/mol, R = 8.314 J/(mol·K)
T2 = 75°C = 75 + 273 = 348K
Substituting these values and solving for k2 (rate constant at 75°C), you get k2 = 0.048 or 4.8 x 10^-2 s^-1.
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A particle has a velocity of v→(t)=5.0ti^+t2j^−2.0t3k^m/s.
(a) What is the acceleration function?
(b) What is the acceleration vector at t = 2.0 s? Find its magnitude and direction.
Answer:
a)[tex]a=5 i+2t j - 6\ t^2k[/tex]
b)[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]
Explanation:
Given that
v(t) = 5 t i + t² j - 2 t³ k
We know that acceleration a is given as
[tex]a=\dfrac{dv}{dt}[/tex]
[tex]\dfrac{dv}{dt}=5 i+2t j - 6\ t^2k[/tex]
[tex]a=5 i+2t j - 6\ t^2k[/tex]
Therefore the acceleration function a will be
[tex]a=5 i+2t j - 6\ t^2k[/tex]
The acceleration at t = 2 s
a= 5 i + 2 x 2 j - 6 x 2² k m/s²
a=5 i + 4 j -24 k m/s²
The magnitude of the acceleration will be
[tex]a=\sqrt{5^2+4^2+24^2}\ m/s^2[/tex]
a= 24.83 m/s²
The direction of the acceleration a is given as
[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]
a)[tex]a=5 i+2t j - 6\ t^2k[/tex]
b)[tex]a=\dfrac{1}{24.83}(5i+4j-24k)\ m/s^2[/tex]
What is the net charge of the Earth if the magnitude of its electric field near the terrestrial surface is 1.08 ✕ 102 N/C? Assume the Earth is a sphere of radius 6.40 ✕ 106 m.
To solve this problem we will apply the concepts related to the electric field based on the laws of Coulomb. Said electric field is equivalent to the product between the Coulomb constant and the rate of change of the charge and the squared distance. Mathematically this is,
[tex]E = \frac{kq}{r^2}[/tex]
Here,
k = Coulomb's constant
q = Charge
r = Distance
Replacing we have that
[tex]E = \frac{kq}{r^2}[/tex]
[tex]1.08*10^2 = \frac{(9*10^{9})q}{(6.4*10^{6})^2}[/tex]
Solving for q,
[tex]q = 491520 C[/tex]
Therefore the net charge of the Earth under the previous condition is 491520 C
Given the following frequencies, calculate the corresponding periods. a. 60 Hz b. 8 MHz c. 140 kHz d. 2.4 GHz
The frequency can be defined as the inverse of the period, that is, it can be expressed as
[tex]T = \frac{1}{f}[/tex]
Here,
T = Period
f = Frequency
For each value we only need to replace the value and do the calculation:
PART A)
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{60Hz}[/tex]
T = 0.0166s
PART B)
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{8*10^6}[/tex]
[tex]T = 1.25*10^{-7} s[/tex]
PART C)
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{140*10^{3}}[/tex]
[tex]T = 7.14*10^{-6}s[/tex]
PART D)
[tex]T = \frac{1}{f}[/tex]
[tex]T = \frac{1}{2.4*10^{9}}[/tex]
[tex]T = 4.166*10^{-10}s[/tex]
Emily challenges her friend David to catch a dollar bill as follows. She holds the bill vertically, with the center of the bill between David's index finger and thumb. David must catch the bill after Emily releases it without moving his hand downward. If his reaction time is 0.2 s, will he succeed? Explain your reasoning.
Answer:
David will not be able to catch the bill .
Explanation:
Reaction time = .2 s .
During this period bill will fall vertically between the fingers.
Distance of fall = 1/2 x g x t²
= .5 x 9.8 x 0.2²
= 19.6 cm or 20 cm .
Generally the bill has size of the order of 25 cm . From central point it requires a fall of 12.5 cm for the bill to escape the catch . Since fall is of 20 cm , that means bill will fall below the level of fingers in .2 s .
So David will not be able to catch the bill.
Canada geese migrate essentially along a north–south direction for well over a thousand kilo-meters in some cases, traveling at speeds up to about 100 km/h. If one goose is flying at 100 km/h relative to the air but a 40-km/h wind is blowing from west to east, (a) at what angle relative to the north–south direction should this bird head to travel directly southward relative to the ground? (b) How long will it take the goose to cover a ground distance of 500 km from north to south? (Note: Even on cloudy nights, many birds can navigate by using the earth’s magnetic field to fix the north–south direction.)
Answer:
a) 66.4 relative to the west in the south-west direction
b) 5.455 hours
Explanation:
a)If the wind is blowing east-ward at a speed of 40km/h, then the west component of the geese velocity must be 40km/h in order to counter balance it. Geese should be flying south-west at an angle of
[tex]cos(\alpha) = 40 / 100 = 0.4[/tex]
[tex]\alpha = cos^{-1}(0.4) = 1.16 rad = 180\frac{1.16}{\pi} = 66.4^0[/tex] relative to the West
b) The south-component of the geese velocity is
[tex]100sin(\alpha) = 100sin(66.4^0) = 91.65 km/h[/tex]
The time it would take for the geese to cover 500km at this rate is
t = 500 / 91.65 = 5.455 hours
A soccer player kicks the ball that travels a distance of 60.0 m on a level field. The ball leaves his foot at an initial speed of (v0) and an angle of 26.0° above the ground. Find the initial speed (v0) of the ball.
Answer:
27.3 m/s
Explanation:
We are given that
Distance travel by ball=x=60 m
[tex]\theta=26^{\circ}[/tex]
We have to find the initial speed([tex]v_0)[/tex] of the ball.
[tex]x=v_0cos\theta t[/tex]
Using the formula
[tex]60=v_0cos 26 t[/tex]
[tex]t=\frac{60}{v_ocos 26}=\frac{60}{v_0\times 0.899}=\frac{66.7}{v_0}[/tex]
The value of y at point of foot of the vertical distance
y=0
[tex]y=v_0sin\theta t-\frac{1}{2}gt^2[/tex]
Using [tex]g=9.8m/s^2[/tex]
Using the formula
[tex]0=v_0sin 26\times \frac{66.7}{v_0}-4.9\times (\frac{66.7}{v_0})^2[/tex]
[tex]4.9\times \frac{(66.7)^2}{v^2_0}=0.44\times 66.7[/tex]
[tex]v^2_0=\frac{4.9\times (66.7)^2}{0.44\times 66.7}[/tex]
[tex]v^2_0=742.8[/tex]
[tex]v_0=\sqrt{742.8}=27.3 m/s[/tex]
Hence, the initial speed of the ball=27.3 m/s
Answer:
27.3 m/s
Explanation:
Horizontal range, R = 60 m
angle of projection, θ = 26°
Let the velocity of projection is vo.
Use the formula of range of the projectile
[tex]R = \frac{u^{2}Sin2\theta} {g}[/tex]
[tex]60 = \frac{v_{0}^{2}Sin52}{9.8}[/tex]
vo = 27.3 m/s
Thus, the velocity of projection is 27.3 m/s.
A has the magnitude 14.4 m and is angled 51.6° counterclockwise from the positive direction of the x axis of an xy coordinate system. Also, B = ( 14.3 m )i + (8.52 m )j on that same coordinate system. We now rotate the system counterclockwise about the origin by 20.0° to form an x'y' system. On this new system, what are (a)Ã and (b) B, both in unit-vector notation? (a) Number i 4.545346 It i 13.66381 Î Units m (b) Number i 10.52359 î+ i 12.89707 Units its
Final answer:
To find the transformed vector representations in a rotated coordinate system, the angle of vector A is adjusted by the rotation angle, and the components are calculated using trigonometric functions. Vector B's components in the rotated system are found using a rotation matrix.
Explanation:
The provided question pertains to transforming the representation of vectors in a rotated coordinate system in the subject of physics. The coordinate system is rotated counterclockwise, and the goal is to find the new representations of vectors A and B in unit-vector notation on the x'y' system. Given the initial magnitude and direction angle of vector A and the Cartesian components of vector B on the xy coordinate system, we can calculate their components on the rotated x'y' coordinate system.
The original vector A has a magnitude of 14.4 m and an angle of 51.6° from the positive x-axis. After rotation by 20°, the new angle becomes 51.6° - 20.0° = 31.6° from the new x'-axis. Using the formulas Ax' = A cos θ' and Ay' = A sin θ', where θ' is the new angle, we can find the rotated components of A.
The vector B is already given in Cartesian coordinates as ( 14.3 m )i + (8.52 m )j. To find the components of B in the rotated system, we use a rotation matrix, giving us new components Bx' and By'.
In conclusion, to find the transformed vectors in the rotated system, we apply the rotation to both the magnitude and angle of A, and use a rotation matrix for the components of B.
A 0.0575 kg ice cube at −30.0°C is placed in 0.557 kg of 35.0°C water in a very well insulated container, like the kind we used in class. The heat of fusion of water is 3.33 x 105 J/kg, the specific heat of ice is 2090 J/(kg · K), and the specific heat of water is 4190 J/(kg · K). The system comes to equilibrium after all of the ice has melted. What is the final temperature of the system?
Answer:
t= 22.9ºC
Explanation:
Assuming no heat exchange outside the container, before reaching to a condition of thermal equilibrium, defined by a common final temperature, the body at a higher temperature (water at 35ºC) must give heat to the body at a lower temperature (the ice), as follows:
Qw = c*m*Δt = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t) (1)
This heat must be the same gained by the ice, which must traverse three phases before arriving at a final common temperature t:
1) The heat needed to reach in solid state to 0º, as ice:
Qi =ci*m*(0ºC-(-30ºC) = 0.0575kg*2090(J/kg.ºC)*30ºC = 3605.25 J
2) The heat needed to melt all the ice, at 0ºC:
Qf = cfw*m = 3.33*10⁵ J/kg*0.0575 kg = 19147.5 J
3) Finally, the heat gained by the mass of ice (in liquid state) in order to climb from 0º to a final common temperature t:
Qiw = c*m*Δt = 4190 (J/kg.ºC)*0.0575 kg*(t-0ºC)
So, the total heat gained by the ice is as follows:
Qti = Qi + Qf + Qiw
⇒Qti = 3605.25 J + 19147.5 J + 240.9*t = 22753 J + 240.9*t (2)
As (1) and (2) must be equal each other, we have:
22753 J + 240.9*t = 4190 (J/kg.ºC)*0.557 kg*(35ºC-t)
⇒ 22753 J + 240.9*t = 81684 J -2334*t
⇒ 2575*t = 81684 J- 22753 J = 58931 J
⇒ [tex]t= \frac{58931J}{2575 J/C} = 22.9C[/tex]
⇒ t = 22.9º C
A plane flies 125 km/hr at 25 degrees north of east with a wind speed of 36 km/hr at 6 degrees south of east. What is the resulting velocity of the plane (in km/hr)?
Answer:
V = 156.85 Km/h
Explanation:
Speed of plane = 125 Km/h
angle of plane= 25° N of E
Speed of wind = 36 Km/h
angle of plane = 6° S of W
Horizontal component of the velocity
V_x = 125 cos 25° + 36 cos 6°
V_x = 149 Km/h
Vertical component of the velocity
V_y = 125 sin 25° - 36 sin 6°
V_y = 49 Km/h
Resultant of Velocity
[tex]V = \sqrt{V_x^2 + V_y^2}[/tex]
[tex]V = \sqrt{149^2 + 49^2}[/tex]
V = 156.85 Km/h
the resulting velocity of the plane is equal to V = 156.85 Km/h
A 22.0-kg child is riding a playground merrygo-round that is rotating at 40.0 rev/min. What centripetal force is exerted if he is 1.25 m from its center?
Answer:
F=480.491 N
Explanation:
Given that
mass ,m = 22 kg
Angular speed ω = 40 rev/min
[tex]\omega=\dfrac{2\pi \times 40}{60}\ rad/s[/tex]
ω =4.18 rad/s
The radius r= 1.25 m
We know that centripetal force is given as
F=m ω² r
Now by putting the values in the above equation we get
[tex]F=22\times 4.18^2\times 1.25\ N[/tex]
F=480.491 N
Therefore the centripetal force on the child will be 480.491 N.
A ray of light is incident on an air/water interface.
The ray makes an angle of θ1 = 32 degrees with respect to the normal of the surface. The index of the air is n1 = 1 while water is n2 = 1.33.
Choose an expression for the angle (relative to the normal to the surface) for the ray in the water, θ2.
a) θ2 = sin (θ1).n1/n2
b) θ2 = asin (n1/n2)
c) θ2 = asin (sin(θ1).n2/n1)
d) θ2 = asin (sin(θ1).n1/n2)
Answer:
[tex]\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})[/tex]
Explanation:
Given that,
The ray makes an angle of 32 degrees with respect to the normal of the surface.
The refractive index of air, [tex]n_1=1[/tex]
The refractive index of water, [tex]n_2=1.33[/tex]
Snell's law is given by :
[tex]n_1\ sin\theta_1=n_2\ sin\theta_2[/tex]
[tex]sin\theta_2=\dfrac{n_1\ sin\theta_1}{n_2}[/tex]
[tex]\theta_2=sin^{-1}(\dfrac{n_1\ sin\theta_1}{n_2})[/tex]
So, option (4) is correct. Hence, this is the required solution.
The answer is option d.
The correct expression for the angle (relative to the normal to the surface) for the ray in the water is d) θ2 = asin (sin(θ1).n1/n2), based on Snell's law of refraction.
Explanation:The question addresses the refraction of light, specifically the change in angle as light moves from air to water. According to Snell's law, which is used to calculate the angle of refraction, the correct expression in your options is d) θ2 = asin (sin(θ1).n1/n2). Here's a step by step process:
First, it's important to understand that light changes direction when it moves from one medium to another, a process known as refraction.Snell's law mathematically expresses this change and is written as n1*sin(θ1) = n2*sin(θ2). In your case, you want to find the angle θ2. So, rearranging Snell's law to solve for θ2 gives you θ2 = asin(n1*sin(θ1)/n2).Learn more about Refraction:https://brainly.com/question/2459833?referrer=searchResults
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Two equally charged tiny spheres of mass 1.0 g are placed 2.0 cm apart. When released, they begin to accelerate away from each other at What is the magnitude of the charge on each sphere, assuming only that the electric force is present? (k = 1/4πε0 = 9.0 × 109 N ∙ m2/C2)
Answer:
[tex]1.36\times 10^{-7} C[/tex]
Explanation:
We are given that
Mass of charged tine spheres=m=1 g=[tex]\frac{1}{1000}=0.001 kg[/tex]
1 kg=1000g
The distance between charged tine spheres=r=2 cm=[tex]\frac{2}{100}=0.02 m[/tex]
1 m=100 cm
Acceleration =[tex]a =414 m/s^2[/tex]
Let q be the charge on each sphere.
[tex]k=9\times 10^9Nm^2/C^2[/tex]
The electric force between two charged particle
[tex]F=\frac{kq_1q_2}{r^2}[/tex]
Using the formula
The force between two charged tiny spheres=[tex]F_e=\frac{kq^2}{(0.02)^2}[/tex]
According to Newton's second law , the net force
[tex]F=ma[/tex]
[tex]F=F_e[/tex]
[tex]0.001\times 414=\frac{9\times 10^9\times q^2}{(0.02)^2}[/tex]
[tex]q^2=\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}[/tex]
[tex]q=\sqrt{\frac{0.001\times 414\times (0.02)^2}{9\times 10^9}}[/tex]
[tex]q=1.36\times 10^{-7} C[/tex]
Hence, the magnitude of charge on each tiny sphere=[tex]1.36\times 10^{-7} C[/tex]
If the pressure of a substance is increased during a boiling process, will the temperature also increase, or will it remain constant? Why?
Answer:
on increasing pressure, temperature will also increase.
Explanation:
Considering the ideal gas equation as:
[tex]PV=nRT[/tex]
where,
P is the pressure
V is the volume
n is the number of moles
T is the temperature
R is Gas constant having value = 0.0821 L.atm/K.mol
Thus, at constant volume and number of moles, Pressure of the gas is directly proportional to the temperature of the gas.
P ∝ T
Also,
Also, using Gay-Lussac's law,
[tex]\frac {P_1}{T_1}=\frac {P_2}{T_2}[/tex]
Thus, on increasing pressure, temperature will also increase.
The ultimate normal stress in members AB and BC is 350 MPa. Find the maximum load P if the factor of safety is 4.5. AB has an outside diameter of 250mm and BC has an outside diameter of 150mm. Both pipes have a wall thickness of 8mm
Answer:
P_max = 278 KN
Explanation:
Given:
- The ultimate normal stress S = 350 MPa
- Thickness of both pipes t = 8 mm
- Pipe AB: D_o = 250 mm
- Pipe BC: D_o = 150 mm
- Factor of safety FS = 4.5
Find:
Find the maximum load P_max
Solution:
- Compute cross sectional areas A_ab and A_bc:
A_ab = pi*(D_o^2 - (D_o - 2t)^2) / 4
A_ab = pi*(0.25^2 - 0.234^2) / 4
A_ab = 6.08212337 * 10^-3 m^2
A_bc = pi*(D_o^2 - (D_o - 2t)^2) / 4
A_bc = pi*(0.15^2 - 0.134^2) / 4
A_bc = 3.568212337 * 10^-3 m^2
- Compute the Allowable Stress for each pipe:
sigma_all = S / FS
sigma_all = 350 / 4.5
sigma_all = 77.77778 MPa
- Compute the net for each member P_net,ab and P_net,bc:
P_net,ab = sigma_all * A_ab
P_net,ab = 77.77778 MPa*6.08212337 * 10^-3
P_net,ab = 473054.0399 N
P_net,bc = sigma_all * A_bc
P_net,bc = 77.77778 MPa*3.568212337 * 10^-3
P_net,bc = 277577.1721 N
- Compute the force P for each case:
P_net,ab = P + 50,000
P = 473054.0399 - 50,000
P = 423 KN
P_net,bc = P = 278 KN
- P_max allowed is the minimum of the two load P:
P_max = min (423, 278) = 278 KN
When looking at the top of a building 450 m away, the angle between the top of the building and your eye level is 30°. If your eyes are 1.5 m above the ground, how tall is the building? ANSWER IN 3 DECIMALS (###.###) You might need to use your calculator's sin,cos or tan
Answer:
261.307 m
Explanation:
b = Base of triangle = 450 m
p = Perpendicular of the triangle
[tex]\theta[/tex] = Angle of the triangle = [tex]30^{\circ}[/tex]
From trigonometry
[tex]tan\theta=\dfrac{p}{b}[/tex]
[tex]\Rightarrow p=btan\theta[/tex]
[tex]\Rightarrow p=450\times tan30[/tex]
[tex]\Rightarrow p=259.807\ m[/tex]
Height of the building = 1.5+259.807 = 261.307 m
You throw a baseball directly upward at time t = 0 at an initial speed of 13.5 m/s.
What is the maximum height the ball reaches above where it leaves your hand? Ignore air resistance and take g = 9.80 m/s².
Answer:
[tex]h=9.30m[/tex]
Explanation:
We have an uniformly accelerated motion, due to the gravitational acceleration. So, we use the kinematic equations, since the ball is throw directly upward, g is negative:
[tex]h=v_0t-\frac{gt^2}{2}[/tex]
First, we need to calculate the time taken by the ball to reach the maximum height, in this point its final speed is zero:
[tex]v_f=v_0-gt\\\\\frac{0-v_0}{-g}=t\\t=\frac{v_0}{g}\\t=\frac{13.5\frac{m}{s}}{9.8\frac{m}{s^2}}\\t=1.38s[/tex]
Now, we can calculate h:
[tex]h=v_0t-\frac{gt^2}{2}\\h=13.5\frac{m}{s}(1.38s)-\frac{9.8\frac{m}{s^2}(1.38s)^2}{2}\\h=9.30m[/tex]
Compute the ratio of the rate of heat loss through a single-pane window with area 0.15 m2 to that for a double-pane window with the same area. The glass of a single pane is 4.5 mm thick, and the air space between the two panes of the double-pane window is 6.60 mm thick. The glass has thermal conductivity 0.80 W/m⋅K. The air films on the room and outdoor surfaces of either window have a combined thermal resistance of 0.15 m2⋅K/W. Express your answer using two significant figures.
Answer:
2.80321285141
Explanation:
[tex]L_g[/tex] = Thickness of glass = 4.5 mm
[tex]k_g[/tex] = Thermal conductivity of glass = 0.8 W/mK
[tex]R_0[/tex] = Combined thermal resistance = [tex]0.15\times m^2K/W[/tex]
[tex]L_a[/tex] = Thickness of air = 6.6 mm
[tex]k_a[/tex] = Thermal conductivity of air = 0.024 W/mK
The required ratio is the inverse of total thermal resistance
[tex]\dfrac{2(L_g/k_g)+R_0+(L_a/k_a)}{(L_g/k_g)+R_0}\\ =\dfrac{2(4.5\times 10^{-3}/0.8)+0.15+(6.6\times 10^{-3}/0.024)}{(4.5\times 10^{-3}/0.8)+0.15}\\ =2.80321285141[/tex]
The ratio is 2.80321285141
Answer:
[tex]\frac{\dot Q}{\dot Q'} =2.6668[/tex]
Explanation:
Given:
area of the each window panes, [tex]A=0.15\ m^2[/tex]thickness of each pane, [tex]t_g=4.5\times 10^{-3}\ m[/tex]air gap between the two pane of a double pane window, [tex]t_a=6.6\times 10^{-3}\ m[/tex]thermal conductivity of glass, [tex]k_g=0.8\ W.m^{-1}.K^{-1}[/tex]thermal resistance of the air on the either sides of double pane window, [tex]R_{th}=0.15\ m^2.K.W^{-1}[/tex]Heat loss through single pane window:
Using Fourier's law of conduction,
[tex]\dot Q=A.dT\div (R_{th}+\frac{t_g}{k} )[/tex]
[tex]\dot Q=0.15\times dT\div (0.15+\frac{4.5\times 10^{-3}}{0.8})[/tex]
[tex]\dot Q=0.9638\ dT\ [W][/tex]
Heat loss through double pane window:
[tex]\dot Q'=dT\times A\div(R_{th}+2\times \frac{t_g}{k}+\frac{t_a}{k_a} )[/tex]
where:
[tex]dT=[/tex] change in temperature
[tex]k_a=[/tex] coefficient of thermal conductivity of air [tex]= 0.026\ W.m^{-1}.K^{-1}[/tex]
[tex]\dot Q'=dT\times 0.15\div (0.15+2\times \frac{4.5\times 10^{-3}}{0.8}+\frac{6.6\times 10^{-3}}{0.026})[/tex]
[tex]\dot Q'=0.3614\ dT\ [W][/tex]
Now the ratio:
[tex]\frac{\dot Q}{\dot Q'} =\frac{0.9638(dT)}{0.3614(dT)}[/tex]
[tex]\frac{\dot Q}{\dot Q'} =2.6668[/tex]
how much work is required to move a 1 microcoulomb charge by a distance of 5 meters along an equipotential line of 6V?
Answer:
The work done is zero.
Solution:
As per the question:
Charge, [tex]q = 1\mu C = 1\times 10^{- 6}\ C[/tex]
Distance moved, d = 5 m
Voltage, V = 6V
Now, we know that an equipotential surface is one where the potential is same everywhere on the surface.
Suppose the the voltage at a distance d = 5 m is V'
Thus
V' = 6 V, (since the surface is equipotential)
Work done in moving a charge is given by:
[tex]W = q\Delta V[/tex]
[tex]W = q(V - V')[/tex]
[tex]W = (1\times 10^{- 6})(V - V')[/tex]
[tex]W = (1\times 10^{- 6})(6 - 6) = 0[/tex]
Thus the work done in moving a charge on an equipotential surface comes out to be zero as the potential difference is zero.
Final answer:
The work required to move a 1 microcoulomb charge by a distance of 5 meters along an equipotential line of 6V is zero because there's no change in potential energy.
Explanation:
The question relates to determining the amount of work needed to move a charge along an equipotential line. When a charge moves along an equipotential, the potential energy of the charge does not change because the voltage (potential difference) across its path remains zero. In other words, the work done on the charge is zero since work is defined as the change in potential energy, which is given by the formula W = qV, where W is work, q is charge in coulombs, and V is potential difference in volts. For movement along an equipotential line, V = 0, hence, Work = 0 Joules.
A driver has a reaction time of 0.50s , and the maximum deceleration of her car is 6.0m/s2 . She is driving at 20m/s when suddenly she sees an obstacle in the road 50m in front of her.
Can she stop the car in time to avoid the collision?
Answer:
given,
reaction time.t_r = 0.50 s
deceleration of the car = 6 m/s²
initial speed,v = 20 m/s
distance at which the car stop = ?
distance travel by the car in reaction time
d= v x t_r
d = 20 x 0.5 = 10 m
using equation of motion
distance travel to stop the car
v² = u² + 2 a s
0² = 20² - 2 x 6 x s
12 s = 400
s = 33.33 m
Total distance travel by the car
D = 10 + 33.33
D = 43.33 m
Hence, the car stops before to avoid collision.
A 200-g block is attached to a horizontal spring and executes simple harmonic motion with a period of 0.250 s. The total energy of the system is 2.00 J. Find (a) the force constant of the spring and (b) the amplitude of the motion.
Answer:
(A) Spring constant will be 126.58 N/m
(B) Amplitude will be equal to 0.177 m
Explanation:
We have given mass of the block m = 200 gram = 0.2 kg
Time period T = 0.250 sec
Total energy is given TE = 2 J
(A) For mass spring system time period is equal to [tex]T=2\pi \sqrt{\frac{m}{K}}[/tex]
So [tex]0.250=2\times 3.14 \sqrt{\frac{0.2}{K}}[/tex]
[tex]0.0398=\sqrt{\frac{0.2}{K}}[/tex]
Now squaring both side
[tex]0.00158=\frac{0.2}{K}[/tex]
K = 126.58 N/m
So the spring constant of the spring will be 126.58 N/m
(B) Total energy is equal to [tex]TE=\frac{1}{2}KA^2[/tex], here K is spring constant and A is amplitude
So [tex]2=\frac{1}{2}\times 126.58\times A^2[/tex]
[tex]A^2=0.0316[/tex]
A = 0.177 m
So the amplitude of the wave will be equal to 0.177 m
A 0.73-kg metal sphere oscillates at the end of a vertical spring. As the spring stretches from 0.12 m to 0.23 m (relative to its unstrained length), the speed of the sphere decreases from 7.2 to 4.5 m/s. What is the spring constant of the spring?
The spring constant (k) can be obtained by employing the principle of conservation of energy. Here, the kinetic energy of the metal sphere at the beginning of the motion equals the potential energy at the maximum stretch of the spring. Substituting the given values into the energy equation, solving for 'k' yields the spring constant.
Explanation:The subject of this question lies within the domain of Physics, specifically the domain of mechanics and dynamics dealing with springs and oscillations. The spring constant (k) can be derived from the principle of conservation of energy. Here, we are ignoring friction and air resistance, meaning that the sum of kinetic energy and potential energy remains constant throughout the motion of the metal sphere.
At the beginning, all the energy is kinetic, and at the maximum stretch, all the energy is potential. This can be represented by the equation 0.5*m*v1^2 = 0.5*k*x2^2. By substituting the given values of m (mass = 0.73 kg), v1 (initial velocity = 7.2 m/s), and x2 (maximum displacement = 0.23 m), we can solve for k (spring constant). Here, the calculation would be as follows: k = m*v1^2/x2^2 = (0.73 kg*(7.2 m/s)^2)/(0.23 m)^2. After performing the required calculations, you can obtain the numerical value of the spring constant.
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How much taller (in m) does the Eiffel Tower become at the end of a day when the temperature has increased by 17°C? Its original height is 324 m and you can assume it is made of steel.
Answer:
324.066096 m.
Explanation:
Given that
height of the tower ,h= 324 m
The increase in temperature ,ΔT = 17°C
We know that coefficient of thermal expansion for steel ,α= 12 x 10⁻⁶ C⁻¹
The increase in height is given as
Δ h = α h ΔT
Now by putting the values in the above equation we get
Δ h= 12 x 10⁻⁶ x 324 x 17 m
Δ h=66096 x 10⁻⁶ m
Δ h=0.066096 m
Therefore the height of the tower become 324.066096 m.
Due to thermal expansion, the Eiffel Tower, made of steel, would increase in height by approximately 0.066 meters or 6.6 cm over a day when the temperature increases by 17°C.
Explanation:The height of the Eiffel Tower increases due to the phenomenon of thermal expansion, which is an increase in volume, including height, in response to an increase in temperature. The amount of expansion can be calculated using this formula: ΔL = α * L_original * ΔT. Let's apply the given values:
'α' the coefficient of linear expansion for steel is approximately 0.000012 per degree Celsius. 'L_original' is the original length in meters, which is 324 m. 'ΔT' is the change in temperature, which is 17°C.
So ΔL = 0.000012 * 324 * 17, which equates to approximately 0.066048 m. Therefore, the Eiffel Tower would increase in height by about 0.066 meters (or 6.6 cm) over the course of a day when the temperature increases by 17°C.
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The surface tension of a liquid is to be measured using a liquid film suspended on a U-shaped wire frame with an 12-cm-long movable side. If the force needed to move the wire is 0.096 N, determine the surface tension of this liquid in air.
The surface tension of the liquid in air is 0.8 N/m.
Explanation:To determine the surface tension of the liquid, we need to use the formula F = yL, where F is the force needed to move the wire, y is the surface tension, and L is the length of the wire. In this case, F = 0.096 N and L = 12 cm. We can rearrange the formula to solve for y: y = F / L. Plugging in the values, we get y = 0.096 N / 0.12 m = 0.8 N/m. So, the surface tension of the liquid in air is 0.8 N/m.
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In a phasor representation of a transverse wave on a string, what does the length of the phasor represent?
Final answer:
The length of the phasor in a phasor diagram representing a transverse wave on a string indicates the wave's amplitude, which corresponds to the maximum displacement of the medium's particles from their equilibrium position.
Explanation:
In the phasor representation of a transverse wave on a string, the length of the phasor corresponds to the amplitude of the wave. In a phasor diagram, this amplitude represents the maximum displacement of the wave particles from the equilibrium position as the wave propagates through the medium. The phasor's length will rotate in a circular motion at a rate determined by the wave's frequency, and this motion represents the oscillatory nature of the wave at a certain point in space over time. The amplitude is a crucial parameter as it determines the energy carried by the wave, with a larger amplitude indicating a greater energy transfer.
The phasor length is particularly important when analyzing multiple wave forms together, such as voltage and current in electrical circuits, where the ratio of their lengths can denote relative magnitudes, such as resistance in the circuit. In this context, however, we focus on mechanical waves on a string, and the length of the phasor would only represent the wave amplitude, not voltage or current.
A boy throws a ball upward with a speed v0 = 12 m/s. The wind imparts a horizontal acceleration of 0.4 m/s2 to the left. At what angle θ must the ball be thrown so that it returns to the point of release? Assume that the wind does not affect the vertical motion.
Answer:
The angle is 2.33°.
Explanation:
Given that,
Speed of ball = 12 m/s
Acceleration = 0.4 m/s²
We need to calculate the time
Using formula of time of flight
[tex]t=\dfrac{2u}{g}[/tex]
[tex]t=\dfrac{2v\cos\theta}{g}[/tex]
Put the value into the formula
[tex]t=\dfrac{2\times12\cos\theta}{9.8}[/tex]
[tex]t=2.44\cos\theta[/tex]
We need to calculate the angle
Using equation of motion along vertical direction
[tex]s=ut-\dfrac{1}{2}at^2[/tex]
[tex]s=v\sin\theta\times t-\dfrac{1}{2}at^2[/tex]
Put the value in the equation
[tex]0=12\sin\theta\times2.44\cos\theta-\dfrac{1}{2}\times0.4\times(2.44\cos\theta)^2[/tex]
[tex]2\times12\sin\theta\times2.44=0.4\times(2.44)^2\cos\theta[/tex]
[tex]\tan\theta=\dfrac{0.4\times2.44}{2\times12}[/tex]
[tex]\theta=\tan^{-1}(0.04066)[/tex]
[tex]\theta=2.33^{\circ}[/tex]
Hence, The angle is 2.33°.
Driving along a crowded freeway, you notice that it takes a time tt to go from one mile marker to the next. When you increase your speed by 7.4 mi/hmi/h , the time to go one mile decreases by 15 ss . What was your original speed?
Answer:
38.6 mi/h
Explanation:
7.4 mi/h = 7.4mi/h * (1/60)hour/min * (1/60) min/s = 0.00206 mi/s
Let v (mi/s) be your original speed, then the time t it takes to go 1 mi/s is
t = 1/v
Since you increase v by 0.00206 mi/s, your time decreases by 15 s, this means
t - 15 = 1/(v+0.00206)
We can substitute t = 1/v to solve for v
[tex]\frac{1}{v} - 15 = \frac{1}{v + 0.00206}[/tex]
We can multiply both sides of the equation with v(v+0.00206)
v+0.00206 - 15v(v+0.00206) = v
[tex]-15v^2 - 0.0308v + 0.00206 = 0[/tex]
[tex]v= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}[/tex]
[tex]v= \frac{0.03083\pm \sqrt{(-0.03083)^2 - 4*(-15)*(0.00205)}}{2*(-15)}[/tex]
[tex]v= \frac{0.03083\pm0.35}{-30}[/tex]
v = -0.01278 or v = 0.01 0724 mi/s
Since v can only be positive we will pick v = 0.010724 mi/s or 0.010724*3600 = 38.6 mi/h
If the center atom has three groups of electrons around it, what type of electron geometry is present?
Answer:
Trigonal planar
Explanation:
Trigonal planar - it is referred to the molecular shape of atom in which three bonds exist around any central atom. As there is no lone pair at the center hence all three atoms have taken the form of a triangle. All three atom lies at same plane and known as peripheral atoms
The angle between all the three atoms is 120 degree
Explain how astronomers might use spectroscopy to determine the composition and temperature of a star.
Final answer:
Astronomers utilize spectroscopy to analyze the spectrum of a star, identifying unique absorption lines corresponding to different elements, which reveals the star's composition. Spectral lines' broadening indicates temperature and pressure, and shifts in these lines help measure a star's motion, including radial and rotational velocities.
Explanation:
Understanding Stellar Spectroscopy
Astronomers use spectroscopy as a powerful tool to determine various characteristics of stars, including their composition and temperature. When light from a star passes through a prism or diffraction grating, it spreads out into a spectrum of colors. This spectrum contains dark lines known as absorption lines, which are unique to the elements present in the star's atmosphere, as different chemical elements absorb light at specific wavelengths. Therefore, by analyzing these lines, astronomers can identify the elements that make up a star.
Analyzing the broadening of spectral lines can inform us about a star's temperature and pressure. Warmer temperatures and higher pressures in a star's atmosphere tend to broaden the spectral lines. Additionally, the pressure can give clues about the star's size, as stars with lower atmospheric pressure tend to be larger, or giant stars.
Motions of the Stars are also revealed through spectroscopy. The Doppler effect causes spectral lines to shift towards the red end of the spectrum if the star is moving away from us (redshift) or towards the blue end if it is approaching (blueshift). This allows astronomers to measure the star's radial velocity. Spectral line broadening can also indicate the star's rotational velocity, while proper motion is deduced from the movement of the lines over time across the spectrum.