Answer:
Explanation:
Given
acceleration is given by
[tex]a=-g-c_Dv^2[/tex]
where [tex]\ddot{y}=a[/tex]
[tex]\dot{y}=v[/tex]
Also acceleration is given by
[tex]a=v\frac{\mathrm{d} v}{\mathrm{d} s}[/tex]
[tex]ds=\frac{v}{a}dv[/tex]
[tex]\int ds=\int \frac{v}{-g-0.001v^2}dv[/tex]
[tex]\Rightarrow Let -g-0.001v^2=t[/tex]
[tex]-0.001\times 2vdv=dt[/tex]
[tex]vdv=-\frac{dt}{0.002}[/tex]
[tex]at\ v_0=50\ m/s,\ t=-g-0.001(50)^2[/tex]
[tex]t=-g-2.5[/tex]
at [tex]v=0,\ t=-g[/tex]
[tex]\int_{0}^{s}ds=\int_{-g}^{-g-2.5}\frac{-dt}{0.002t}[/tex]
[tex]\int_{0}^{s}ds=\int^{-g}_{-g-2.5}\frac{dt}{0.002t}[/tex]
[tex]s=\frac{1}{0.002}lnt|_{-g}^{-g-2.5}[/tex]
[tex]s=\frac{1}{0.002}\ln (\frac{g+2.5}{g})[/tex]
[tex]s=113.608\ m[/tex]
when air drag is neglected maximum height reached is
[tex]h=\frac{v_0^2}{2g}[/tex]
[tex]h=\frac{50^2}{2\times 9.8}[/tex]
[tex]h=127.55\ m[/tex]
A space-based telescope can achieve a diffraction-limited angular resolution of 0.05″ for red light (wavelength 700 nm). What would the resolution of the instrument be (a) in the infrared, at 3.5 µm, and (b) in the ultraviolet, at 140 nm?
Answer:
a) [tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]
b) [tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]
Explanation:
We are comparing two wavelengths with the radius and diameter constant, and if we want to compare it, we need to use the following formula:
[tex]\frac{\theta_1}{\theta_2}= \frac{\lambda_1}{\lambda_2}[/tex]
Where [tex] \theta[/tex] represent the angular resolution and [tex]\lambda[/tex] the wavelength.
So if we have a fixed resolution and wavelength 1 and we want to find the resolution for a new condition we can solve for [tex] \theta_2[/tex] and we got
[tex] \theta_2 = \theta_1 \frac{\lambda_2}{\lambda_1}[/tex]
Part a
For this case the subindex 1 is for the color red and we know that:
[tex] \lambda_1 = 700 nm *\frac{1 \mu m}{1000 nm} = 0.7 \mu m[/tex]
And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]
And for the infrared case we know that [tex] \lambda_2 = 3.5 \mu m[/tex], so if we replace we got:
[tex] \theta_2 = 0.05 * \frac{3.5}{0.7} = 0.25[/tex]
Part b
For this case the subindex 1 is for the color red and we know that:
[tex] \lambda_1 = 700 nm[/tex]
And the angular resolution for the color red is specified as [tex] \theta_1 = 0.05[/tex]
And for the ultraviolet case we know that [tex] \lambda_2 = 140 nm[/tex], so if we replace we got:
[tex] \theta_2 = 0.05 * \frac{140}{700} = 0.01[/tex]
An object is released from rest near and above Earth’s surface from a distance of 10m. After applying the appropriate kinematic equation, a student predicts that it will take 1.43s for the object to reach the ground with a speed of 14.3m/s . After performing the experiment, it is found that the object reaches the ground after a time of 3.2s. How should the student determine the actual speed of the object when it reaches the ground? Assume that the acceleration of the object is constant as it falls.
The student determine the actual speed of the object when it reaches the ground as 12.52 m/s.
Given data:
The distance from the Earth's surface is, = 10 m.
Time taken to reach the ground is, t = 1.43 s.
The speed of object is, v = 14.3 m/s.
Experimental value of time interval is, t' = 3.2 s.
Use kinematic equation of motion to compute true value for acceleration of the ball as it reaches the ground:
[tex]h=ut+\dfrac{1}{2}a't'^{2} \\\\10=0 \times t+\dfrac{1}{2} \times a' \times 3.2^{2} \\\\a'=\dfrac{20}{3.2^{2}}\\\\a'= 1.95 \;\rm m/s^{2}[/tex]
Now, use the principle of conservation of total energy of system:
Potential energy - work done by air resistance = Kinetic energy
[tex]mgh-(ma) \times h=\dfrac{1}{2}mv^{2} \\\\gh-(a) \times h=\dfrac{1}{2}v^{2} \\\\v=\sqrt{2h(g-a)}[/tex]
Here, v is the actual speed of object while reaching the ground.
Solving as,
[tex]v=\sqrt{2 \times 10(9.8-1.95)}\\\\v=12.52 \;\rm m/s[/tex]
Thus, we can conclude that the student determine the actual speed of the object when it reaches the ground as 12.52 m/s.
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The actual speed of the object when it reaches the ground is approximately [tex]31.392 \, m/s.[/tex]
The actual speed of the object when it reaches the ground can be determined using the kinematic equation that relates initial velocity, acceleration, and time to the final velocity. Since the object is released from rest, the initial velocity [tex]\( u \)[/tex] is 0 m/s, and the acceleration [tex]\( a \)[/tex] is due to gravity, which is approximately [tex]\( 9.81 \, m/s^2 \)[/tex] near the Earth's surface.
The kinematic equation that relates these quantities to the final velocity [tex]\( v \)[/tex] is:
[tex]\[ v = u + at \][/tex]
Given that [tex]\( u = 0 \, m/s \)[/tex] and [tex]\( a = 9.81 \, m/s^2 \)[/tex], and the time [tex]\( t \)[/tex] to reach the ground is [tex]\( 3.2 \, s \)[/tex], we can substitute these values into the equation to find the actual final velocity:
[tex]\[ v = 0 + (9.81 \, m/s^2)(3.2 \, s) \] \[ v = (9.81)(3.2) \, m/s \] \[ v = 31.392 \, m/s \][/tex]
Therefore, the actual speed of the object when it reaches the ground is approximately [tex]31.392 \, m/s.[/tex]
"Two point masses m and M are separated by a distance d. If the separation d remains fixed and the masses are increased to the values 3 m and 3 M respectively,
how does the gravitational force between them change?
Answer:
The force of gravitational attraction increases by 9 as the two point masses increase by 3.
Explanation:
Gravitational force of attraction, F is the force that pulls two point masses, m and M which are separated by a distance, d.
Mathematically,
Fg = GMm/r^2
Initially,
M1 = M1
M2 = M2
The remaining parameters are unchanged.
Fg1 = G * M1 * m1/(d/2)^2
Then,
M1 = 3M1
M2 = 3M2
Fg2 = G * 3M1 * 3M2/(d/2)^2
Making the constants G/(d/2)^2 the subject of formula and then comparing both equations,
= Fg1 = (M1 * M2); Fg2 = (9 * M1 * M2)
= Fg2 = 9 * Fg1
The force of gravitational attraction increases by 9 as the two point masses increase by 3.
Consider a portion of a cell membrane that has a thickness of 7.50nm and 1.3 micrometers x 1.3 micrometers in area. A measurement of the potential difference across the inner and outer surfaces of the membrane gives a reading of 92.2mV. The resistivity of the membrane material is 1.30 x 10^7 ohms*m
PLEASE SHOW WORK!
a) Determine the amount of current that flows through this portion of the membrane
Answer: _____A
b) By what factor does the current change if the side dimensions of the membrane portion is halved? The other values do no change
increase by factor of 2
decrease by factor of 8
decrease by factor of 2
decrease by a factor of 4
increase by factor of 4
The amount of current that flows through this given portion of a cell membrane, calculated using Ohm's law and the properties of the membrane, is 1.60 µA. If the side dimensions of the membrane are halved, the current will decrease by a factor of 4.
Explanation:The relevant concept needed to answer these questions is Ohm's Law, defined as Voltage = Current x Resistance. In this context, Resistance = Resistivity x (Thickness/Area) and the area is a square.
a) Determine the amount of current that flows through this portion of the membrane:
First, calculate the resistance: R = ρ x (Thickness/ Area)
Remove the micrometers units of the area and convert it into meters to match the ρ units. So, you get an area of 1.3 x 10^-6 m x 1.3 x 10^-6 m = 1.69 x 10^-12 m^2. Then, R = 1.30 x 10^7 Ω*m x (7.50 x 10^-9 m / 1.69 x 10^-12 m^2) = 57.404 Ω.
By plugging the calculated resistance and given voltage into Ohm's Law, we can find the current: I = V/R = 92.2 x 10^-3 V / 57.4 Ω = 1.60 μA
b) By what factor does the current change if the side dimensions of the membrane portion is halved:If the side dimensions are halved, the area of the membrane becomes one-fourth of the original, thus the resistance increases by a factor of 4. According to Ohm's Law, as resistance increases, the current decreases, meaning that if the resistance is multiplied by 4, the current will decrease by a factor of 4.
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a) Therefore, the amount of current that flows through this portion of the membrane is approximately [tex]\({1.60 \times 10^{-6} \, \text{A}} \)[/tex]. b) The correct answer is decrease by a factor of 5208. The current decreases by a factor of approximately 5208.
Part (a): Determine the amount of current that flows through this portion of the membrane
To find the current ( I ) flowing through the membrane portion, we use Ohm's law and the given potential difference ( V ) across the membrane.
1. Calculate the resistance ( R ) of the membrane:
The resistivity [tex](\( \rho \))[/tex] is given as [tex]\( 1.30 \times 10^7 \) ohms\·m.[/tex]
First, calculate the cross-sectional area ( A ) of the membrane portion:
[tex]\[ A = 1.3 \, \mu \text{m} \times 1.3 \, \mu \text{m} = (1.3 \times 10^{-6} \, \text{m})^2 = 1.69 \times 10^{-12} \, \text{m}^2 \][/tex]
Then, calculate the resistance ( R ):
[tex]\[ R = \frac{\rho \cdot L}{A} \][/tex]
[tex]\[ R = \frac{1.30 \times 10^7 \, \text{ohm} \cdot \text{m} \cdot 7.50 \times 10^{-9} \, \text{m}}{1.69 \times 10^{-12} \, \text{m}^2} \][/tex]
[tex]\[ R = \frac{9.75 \times 10^{-2}}{1.69 \times 10^{-12}} \approx 5.77 \times 10^7 \, \text{ohms} \][/tex]
2. Calculate the current ( I ):
Ohm's law states [tex]\( I = \frac{V}{R} \).[/tex]
Given potential difference [tex]\( V = 92.2 \, \text{mV} = 92.2 \times 10^{-3} \, \text{V} \):[/tex]
[tex]\[ I = \frac{92.2 \times 10^{-3} \, \text{V}}{5.77 \times 10^7 \, \text{ohms}} \approx 1.60 \times 10^{-6} \, \text{A} \][/tex]
Part (b): By what factor does the current change if the side dimensions of the membrane portion is halved?
If the side dimensions of the membrane portion are halved, the cross-sectional area ( A ) of the membrane will decrease by a factor of ( 4 ) (since both length and width are halved).
1. New cross-sectional area ( A' ):
[tex]\[ A' = \left( \frac{1.3 \, \mu \text{m}}{2} \right) \times \left( \frac{1.3 \, \mu \text{m}}{2} \right) = \left( \frac{1.3}{2} \times 10^{-6} \, \text{m} \right)^2 = 0.325 \times 10^{-12} \, \text{m}^2 \][/tex]
2. New resistance ( R' ):
Using the same resistivity [tex]\( \rho \)[/tex] and thickness ( L ):
[tex]\[ R' = \frac{\rho \cdot L}{A'} = \frac{1.30 \times 10^7 \cdot 7.50 \times 10^{-9}}{0.325 \times 10^{-12}} \approx 3.00 \times 10^8 \, \text{ohms} \][/tex]
3. New current ( I' ):
[tex]\[ I' = \frac{V}{R'} = \frac{92.2 \times 10^{-3}}{3.00 \times 10^8} \approx 3.07 \times 10^{-10} \, \text{A} \][/tex]
4. Calculate the factor by which the current changes:
[tex]\[ \frac{I'}{I} = \frac{3.07 \times 10^{-10}}{1.60 \times 10^{-6}} \approx 1.92 \times 10^{-4} \][/tex]
Since the current decreases, we consider the reciprocal:
[tex]\[ \frac{I}{I'} \approx \frac{1}{1.92 \times 10^{-4}} \approx 5208 \][/tex]
A block of mass M M is placed on a semicircular track and released from rest at point P P , which is at vertical height H 1 H1 above the track’s lowest point. The surfaces of the track and block are considered to be rough such that a coefficient of friction exists between the track and the block. The block slides to a vertical height H 2 H2 on the other side of the track. How does H 2 H2 compare to H 1 H1 ?
Answer:
Explanation:
A block of mass M is placed on a semicircular track and released from rest at point P , which is at vertical height H₁ above the track’s lowest point.
Its initial potential energy = mgH₁
Kinetic energy = 0
Total energy = mgH₁
When block slides to a vertical height H₂ on the other side of the track
Its final potential energy = mgH₂
Kinetic energy = 0
Total final energy = mgH₂
As negative work is done by frictional force while block moves ,
final energy < initial energy
mgH₂ < mgH₁
H₂ < H₁
H₂ will be less than H₁ .
If the frequency of an electromagnetic wave increases, does the number of waves passing by you increase, decrease, or stay the same?
Answer:
If the frequency of an electromagnetic wave increases, the number of waves passing by you increase.
Explanation:
The total number of vibration or oscillation per unit time is called frequency of a waver. It is given by :
[tex]f=\dfrac{n}{t}[/tex]
n is the number of waves passing
t is the time taken
It is clear that the frequency is directly proportional to the number of waves. Hence, if the frequency of an electromagnetic wave increases, the number of waves passing by you increase.
How much work does it take to slide a box 37 meters along the ground by pulling it with a 217 N force at an angle of 19° from the horizontal?
Answer:
W = 7591.56 J
Explanation:
given,
distance of the box, d = 37 m
Force for pulling the box, F = 217 N
angle of inclination with horizontal,θ = 19°
We know,
Work done is equal to product of force and the displacement.
W = F.d cos θ
W = 217 x 37 x cos 19°
W = 7591.56 J
Hence, the work done to pull the box is equal to W = 7591.56 J
Final answer:
The work done to slide the box is 7586.09 Joules.
Explanation:
To calculate the work done to slide a box along the ground, we can use the formula:
Work = Force x Distance x cos(theta)
Where:
Force = 217 N (the force applied to pull the box)
Distance = 37 meters (the distance the box is being slid)
theta = 19° (the angle between the applied force and the horizontal)
Plugging in these values into the formula, we get:
Work = 217 N x 37 m x cos(19°)
Calculating this using a calculator, we find that the work done to slide the box is approximately 7586.09 Joules.
Two children of mass 20 kg and 30 kg sit balanced on a seesaw with the pivot point located at the center of the seesaw. If the children are separated by a distance of 3 m, at what distance from the pivot point is the small child sitting in order to maintain the balance?
Answer:
Explanation:
Given
mass of first child [tex]m_1=20\ kg[/tex]
mass of second child [tex]m_2=30\ kg[/tex]
Distance between two children is [tex]d=3\ m[/tex]
Suppose light weight child is placed at a distance of x m from Pivot point
therefore
Torque due to heavy child [tex]T_1=m_2g\times (3-x)[/tex]
Torque due to small child [tex]T_2=m_1g\times x[/tex]
Net Torque about Pivot must be zero
Therefore [tex]T_1=T_2[/tex]
[tex]30\times g\times (3-x)=20\times g\times x[/tex]
[tex]9-3x=2x[/tex]
[tex]9=5x[/tex]
[tex]x=\frac{9}{5}[/tex]
[tex]x=1.8\ m[/tex]
How does a person become "charged" as he or she shuffles across a carpet with bare feet on a dry winter day?
This process occurs because there is a contact between the carpet and the person's feet. Basically that contact generates the transfer of some electrons to the carpet on dry winter days.
In this way a person is charged when dragging bare feet on the carpet on a dry winter day.
Therefore, the net positive charge occurs on the surface of the carpet.
A person becomes charged when shuffling across a carpet due to the transfer of electrons from the feet to the carpet, leaving a net positive charge. The lack of humidity on a dry winter day allows the static charge to build up, leading to noticeable static shocks when touching a metal object. Humidity helps in dissipating the charge, making shocks less common on humid days.
When a person shuffles across a carpet with bare feet, they can become "charged" through a process known as charging by friction. This occurs when electrons are transferred from one surface to another due to the contact and relative motion between them. In this case, electrons move from the person's feet to the carpet, leaving the feet with a net positive charge.
Materials have different affinities for electrons, and when they come into close contact, the one with the higher affinity will take on electrons from the other. Since a dry winter day has low humidity, there is less moisture in the air to carry away the excess electrons. Therefore, the static charge you accumulate is less likely to be neutralized by the surrounding air, making static shocks more frequent and noticeable when you touch a metal object like a doorknob.
The reason for the shock is the rapid movement of electrons as they try to redistribute themselves to reach a state of electrical neutrality. When you touch a metal object, the excess electrons on your body rapidly transfer to the metal, causing the shock. On a humid day, the air's moisture helps electrons move away from your body more easily, preventing the build-up of a significant static charge.
If the wind velocity is 43 km/h due south, in what direction should the pilot set her course to travel due west? Use the same airspeed of 212 km/h
Answer:
11.48 degree N of W
Explanation:
We are given that
Wind velocity=[tex]v_w=-43[/tex]km/h
Because wind is blowing towards south
Air speed=[tex]v_a=-212km/h[/tex]
Because the captain want to move with air speed in west direction.
x component of relative velocity=-212 km/h
y-Component of relative velocity=-43km/h
Direction=[tex]\theta=tan^{-1}(\frac{y}{x})[/tex]
[tex]\theta=tan^{-1}(\frac{-43}{-212}=11.48^{\circ}[/tex]N of W
Hence, the direction in which the pilot should set her course to travel due west=11.48 degree N of W
At 47 °C, what is the fraction of collisions with energy equal or greater than an activation energy of 88.20 kJ/mol?
Answer:
The fraction of collision is [tex]4.00\times10^{-15}[/tex]
Explanation:
Given that,
Temperature = 47°C
Activation energy = 88.20 KJ/mol
From Arrhenius equation,
[tex]k=Ae^{-\dfrac{E_{a}}{RT}}[/tex]
Here, [tex]e^{-\dfrac{E_{a}}{RT}}[/tex]=fraction of collision
We need to calculate the fraction of collisions
Using formula of fraction of collisions
[tex]f=e^{-\dfrac{E_{a}}{RT}}[/tex]
Where f = fraction of collision
E = activation energy
R = gas constant
T = temperature
Put the value into the formula
[tex]f=e^{-\dfrac{88.20}{8.314\times10^{-3}\times(47+273)}}[/tex]
[tex]f=4.00\times10^{-15}[/tex]
Hence, The fraction of collision is [tex]4.00\times10^{-15}[/tex]
You run due east at a constant speed of 3.00 m/s for a distance of 120.0 m and then continue running east at a constant speed of 5.00 m/s for another 120.0 m. For the total 240.0-m run, is your average velocity 4.00 m/s, greater than 4.00 m/s, or less than 4.00 m/s? Explain.
Answer:
Explanation:
Given
Speed while running towards east is [tex]v_1=3\ m/s[/tex]
Distance traveled in east direction [tex]x_1=120\ m[/tex]
For Another interval you run with velocity
[tex]v_2=5\ m/s[/tex]
[tex]x_2=240\ m[/tex]
Total displacement[tex]=x_1+x_2[/tex]
[tex]=120+120=240\ m[/tex]
Time for first interval
[tex]t_1=\frac{x_1}{v_1}=\frac{120}{3}[/tex]
[tex]t_1=\frac{120}{3}=40\ s[/tex]
Time for second interval
[tex]t_2=\frac{x_2}{v_2}=\frac{120}{5}=24\ s[/tex]
total time [tex]t=t_1+t_2[/tex]
[tex]t=40+24=64\ s[/tex]
average velocity [tex]v_{avg}=\frac{x_1+x_2}{t}[/tex]
[tex]v_{avg}=\frac{240}{64}=3.75\ m/s[/tex]
Therefore average velocity is less than [tex]4 m/s[/tex]
What is the electric potential V V due to the nucleus of hydrogen at a distance of 5.292 × 10 − 11 m 5.292×10−11 m ?
Answer:
27.1806500378 V
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
q = Charge = [tex]1.6\times 10^{-19}\ C[/tex]
r = Distance = [tex]5.292\times 10^{-11}\ m[/tex]
Voltage is given by
[tex]V=k\dfrac{q}{r}\\\Rightarrow V=8.99\times 10^9\dfrac{1.6\times 10^{-19}}{5.292\times 10^{-11}}\\\Rightarrow V=27.1806500378\ V[/tex]
The potential difference is 27.1806500378 V
A rescue airplane is diving at an angle of 37º below the horizontal with a speed of 250 m/s. It releases a survival package when it is at an altitude of 600 m. If air resistance is ignored, the horizontal distance of the point of impact from the plane at the moment of the package's release is what? 1. 720 m.
2. 420 m.3. 2800 m.
4. 6800 m
5. 5500 m
Answer:
The correct option is 1. 720 m
Explanation:
Projectile Motion
When an object is launched in free air (no friction) with an initial speed vo at an angle [tex]\theta[/tex], it describes a curve which has two components: one in the horizontal direction and the other in the vertical direction. The data provided gives us the initial conditions of the survival package's launch.
[tex]\displaystyle V_o=250\ m/s[/tex]
[tex]\displaystyle \theta =-37^o[/tex]
The initial velocity has these components in the x and y coordinates respectively:
[tex]\displaystyle V_{ox}=250\ cos(-37^o)=199.7\ m/s[/tex]
[tex]\displaystyle V_{oy}=250\ sin(-37^o)=-150.5\ m/s[/tex]
And we know the plane has an altitude of 600 m, so the package will reach ground level when:
[tex]\displaystyle y=-600\ m[/tex]
The vertical distance traveled is given by:
[tex]\displaystyle y=V_{oy}\ t-\frac{g\ t^2}{2}=-600[/tex]
We'll set up an equation to find the time when the package lands
[tex]\displaystyle -150.5t-4.9\ t^2=-600[/tex]
[tex]\displaystyle -4.9\ t^2-150.5\ t+600=0[/tex]
Solving for t, we find only one positive solution:
[tex]\displaystyle t=3.6\ sec[/tex]
The horizontal distance is:
[tex]\displaystyle x=V_{ox}.t=199.7\times3.6=720\ m[/tex]
The correct option is 1. 720 m
The horizontal distance of the point of impact from the plane at the moment of the package's release is approximately 1760 m.
Explanation:The time taken for the package to reach the ground can be found using the equation y = v0y * t + (1/2) * g * t2, where y is the initial altitude, v0y is the vertical component of the initial velocity, t is the time taken, and g is the acceleration due to gravity. Solving for t gives us a value of approximately 5 seconds. The horizontal distance traveled by the package can be found using the equation x = v0x * t, where x is the horizontal distance, v0x is the horizontal component of the initial velocity, and t is the time taken. Plugging in the values gives us x = 250 m/s * cos(37º) * 5 s, which simplifies to approximately 1760 m. So the horizontal distance of the point of impact is approximately 1760 m.
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P3.43 Water at 20 C flows through a 5-cm-diameter pipe that has a 180 vertical bend, as in Fig. P3.43. The total length of pipe between flanges 1 and 2 is 75 cm. When the weight flow rate is 230 N/s, p1
Answer:
F_x = 750.7 N
Explanation:
Given:
- Length of the pipe between flanges L = 75 cm
- Weight Flow rate is flow(W) = 230 N/c
- P_1 = 165 KPa
- P_2 = 134 KPa
- P_atm = 101 KPa
- Diameter of pipe D = 0.05 m
Find:
The total force that the flanges must withstand F_x.
Solution:
- Use equation of conservation of momentum.
(P_1 - P_a)*A + (P_2 - P_a)*A - F_x = flow(m)*( V_2 - V_1)
- From conservation of mass:
A*V_1 = A*V_2
V_1 = V_2 ( but opposite in directions)
- Hence,
(P_1 - P_a)*A + (P_2 - P_a)*A - F_x = - 2*flow(m)*V_1
flow(m) = flow(W) / g
p*A*V_1 = flow(W) / g
V_1 = flow(W) / g*p*A
Hence,
(P_1 - P_a)*A + (P_2 - P_a)*A - F_x = - 2*flow(W)^2 / g^2*p*A
Hence, compute:
64*10^3 *pi*0.05^2 /4 + 33*10^3 *pi*0.05^2 /4 - F_x = - 2*(230/9.81)^2 / 997*pi*0.05^2 /4
125.6 + 64.7625 - F_x = -560.33
F_x = 750.7 N
A heat engine operates at 30% of its maximum possible efficiency and needs to do 995 J of work. Its cold reservoir is at 22 ºC and its hot reservoir is at 610 ºC. (a) How much energy does it need to extract from the hot reservoir? (b) How much energy does it deposit in the cold reservoir?
Answer:
(a) The energy extracted from the hot reservoir (Qh) is 3316.67J
(b) The energy deposited in the cold reservoir (Qc) is 2321.67J
Explanation:
Part (a) The energy extracted from the hot reservoir (Qh)
e = W/Qh
where;
e is the maximum efficiency of the system = 30% = 0.3
W is the the work done on the system = 995 J
Qh is the heat absorbed from the hot reservoir
Qh = W/e
Qh = 995/0.3
Qh = 3316.67J
Part (b) The energy deposited in the cold reservoir (Qc)
e = W/Qh
W = Qh - Qc
where;
Qc is the heat deposited in the cold reservoir
e = (Qh - Qc)/Qh
Qh - Qc = e*Qh
Qc = Qh - e*Qh
Qc = 3316.67J - 0.3*3316.67J
Qc = 3316.67J - 995J
Qc = 2321.67J
A pressure gage connected to a tank reads 55 kPa at a location where the atmospheric pressure is 72.1 cmHg. The density of mercury is 13,600 kg/m3 . Calculate the absolute pressure in the tank.
Answer:
Explanation:
Given
Gauge Pressure of a tank is
[tex]P_{gauge}=55\ kPa[/tex]
at that place atmospheric Pressure is [tex]h=72.1\ cm\ of\ Hg[/tex]
Density of mercury [tex]\rho _{Hg}=13600\ kg/m^3[/tex]
Atmospheric Pressure in kPa is given by
[tex]P_{atm}=\rho _{Hg}\times g\times h[/tex]
[tex]P_{atm}=13600\times 9.8\times 0.721[/tex]
[tex]P_{atm}=96.09\ kPa[/tex]
and Absolute Pressure is summation of gauge pressure and atmospheric Pressure
[tex]P_{abs}=P_{gauge}+P_{atm}[/tex]
[tex]P_{abs}=55+96.09=151.09\ kPa[/tex]
A uniformly dense solid disk with a mass of 4 kg and a radius of 2 m is free to rotate around an axis that passes through the center of the disk and perpendicular to the plane of the disk. The rotational kinetic energy of the disk is increasing at 20 J/s. If the disk starts from rest through what angular displacement (in rad) will it have rotated after 5 s
The angular displacement of the solid disk after 5 seconds is 0 rad.
Explanation:To determine the angular displacement of the solid disk after 5 seconds, we can use the formula:
Δθ = ΔErot / (I * ω)
where Δθ is the angular displacement, ΔErot is the change in rotational kinetic energy, I is the moment of inertia of the disk, and ω is the angular velocity of the disk.
The moment of inertia of a solid disk rotating around an axis through its center perpendicular to its plane is given by:
I = (1/2) * m * r2
where m is the mass of the disk and r is the radius.
Given that ΔErot = 20 J/s, m = 4 kg, r = 2 m, and the disk starts from rest, we can calculate the angular displacement:
Δθ = ΔErot / (I * ω) = 20 / [(1/2) * 4 * 22 * ω]
Since the disk starts from rest, the initial angular velocity ω is 0. Therefore, the angular displacement after 5 seconds is:
Δθ = 20 / [(1/2) * 4 * 22 * 0] = 0 rad
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A sled starts from rest at the top of a hill and slides down with a constant acceleration. At some later time it is 14.4 m from the top; 2.00 s after that it is 25.6 m from the top, 2.00 slater 40.0 m from the top, and 2.00 s later it is 57.6 m from the top.
(a) What is the magnitude of the average velocity of the sled during each of the 2.00-s intervals after passing the 14.4-m point?
(b) What is the acceleration of the sled?
(c) What is the speed of the sled when it passes the 14.4-m point?
(d) How much time did it take to go from the top to the 14.4-m point?
(e) How far did the sled go during the first second after passing the 14.4-m point?
Answer:
(a) 5.6 m/s, 7.2 m/s, and 8.8 m/s, respectively.
(b) 0.8 m/s^2
(c) 4.8 m/s
(d) 6 s
(e) 5.2 m
Explanation:
(a) The average velocity is equal to the total displacement divided by total time.
For the first 2s. interval:
[tex]V_{\rm avg} = \frac{\Delta x }{\Delta t} = \frac{25.6 - 14.4}{2} = 5.6~{\rm m/s}[/tex]
For the second 2s. interval:
[tex]V_{\rm avg} = \frac{40 - 25.6}{2} = 7.2~{\rm m/s}[/tex]
For the third 2s. interval:
[tex]V_{\rm avg} = \frac{57.6 - 40}{2} = 8.8~{\rm m/s}[/tex]
(b) Every 2 s. the velocity increases 1.6 m/s. Therefore, for each second the velocity increases 0.8 m/s. So, the acceleration is 0.8 m/s2.
(c) The sled starts from rest with an acceleration of 0.8 m/s2.
[tex]v^2 = v_0^2 + 2ax\\v^2 = 0 + 2(0.8)(14.4)\\v = 4.8~{\rm m/s}[/tex]
(d) The following kinematics equation will yield the time:
[tex]\Delta x = v_0 t + \frac{1}{2}at^2\\14.4 = 0 + \frac{1}{2}(0.8)t^2\\t = 6~{\rm s}[/tex]
(e) The same kinematics equation will yield the displacement:
[tex]\Delta x = v_0t + \frac{1}{2}at^2\\\Delta x = (4.8)(1) + \frac{1}{2}(0.8)1^2\\\Delta x = 5.2~{\rm m}[/tex]
An initially stationary 2.7 kg object accelerates horizontally and uniformly to a speed of 13 m/s in 4.0 s. (a) In that 4.0 s interval, how much work is done on the object by the force accelerating it? What is the instantaneous power due to that force (b) at the end of the interval and (c) at the end of the first half of the interval?
Explanation:
A.
Given:
V = 13 m/s
t = 4 s
Constant acceleration, a= (V-Vi)/t
= 13/4
= 3.25 m/s^2
F = mass * acceleration
= 2.7 * 3.25
= 8.775 N.
Using equations of motion,
distance,S = (13 * 4) - (1/2)(3.25)(4^2)
= 26 m
Workdone, W = force * distance
= 8.775 * 26
= 228.15 J
B.
Instantaneous power, P = Force *Velocity
= 8.775 * 13
= 114. 075 W
C.
t = 2 s,
Constant acceleration, a= (V-Vi)/t
= 13/2
= 6.5 m/s^2
Force = mass * acceleration
= 2.7 * 6.5
= 17.55 N
Instantaneous power, P = Force *Velocity
= 17.55 * 13
= 228.15 W.
= 114. 075 W.
The solar system is of largely uniform composition. (T/F)
Answer:
False
Explanation:
Sun mass is dominating in Solar system as compared to other planets, asteroids and comets. Sun itself accounting for the 99.9% of the mass of the solar system. Hence the gravitational force exerted by the Sun dominates the other objects in the solar system. So we can conclude that solar system has non-uniform composition. The given statement is false
The plates of a parallel-plate capacitor are 3.50 mm apart, and each carries a charge of magnitude 75.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 5.00×10^6 V/m.a. What is the potential difference between the plates?b. What is the area of each plate?c. What is the capacitance?
Answer:
Vab =17.5kV
A = 16.9 cm2
C = 4.27pF
Explanation:
a) Find the voltage difference:
Vab = Ed
E Electric field
d distance between plates
Vab potential difference
d = 3.5mm
= 3.5 * 10^(-3) m
Q = 75.0nC
= 75 * 10^(-9)
E = 5.00 * 10^6 V/m
Vab = (5.00 * 10^6) * (3.5 * 10^(-3))
= 17.5 * 10^3 V
=17.5kV
b. What is the area of the plate?
The relation between the electric field and area is given as:
E = Q/(ϵ0 * A)
A = Q/(ϵ0 *E)
Where ϵ0 is the electric constant and equals 8.854 × 10^ (-12) C2/N•m2
A = 75 * 10^ (-9) / (8.854 × 10^ (-12) (5.00 * 10^6)
= 1.69 X 10^ (-3) m2
= 16.9 cm2
c. Find the capacitance
The equation relating capacitance, area of plate and plate distance is given by:
C = ϵ0 A/d
plug in the values of d, ϵ0 and A above to get the capacitance:
C = (8.854 × 10^ (-12) * 1.69 X 10^ (-3) / 3.5 * 10^ (-3)
= 4.27 * 10^ (-12) F
= 4.27pF
A swimming pool has the shape of a box with a base that measures 30 m by 12 m and a uniform depth of 2.2 m. How much work is required to pump the water out of the pool when it is full? Use 1000 kg divided by m cubed for the density of water and 9.8 m divided by s squared for the acceleration due to gravity.
Final answer:
The problem requires calculating the work done to pump water out of a full swimming pool using given dimensions, the density of water, and gravity.
Explanation:
The question involves finding the amount of work required to pump the water out of a swimming pool when it is full. The dimensions of the pool are given, along with the density of water and the acceleration due to gravity. Using the density of water (1000 kg/m3), the volume of the pool can be calculated to determine the total mass of the water. The work done in pumping the water is found by multiplying the mass by the gravitational constant (9.8 m/s2) and the vertical distance the water needs to be moved (2.2 m, which is the uniform depth of the pool). This distance can be different depending on the location of the pump, but for this problem, we assume the water is being pumped from the very bottom.
True or False? The superposition or overlapping of two waves always results in destructive interference between the different waves.
Answer:
Explanation:
False
When two waves overlap or superimpose over each other then they either undergo Constructive or destructive interference.
waves are the disturbance created by a force and add up to gives constructive interference when they are in the same line i.e. in the same phase.
When these disturbances are in the opposite phase then they superimpose to give destructive interference where the amplitude of the resulting wave will be much smaller as compared to original waves.
9) A balloon is charged with 3.4 μC (microcoulombs) of charge. A second balloon 23 cm away is charged with -5.1 μC of charge. The force of attraction / repulsion between the two charges will be: ______________________ 10) If one of the balloons has a mass of 0.084 kg, with what acceleration does it move toward or away from the other balloon? (calculate both magnitude AND direction) ________________________________________
Answer:
9. The force is a force of attraction and it is 2.95N
10. The magnitude of acceleration 35.12m/s^2 and the direction of this acceleration is away from the other balloon.
Explanation:
Parameters given:
Q1 = 3.4 * 10^-6C
Q2 = - 5.1 * 10^-6C
Distance between the two balloons = 23cm = 0.23m
9. Force acting between the two balloons is a force of attraction because they are unlike charges. Hence, the force between them is:
F = kQ1Q2/r^2
F = (9 *10^9 * 3.4 * 10^-6 * -5.1 * 10^-6)/(2.3 * 10^-1)^2
F = (1.56 * 10^-1)/(5.29 * 10^-2)
F = - 2.95N
10. Assuming that Balloon A has a mass, m, of 0.084kg, then:
F = ma
Where a = acceleration
a = F/m
a = -2.95/0.084
a = - 35.12m/s^2
The acceleration has a magnitude of 35.12m/s^2 and its direction is away from balloon B.
The negative sign shows that the balloon A is slowing down as it moves towards balloon B. Hence, it's velocity is reducing slowly.
What is the strength of an electric field that will balance the weight of a 9.6 g plastic sphere that has been charged to -9.2 nC ? Express your answer to two significant figures and include the appropriate units.
The strength of an electric field that will balance the weight is 1.023 × 10⁷ N/C.
What is electric field?An electric field is a physical field that surrounds electrically charged particles and acts as an attractor or repellent to all other charged particles in the vicinity. Additionally, it refers to a system of charged particles' physical field.
Electric charges and time-varying electric currents are the building blocks of electric fields.
The strength of an electric field that will balance the plastic sphere is = weight of the object/charge on the object
= ( 9.6 ×10⁻³×9.8)/(9.2×10⁻⁹) N/C
= 1.023 × 10⁷ N/C
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How would the period of a simple pendulum be affected if it were located on the moon instead of the earth?
Answer:
On moon time period will become 2.45 times of the time period on earth
Explanation:
Time period of simple pendulum is equal to [tex]T=2\pi \sqrt{\frac{l}{g}}[/tex] ....eqn 1 here l is length of the pendulum and g is acceleration due to gravity on earth
As when we go to moon, acceleration due to gravity on moon is [tex]\frac{1}{6}[/tex] times os acceleration due to gravity on earth
So time period of pendulum on moon is equal to
[tex]T_{moon}=2\pi \sqrt{\frac{l}{\frac{g}{6}}}=2\pi \sqrt{\frac{6l}{g}}[/tex] --------eqn 2
Dividing eqn 2 by eqn 1
[tex]\frac{T_{moon}}{T}=\sqrt{\frac{6l}{g}\times \frac{g}{l}}[/tex]
[tex]T_{moon}=\sqrt{6}T=2.45T[/tex]
So on moon time period will become 2.45 times of the time period on earth
Final answer:
The period of a pendulum on the Moon would be longer because the Moon's gravity is weaker. To achieve the same one-second period as on Earth, a pendulum needs to be much shorter due to the Moon's 1/6th gravitational acceleration. Consequently, a pendulum's frequency would decrease if taken from Earth to the Moon.
Explanation:
The period of a simple pendulum is affected by the acceleration due to gravity, which is less on the Moon than on Earth. Hence, if you took a pendulum clock, like a grandfather clock, to the Moon, its pendulum would swing more slowly because of the Moon's weaker gravity. To maintain a steady tick-tock of one second per period on the Moon, the pendulum would need to be much shorter. A grandfather clock pendulum designed to have a two-second period on Earth with a length of 50 cm would need to be only 8.2 cm long on the Moon to achieve the same period, since the Moon's gravity is 1/6th that of Earth. Therefore, if a pendulum from Earth was taken to the Moon, its frequency would decrease because the acceleration due to gravity on the Moon is less than that on Earth.
A square steel bar has a length of 9.7 ft and a 2.9 in by 2.9 in cross section and is subjected to axial tension. The final length is 9.70710 ft . The final side length is 2.89933 in . What is Poisson's ratio for the material? Express your answer to three significant figures.
The Poisson's ratio definition is given as the change in lateral deformation over longitudinal deformation. Mathematically it could be expressed like this,
[tex]\upsilon = \frac{\text{Lateral Strain}}{\text{Longitudinal Strain}}[/tex]
[tex]\upsilon = \frac{(\delta a/a)}{(\delta l/l)}[/tex]
Replacing with our values we would have to,
[tex]\upsilon = \frac{(2.89933-2.9/2.9)}{(9.70710-9.7/97)}[/tex]
[tex]\upsilon = 0.1977[/tex]
Therefore Poisson's ratio is 0.1977.
Equations E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 are approximations of the magnitude of the electric field of an electric dipole, at points along the dipole axis. Consider a point P on that axis at distance z = 4.50d from the dipole center (where d is the separation distance between the particles of the dipole). Let Eappr be the magnitude of the field at point P as approximated by E = 1 2πε0 qd z3 and E = 1 2πε0 P z3 (electric dipole). Let Eact be the actual magnitude. By how much is the ratio Eappr/Eact less than 1?
Answer:
The ratio of [tex]E_{app}[/tex] and [tex]E_{act}[/tex] is 0.9754
Explanation:
Given that,
Distance z = 4.50 d
First equation is
[tex]E_{act}=\dfrac{qd}{2\pi\epsilon_{0}\times z^3}[/tex]
[tex]E_{act}=\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}[/tex]
Second equation is
[tex]E_{app}=\dfrac{P}{2\pi\epsilon_{0}\times z^3}[/tex]
We need to calculate the ratio of [tex]E_{act}[/tex] and [tex]E_{app}[/tex]
Using formula
[tex]\dfrac{E_{app}}{E_{act}}=\dfrac{\dfrac{P}{2\pi\epsilon_{0}\times z^3}}{\dfrac{Pz}{2\pi\epsilon_{0}\times (z^2-\dfrac{d^2}{4})^2}}[/tex]
[tex]\dfrac{E_{app}}{E_{act}}=\dfrac{(z^2-\dfrac{d^2}{4})^2}{z^3(z)}[/tex]
Put the value into the formula
[tex]\dfrac{E_{app}}{E_{act}}=\dfrac{((4.50d)^2-\dfrac{d^2}{4})^2}{(4.50d)^3\times4.50d}[/tex]
[tex]\dfrac{E_{app}}{E_{act}}=0.9754[/tex]
Hence, The ratio of [tex]E_{app}[/tex] and [tex]E_{act}[/tex] is 0.9754
What is the relationship between wavelength, wave frequency, and wave velocity?
Relation Between Velocity And Wavelength
Wavelength is the measure of the length of a complete wave cycle. The velocity of a wave is the distance travelled by a point on the wave. In general, for any wave the relation between Velocity and Wavelength is proportionate. It is expressed through the wave velocity formula.
Velocity And Wavelength
For any given wave, the product of wavelength and frequency gives the velocity. It is mathematically given by wave velocity formula written as-
V=f×λ
Where,
V is the velocity of the wave measure using m/s.
f is the frequency of the wave measured using Hz.
λ is the wavelength of the wave measured using m.
Velocity and Wavelength Relationtion
Amplitude, Frequency, wavelength, and velocity are the characteristic of a wave. For a constant frequency, the wavelength is directly proportional to velocity.
Given by:
V∝λ
Example:
For a constant frequency, If the wavelength is doubled. The velocity of the wave will also double.
For a constant frequency, If the wavelength is made four times. The velocity of the wave will also be increased by four times.
Hope you understood the relation between wavelength and velocity of a wave. You may also want to check out these topics given below!
Relation between phase difference and path difference
Relation Between Frequency And Velocity
Relation Between Escape Velocity And Orbital Velocity
Relation Between Group Velocity And Phase Velocity
The relationship between wavelength, wave frequency, and wave velocity is described by the equation v = fλ. Wavelength and frequency are inversely proportional given a constant wave velocity – high frequency correlates with short wavelength and vice versa.
Explanation:In Physics, there's a mathematical relationship between wavelength, wave frequency, and wave velocity for any type of wave motion. This relationship is often stated as v = fλ, where v is wave velocity, f is the frequency of the wave, and λ is the wavelength. The wavelength is the distance between identical parts of the wave, while the velocity is the speed at which the disturbance moves, and the frequency is the rate of oscillation of the wave.
When you look at this formula, it becomes clear that if the wave velocity (v) is constant, a wave with a longer wavelength (λ) will have lower frequency (f). On the other side, higher frequency means shorter wavelength. This is because frequency and wavelength are inversely proportional in the given formula.
Example
For instance, the speed of light in vacuum is a constant value (approximately 3.00×108 m/s). So, if a certain light wave has a larger wavelength, its frequency will be lower to ensure this speed remains consistent.
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