Answer:
83816746.4254 m/s
Explanation:
m = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
q = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
V = Voltage = [tex]2\times 10^4\ V[/tex]
The kinetic energy of the electron is
[tex]K=\dfrac{1}{2}mv^2[/tex]
Energy is given by
[tex]E=qV[/tex]
Balancing the energy
[tex]qV=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{\dfrac{2qV}{m}}\\\Rightarrow v=\sqrt{\dfrac{2\times 1.6\times 10^{-19}\times 2\times 10^4}{9.11\times 10^{-31}}}\\\Rightarrow v=83816746.4254\ m/s[/tex]
The velocity of the electrons is 83816746.4254 m/s
Consider a Boeing 777 flying at a standard altitude of 11 km with a cruising velocity of 935 km/h. At a point on the wing, the velocity is 280 m/s. Calculate the temperature and pressure at this point.
Answer:
ΔP = 97.93 Pa , T₂-T₁ = 71.5° C
Explanation:
For this problem let's use Bernoulli's relationship, as point 1 we will take the plane and as point 2 the air
P₁ + ½ ρ g v₁² + ρ g y₁ = P2 + ½ ρ v₂² + ρ g y₂
As the whole process occurs at the same height y₁ = y₂ = 11 km. We will consider that the air goes in the opposite direction to the plane
P₂ –P₁ = ½ ρ (v₁² - v₂²)
Let's reduce the magnitudes to the SI system
v₁ = 935 km / h (1000 m / 1 km) (1h / 3600s) = 259.72 m / s
v₂ = 280 m / s
The density of air at 11000 m is
Rho = 0.3629 kg / m
P₂-P₁ = ½ 0.3629 (259.72 + 280)
ΔP = 97.93 Pa
The variation of the temperature with the altitude is 0.65 per 100 m
T₂ –T₁ = (0.65 / 100) 1000
T₂-T₁ = 71.5° C
The temperature has decreased this value
what is the voltage in mv across an 5.5 nm thick membrane if the electric field strength across it is 5.75 mv/m? you may assume a uniform electric field.
Answer:
V = 31.62 m V
Explanation:
Given,
Voltage across thick membrane = ?
Thickness of the membrane, d = 5.5 n m
Electric field = 5.75 M V/m
we know
[tex]E = \dfrac{V}{d}[/tex]
V = E d
V = 5.75 x 10⁶ x 5.5 x 10⁻⁹
V = 31.62 x 10⁻³ V
V = 31.62 m V
Hence, The Voltage across the membrane is equal to 31.62 m V.
The voltage across the membrane is dependent on the electric field and thickness of the membrane. The voltage across the membrane is 0.0316 V.
What is the voltage?The voltage is defined as the difference in electric potential between two points.
Given that the electric field E is 5.75 mv/m and the thickness of the membrane d is 5.5 nm.
The voltage across the membrane is calculated as given below.
[tex]E = \dfrac {V}{d}[/tex]
[tex]V = Ed[/tex]
[tex]V = 5.75 \times 10^6 \times 5.5 \times 10^{-9}[/tex]
[tex]V = 0.0316 \;\rm V[/tex]
Hence we can conclude that the voltage across the membrane is 0.0316 V.
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A car’s velocity as a function of time is given by vx(t) = α + βt2, where α = 3.00 m/s and β = 0.100 m/s3. (a) Calculate the average acceleration for the time interval t = 0 to t = 5.00 s. (b) Calculate the instantaneous acceleration for t = 0 and t = 5.00 s. (c) Draw vx-t and ax-t graphs for the car’s motion between t = 0 and t = 5.00 s.
The average acceleration from t=0 to t=5.00 s is 0.5 m/s^2. instantaneous acceleration at t=0 is 0 m/s^2 and at t=5.00 s is 1.0 m/s^2. The velocity-time graph is parabolic, and the acceleration-time graph shows linear growth.
The velocity of a car as a function of time is given by vx(t) = \\(\alpha + \beta t^2\\), where \\(\alpha = 3.00 m/s\\) and \beta = 0.100 m/s^3. To find the average acceleration for the time interval from t = 0 to t = 5.00 s, we need the change in velocity over the time period. Using the given function, we can calculate the initial velocity at t = 0, which is v(0) = 3.00 m/s, and the final velocity at t = 5.00 s, which is v(5) = 3.00 + 0.100(5)^2 = 3.00 + 2.5 = 5.5 m/s. The average acceleration is then (5.5 m/s - 3.0 m/s) / (5.0 s - 0 s) = 0.5 m/s^2.
For instantaneous acceleration at any given time, we differentiate the velocity function with respect to time to obtain a(t) = dvx/dt = 2\beta t. Therefore, the instantaneous acceleration at t = 0 is a(0) = 2(0.100)(0) = 0 m/s^2, and at t = 5.00 s is a(5) = 2(0.100)(5) = 1.0 m/s^2.
When drawing the vx-t and ax-t graphs, for the velocity-time graph, the curve will start at 3.00 m/s and rise parabolically due to the \(t^2\) term in the function. For the acceleration-time graph, the line will start at 0 m/s^2 and increase linearly with time, passing through 1.0 m/s^2 at 5 seconds.
A 238 92U nucleus is moving in the x-direction at 5.0 × 105 m/s when it decays into an alpha particle 4 2He and a 234 90Th. If the alpha particle moves off at 25.4° above the x axis with a speed of 1.4 × 107 m/s, what is the recoil velocity of the thorium nucleus? Assume the uranium-thorium-alpha system is isolated; you may also assume the particles are pointlike.
Answer:
v_th = 3.1 * 10^5 m/s
Explanation:
Given:
- mass of 238-Uranium m_u = 3.952 *10^-25 kg
- mass of alpha particle m_a = 6.64 * 10^-27 kg
- mass of thorium particle m_th = 3.885*10^-25 kg
- velocity of 238-Uranium v_u = 5.0 *10^5 m/s
- velocity of alpha particle v_a = 1.4 *10^7 m/s
Find:
- The recoil velocity of the thorium particle.
Solution:
- To solve this problem we will use conservation of momentum in both x and y direction.
- Momentum conservation in x-direction:
P_i = P_f
m_u*v_u = m_a*v_a*cos(Q) + m_th*v_th,x
where v_th,x is the x component of thorium velocity:
P_i = 3.952 *10^-25 * 5.0 *10^5 = 1.976*10^-19
P_f = 6.64 * 10^-27*1.4 *10^7*cos(25.4) + 3.885*10^-25*v_th,x
P_f = 8.3974*10^-20 + 3.885*10^-25*v_th,x
Hence,
1.976*10^-19 = 9.296*10^-20 + 3.885*10^-25*v_th,x
v_th,x = 2.92473 * 10^5 m/s
- Momentum conservation in y-direction:
P_i = P_f
0 = m_a*v_a*sin(Q) + m_th*v_th,y
where v_th,x is the x component of thorium velocity:
v_th,y = m_a*v_a*sin(Q) / m_th
v_th,y = 6.64 * 10^-27*1.4 *10^7*sin(25.4) / 3.885*10^-25
Hence,
v_th,y = 1.02635 * 10^5 m/s
- The magnitude of recoil velocity:
v_th = sqrt ( v_th,x ^2 + v_th,y ^2 )
v_th = sqrt ( (2.92473 * 10^5)^2 + (1.02635 * 10^5)^2 )
v_th = 3.1 * 10^5 m/s
F1 = (3.3,-0.5) and F2 = (-3.8,-0.3) where all components are in newtons. What angle does the vector F1 + F2 make with the positive x-axis? The angle is measured counterclockwise from the positive x-axis and must be in the range from 0 to 360 degrees.
Answer:
238 Degree
Explanation:
Data given
F1=(3.3,-0.5) and F2=(-3.8,-0.3).
To determine the angle F1+F2 makes with the positive x-axis, we need to determine the magnitude of the force F1+F2.
Since force is a vector quantity, we add the vectors component by component
[tex]F_{1}+F_{2}=<3.3,-0.5> +<-3.8,-0.3>\\F_{1}+F_{2}=<3.3+(-3.8), -0.5+(-0.3)>\\ F_{1}+F_{2}=<-0.5,-0.8>\\F_{1}+F_{2}=(-0.5,-0.8)[/tex]
To determine the angle, we use
[tex]F=(x,y)\\\alpha=arctan(\frac{y}{x} )\\Hence for \\F_{1}+F_{2}=(-0.5,-0.8)\\\alpha=arctan(\frac{-0.8}{-0.5} )\\\alpha=58^{0}[/tex]
Since the component of the force F1+F2 is a negative y and negative x which are located in the 3rd quadrant, the angle can be calculated as
∝=58+180=238 degree
Hence The angle is measured counterclockwise from the positive x-axis and must be in the range from 0 to 360 degrees is 238 Degree
The angle measured counterclockwise from the positive x-axis is θ = 57.99°
Finding the direction of the force.
Here we know that:
F1 = (3.3, -0.5)F2 = (-3.8, -0.3)First, we need to add the forces, we will get:
F1 + F2 = (3.3, -0.5) + (-3.8, -0.3) = (3.3 - 3.8, -0.5 - 0.3))
F1 + F2 = (-0.5, -0.8)
Now, the angle measured counterclockwise from the positive x-axis of a vector
(a, b) is given by:
θ = Atan(b/a).
Where Atan(x) is the inverse tangent function.
So in this case the angle will be:
θ = Atan(-0.8/-0.5) = 57.99°
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A half-full recycling bin has mass 3.0 kg and is pushed up a 40.0^\circ40.0 ∘ incline with constant speed under the action of a 26-N force acting up and parallel to the incline. The incline has friction. What magnitude force must act up and parallel to the incline for the bin to move down the incline at constant velocity?
Final answer:
A 26-N force must act down and parallel to the incline for the bin to move down at constant velocity since this force would balance out the 26-N frictional force acting up the incline.
Explanation:
The subject of this question is Physics, specifically the concepts related to mechanics and forces on inclines with friction. The student is in High School level, most likely studying the fundamentals of Newtonian mechanics.
Given that the bin is already being pushed up the incline with a 26-N force and moves at a constant speed, this means the net force on the bin is zero, therefore the force of friction must also be 26 N but acting downwards along the incline. When the bin moves down at a constant velocity, the force of friction still acts up the incline (opposite to the bin's movement) and therefore, to maintain a constant velocity, the force applied must be equal in magnitude to the frictional force but directed down the incline. Hence, a 26-N force must be applied down and parallel to the incline for the bin to move down the incline at constant velocity.
A vessel, divided into two parts by a partition, contains 4 mol of nitrogen gas at 75°C and 30 bar on one side and 2.5 mol of argon gas at 130°C and 20 bar on the other. If the partition is removed and the gases mix adiabatically and completely, what is the change in entropy? Assume nitrogen to be an ideal gas with CV = (5/2)R and argon to be an ideal gas with CV = (3/2)R.
Answer:
assume nitrogen is an ideal gas with cv=5R/2
assume argon is an ideal gas with cv=3R/2
n1=4moles
n2=2.5 moles
t1=75°C in kelvin t1=75+273
t1=348K
T2=130°C in kelvin t2=130+273
t2=403K
u=пCVΔT
U(N₂)+U(Argon)=0
putting values:
=>4x(5R/2)x(Tfinal-348)=2.5x(3R/2)x(T final-403)
by simplifying:
Tfinal=363K
To determine the change in entropy when the gases mix adiabatically and completely, we'll follow these steps:
1. Calculate the initial entropy of each gas before mixing.
2. Calculate the final entropy of the combined gases after mixing.
3. Calculate the change in entropy.
Given information:
For nitrogen gas:
- Moles of nitrogen gas, [tex]n_N2[/tex] = 4 mol
- Temperature of nitrogen gas, [tex]T_N2[/tex] = 75°C = 75 + 273.15 K
- Pressure of nitrogen gas, [tex]P_N2[/tex] = 30 bar = 30 * 100 kPa
For argon gas:
- Moles of argon gas, [tex]n_Ar[/tex] = 2.5 mol
- Temperature of argon gas, [tex]T_Ar[/tex] = 130°C = 130 + 273.15 K
- Pressure of argon gas, [tex]P_VR[/tex] = 20 bar = 20 * 100 kPa
Given specific heat capacities :
- [tex]CV_N2[/tex] = (5/2)R for nitrogen gas
- [tex]CV_Ar[/tex] = (3/2)R for argon gas
The change in entropy, ΔS, can be calculated using the equation:
[tex]\[ΔS = n_{\text{N2}} \cdot C_{V_{\text{N2}}} \cdot \ln\left(\frac{T_f}{T_{\text{N2}}}\right) + n_{\text{Ar}} \cdot C_{V_{\text{Ar}}} \cdot \ln\left(\frac{T_f}{T_{\text{Ar}}}\right)\][/tex]
where [tex]\(T_f\)[/tex] is the final temperature after mixing.
1. Calculate the initial entropy of each gas:
[tex]\[S_{\text{N2}} = n_{\text{N2}} \cdot C_{V_{\text{N2}}} \cdot \ln\left(\frac{T_{\text{N2}}}{T_0}\right)\][/tex]
[tex]\[S_{\text{Ar}} = n_{\text{Ar}} \cdot C_{V_{\text{Ar}}} \cdot \ln\left(\frac{T_{\text{Ar}}}{T_0}\right)\][/tex]
where [tex]\(T_0\)[/tex] is the reference temperature (usually taken as 298.15 K).
2. Calculate the final temperature after mixing using the adiabatic process equation:
[tex]\[T_f = \left(\frac{n_{\text{N2}} \cdot C_{V_{\text{N2}}} \cdot T_{\text{N2}} + n_{\text{Ar}} \cdot C_{V_{\text{Ar}}} \cdot T_{\text{Ar}}}{n_{\text{N2}} \cdot C_{V_{\text{N2}}} + n_{\text{Ar}} \cdot C_{V_{\text{Ar}}}}\right)\][/tex]
3. Calculate the change in entropy using the formula above.
Let's plug in the values and calculate the change in entropy. First, we'll calculate the initial entropies of each gas:
[tex]\[S_{\text{N2}} = 4 \times \left(\frac{5}{2}\right)R \times \ln\left(\frac{75 + 273.15}{298.15}\right)\][/tex]
[tex]\[S_{\text{Ar}} = 2.5 \times \left(\frac{3}{2}\right)R \times \ln\left(\frac{130 + 273.15}{298.15}\right)\][/tex]
Now, calculate [tex]\(T_f\)[/tex] using the adiabatic process equation:
[tex]\[T_f = \left(\frac{4 \times \left(\frac{5}{2}\right)R \times 75 + 2.5 \times \left(\frac{3}{2}\right)R \times 130}{4 \times \left(\frac{5}{2}\right)R + 2.5 \times \left(\frac{3}{2}\right)R}\right)\][/tex]
Finally, calculate the change in entropy:
[tex]\[ΔS = 4 \times \left(\frac{5}{2}\right)R \times \ln\left(\frac{T_f}{75 + 273.15}\right) + 2.5 \times \left(\frac{3}{2}\right)R \times \ln\left(\frac{T_f}{130 + 273.15}\right)\][/tex]
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Consider two identical and symmetrical wave pulses on a string. Suppose the first pulse reaches the fixed end of the string and is reflected back and then meets the second pulse. When the two pulses overlap exactly, the superposition principle predicts that the amplitude of the resultant pulses, at that moment, will be what factor times the amplitude of one of the original pulses?a. 0b. 1c. -2d. -1
Answer:
The amplitude of the resultant wave will be 0.
Explanation:
Suppose the first wave has an amplitude of A. Its angle is given as wt.
The second way will also have the same amplitude as that of first.
After the reflection, a phase shift of π is added So the wave is given as
[tex]W_1=W_2=Acos(\omega t)\\W_1^{'}=Acos(\omega t+ \pi)[/tex]
Adding the two waves give
[tex]W_1'+W_2=Acos(\omega t+ \pi)+Acos(\omega t)\\W_1'+W_2=-Acos(\omega t)+Acos(\omega t)\\W_1'+W_2=0[/tex]
So the amplitude of the resultant wave will be 0.
"Sarah is out playing fetch with her dog, and throws a tennis ball as far as she can. At the moment the ball reaches its maximum height, which of the following are true. You may neglect air resistance."a-You may neglect air resistance.
b-The ball's vertical acceleration is downwards
c-The ball's vertical acceleration is zero
d-The ball's vertical acceleration is upwards
e-The ball's horizontal acceleration is zero
f-The ball's horizontal acceleration is in the direction of the throw
g-The ball's horizontal acceleration is oppoite the direction of the throw
h-The ball's vertical acceleration is upwards
Answer:
a.The ball's vertical acceleration is downwards.e.The ball's horizontal acceleration is zeroExplanation:
We are given that Sarah throws a tennis ball as far as she can.
At the moment the ball reaches its maximum height.
We have to find the true statement if air resistance is neglect.
When air resistance is negligible then the force act on the ball is force due to gravity.
The ball throw vertically then the acceleration act on the ball is acceleration due to gravity.
The value of g=-9.8 m/square sec
It acts on the ball in downward direction .
Therefore, the ball's vertical acceleration is downwards.
The horizontal acceleration is zero because the ball reaches at maximum height then there is no force which act in horizontal direction on the ball.
Therefore, horizontal acceleration of the ball is zero.
Hence, option a and e are true.
The statement B is correct which states that "The ball's vertical acceleration is downwards".
The statement E is correct which states that "The ball's horizontal acceleration is zero".
How do you find out which statement is true?The given condition is that Sarah throws a tennis ball as far as she can. Also given that the air resistance is negligible.
Let consider that the ball is thrown vertically. In this case, the force on the ball is the force due to gravitational acceleration on the ball.
Hence the vertical acceleration on the ball is equivalent to the gravitational acceleration of 9.8 m/s2. The vertical force acting on the ball will be in a downward direction because this is the force due to gravity.
Hence the statement B is correct which states that "The ball's vertical acceleration is downwards".
When the ball reaches its maximum height, the horizontal acceleration will be zero on the ball, as there is no force acting on the ball in the horizontal direction.
Hence the statement E is correct which states that "The ball's horizontal acceleration is zero".
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In one of the classic nuclear physics experiments at the beginning of the 20th century, an alpha particle was accelerated toward a gold nucleus, and its path was substantially deflected by the Coulomb interaction. If the energy of the doubly charged alpha nucleus was 5.97 MeV, how close (in m) to the gold nucleus (79 protons) could it come before being deflected?
Answer:
3.8 × 10 ⁻¹⁴ m
Explanation:
The alpha particle will be deflected when its kinetic energy is equal to the potential energy
Charge of the alpha particle q₁= 2 × 1.6 × 10⁻¹⁹ C = 3.2 × 10⁻¹⁹ C
Charge of the gold nucleus q₂= 79 × 1.6 × 10⁻¹⁹ = 1.264 × 10⁻¹⁷C
Kinetic energy of the alpha particle = 5.97 × 10⁶ × 1.602 × 10⁻¹⁹ J ( 1 eV) = 9.564 × 10⁻¹³
k electrostatic force constant = 9 × 10⁹ N.m²/c²
Kinetic energy = potential energy = k q₁q₂ / r where r is the closest distance the alpha particle got to the gold nucleus
r = ( 9 × 10⁹ N.m²/c² × 3.2 × 10⁻¹⁹ C × 1.264 × 10⁻¹⁷C) / 9.564 × 10⁻¹³ = 3.8 × 10 ⁻¹⁴ m
In fighting forest fires, airplanes work in support of ground crews by dropping water on the fires. For practice, a pilot drops a canister of red dye, hoping to hit a target on the ground below. If the plane is flying in a horizontal path 90.0 m above the ground and has a speed of 64.0 m/s (143 mi/h), at what horizontal distance from the target should the pilot release the canister? Ignore air resistance.
Answer:
[tex]s=274.2857\ m[/tex] is the distance from the target before which the pilot must release the canister.
Explanation:
Given:
height of the plane, [tex]h=90\ m[/tex]horizontal speed of plane, [tex]v_x=64\ m.s^{-1}[/tex]Time taken by the canister to hit the ground:
using equation of motion
[tex]h=u_y.t+\frac{1}{2} g.t^2[/tex]
where:
[tex]u_y=[/tex] initial vertical velocity of the canister = 0 (since the the object is dropped from a horizontally moving plane)
[tex]t=[/tex] time taken to hit the ground
[tex]90=0+0.5\times 9.8\times t^2[/tex]
[tex]t=4.2857\ s[/tex]
Now the horizontal distance travelled by the canister after dropping:
[tex]s=v_x\times t[/tex]
[tex]s=64\times 4.2857[/tex]
[tex]s=274.2857\ m[/tex] is the distance from the target before which the pilot must release the canister.
A rock is thrown horizontally from a tower 4.9 m above the ground, and the rock strikes the ground after travelling 20 m horizontally. What was the rock's initial launch speed (in m/s)?
We have that the the rock's initial launch speed is mathematically given as
u=20m/sFrom the question we are told
A rock is thrown horizontally from a tower 4.9 m above the ground, and the rock strikes the ground after travelling 20 m horizontally. What was the rock's initial launch speed (in m/s)?SpeedGenerally the equation for the Motion is mathematically given as
[tex]H=ut+0.5gt^2\\\\Therefore\\\\4.9=0*t+0.5*9.8*t^2\\\\t=1s\\\\Therefore\\\\sx=uxt\\\\20=ux*1\\\\[/tex]
u=20m/sFor more information on speed visit
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To determine the initial launch speed of the rock, we can use the equations of projectile motion. Since the rock is thrown horizontally, and it travels 20 m horizontally in 1 second, the initial launch speed can be calculated as 20 m/s.
Explanation:To determine the initial launch speed of the rock, we can use the equations of projectile motion. Since the rock is thrown horizontally, its initial vertical velocity is zero. The only force acting on the rock in the vertical direction is gravity, which causes it to accelerate downward at a rate of 9.8 m/s². Using the formula:
-yf = yi + vit - 1/2gt²
where yi represents the initial height and yf represents the final height, we can solve for the initial downward velocity. Since the rock reaches the ground (yf = 0) and starts at a height of 4.9 m, the equation becomes:
0 = 4.9 + 0 - 1/2(9.8)t²
Simplifying the equation gives:
t² = 4.9/4.9
t = 1 second.
Since the horizontal distance traveled by the rock is 20 m, and it takes 1 second to reach the ground, the horizontal speed (initial launch speed) of the rock can be calculated using the formula:
v = d/t = 20 m/1 s = 20 m/s.
Therefore, the rock's initial launch speed is 20 m/s.
By what factor must we increase the amplitude of vibration of an object at the end of a spring in order to double its maximum speed during a vibration? A is the old amplitude and A′ is the new one.
Answer:
[tex]A'=2A[/tex]
Explanation:
According to the law of conservation of energy, the total energy of the system can be expresed as the sum of the potential energy and kinetic energy:
[tex]E=U+K=\frac{kA^2}{2}\\E=\frac{kx^2}{2}+\frac{mv^2}{2}=\frac{kA^2}{2}[/tex]
When the spring is in its equilibrium position, that is [tex]x=0[/tex], the object speed its maximum. So, we have:
[tex]\frac{k(0)^2}{2}+\frac{mv_{max}^2}{2}=\frac{kA^2}{2}\\A^2=\frac{mv_{max}^2}{k}\\A=\sqrt{\frac{mv_{max}^2}{k}}[/tex]
In order to double its maximum speed, that is [tex]v'{max}=2v_{max}[/tex]. We have:
[tex]A'=\sqrt{\frac{m(v'_{max})^2}{k}}\\A'=\sqrt{\frac{m(2v_{max})^2}{k}}\\A'=\sqrt{\frac{4mv_{max}^2}{k}}\\A'=2\sqrt{\frac{mv_{max}^2}{k}}\\A'=2A[/tex]
The factor we must increase the amplitude of the vibration to double its maximum speed is 2.
Apply the principle of conservation of energy to determine the amplitude;
[tex]U = K.E\\\\\frac{1}{2} KA^2 = \frac{1}{2}mv^2\\\\KA^2 = mv^2\\\\A^2 = \frac{m}{K} v^2\\\\\frac{A_1^2}{v_1^2 } = \frac{A_2^2}{v_2^2 } \\\\A_2 = \frac{v_2 A_1}{v_1} \\\\A_2 = \frac{2v_1 \times A_1}{v_1} \\\\A_2 =2A_1[/tex]
Thus, the factor we must increase the amplitude of the vibration to double its maximum speed is 2.
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An amusement park ride consists of a car moving in a vertical circle on the end of a rigid boom of negligible mass. The combined weight of the car and riders is 5.30 kN, and the radius of the circle is 12.0 m. At the top of the circle, (a) what is the force FB on the car from the boom (using the minus sign for downward direction) if the car's speed is v = 3.60 m/s? (b) What is FB if v = 14.0 m/s? Use g=9.80 m/s2.
Explanation:
As the force is given as 5.30 kN or [tex]5.30 \times 1000 N[/tex]. Hence, mass will be calculated as follows.
[tex]F_{w}[/tex] = mg
[tex]5.30 \times 10^{3} = m \times 9.8 m/s^{2}[/tex]
m = 540.816 kg
(a) At the top, centripetal force [tex]F_{c}[/tex] is acting upwards and the weight of the riders and car, [tex]F_{w}[/tex] will be acting downwards.
Therefore, force on the car by the boom will be calculated as follows.
[tex]F_{B} = F_{w} - F_{c}[/tex]
or, [tex]F_{B} = mg - \frac{mv^{2}}{r}[/tex]
= [tex]5.30 \times 1000 N - \frac{540.816 kg \times (3.60)^{2}}{12}[/tex]
= 4715.919 N
Hence, the force [tex]F_{B}[/tex] on the car from the boom is 4715.919 N. This means that the car will be hanging on the boom and the boom will exert an upward force.
(b) Now at the top, centripetal force [tex]F_{c}[/tex] will be acting upwards and the weight of cars and car riders will be acting in the downwards direction.
Hence, we will calculate the force on car by the boom as follows.
[tex]F_{B} = F_{w} - F_{c}[/tex]
or, [tex]F_{B} = mg - \frac{mv^{2}}{r}[/tex]
= [tex]5.30 \times 1000 N - \frac{540.816 kg \times (14.0)^{2}}{12}[/tex]
= -3533.33 N
Therefore, car will be pushing on the boom and the boom will exert a downward force.
Some plants disperse their seeds when the fruit splits and contracts, propelling the seeds through the air. The trajectory of these seeds can be determined with a high-speed camera. In an experiment on one type of plant, seeds are projected at 20 cm above ground level with initial speeds between 2.3 m/s and 4.6 m/s. The launch angle is measured from the horizontal, with + 90° corresponding to an initial velocity straight up and – 90° straight down.
About how long does it take a seed launched at 90° at the highest possible initial speed to reach its maximum height? Ignore air resistance. (a) 0.23 s; (b) 0.47 s; (c) 1.0 s; (d) 2.3 s.
Final answer:
The seed takes approximately 0.47 seconds to reach its maximum height when launched at 90° with the highest possible initial speed of 4.6 m/s.
Explanation:
When a seed is launched with an initial speed of 4.6 m/s and at an angle of 90°, the vertical component of the initial velocity is 4.6 m/s. The time it takes for a projectile to reach its maximum height can be calculated using the equation for vertical motion:
t = (Vf - Vi) / g
Where:
t is the time
Vf is the final velocity (which is 0 m/s at the maximum height)
Vi is the initial velocity
g is the acceleration due to gravity (9.8 m/s²)
Using the given information, we can calculate the time it takes for the seed to reach its maximum height:
t = (0 - 4.6) / -9.8 = 0.469 s
Therefore, the seed takes approximately 0.47 seconds to reach its maximum height.
A very weak pressure wave, i.e., a sound wave, across which the pressure rise is 30 Pa moves through air which has a temperature of 30°C and a pressure of 101 kPa. Find the density change, the temperature change, and the velocity change across this wave
Answer:
Density change, Δρ = 2.4 × 10⁻⁴ kg/m³
Temperature Change, ΔT = 0.0258 K
Velocity Change, Δc = 0.0148 m/s
Explanation:
For sound waves moving through the air,
Pressure and Temperature varies thus
(P₀/P) = (T₀/T)^(k/(k-1))
Where P₀ = initial pressure of air = 101KPa = 101000 Pa
P = final pressure of air due to the change brought about by the moving sound wave = 101000+30 = 101030 Pa
T₀ = initial temperature of air = 30°C = 303.15 K
T = final temperature of air = ?
k = ratio of specific heats = Cp/Cv = 1.4
(101000/101030) = (303.15/T)^(1.4/(1.4-1))
0.9990703 =(303.15/T)^(3.5)
Solving This,
T = 303.1758 K
ΔT = T - T₀ = 303.1758 - 303.15 = 0.0258 K
Density can be calculate in two ways,
First method
Δρ = ρ - ρ₀
P₀ = ρ₀RT₀
ρ₀ = P₀/RT₀
R = gas constant for air = 287 J/kg.k
where all of these are values for air before the wave propagates
P₀ = 101000 Pa, R = 287 J/kg.K, T₀ = 303.15K
ρ₀ = 101000/(287 × 303.15) = 1.1608655 kg/m³
ρ = P/RT
P = 101030 Pa, T = 303.1758K
ρ = 101030/(287×303.1758) = 1.1611115 kg/m³
Δρ = ρ - ρ₀ = 1.1611115 - 1.1608655 = 0.00024 kg/m³ = 2.4 × 10⁻⁴ kg/m³
Second method
(ρ₀/ρ) = (T₀/T)^(1/(k-1))
Where ρ₀ is initially calculated from ρ₀ = P₀/RT₀, then ρ is then computed and the diff taken.
Velocity Change
c₀ = √(kRT₀) = √(1.4 × 287 × 303.15) = 349.00669 m/s
c = √(kRT) = √(1.4 × 287 × 303.1758) = 349.0215415 m/s
Δc = c₀ - c = 349.0215415 - 349.00669 = 0.0148 m/s
QED!
In this exercise we have to use the pressure knowledge to calculate the velocity, temperature and density so we have:
Density: [tex]\Delta \rho = 2.4 * 10^{-4} kg/m^3[/tex]Temperature: [tex]\Delta T = 0.0258 K[/tex]Velocity: Δc = 0.0148 m/sThe variation of temperature and pressure is given by the formula of:
[tex](P_0/P) = (T_0/T)^{(k/(k-1))}[/tex]
From the formula given above we can identify:
P₀ = initial pressure of airP = final pressure of air due to the change brought about by the moving sound wave T₀ = initial temperature of air T = final temperature of airk = ratio of specific heatsSolving the formula for temperature we find:
[tex](101000/101030) = (303.15/T)^{(1.4/(1.4-1))}\\0.9990703 =(303.15/T)^{(3.5)}\\T = 303.1758 K\\\Delta T = T - T_0 = 303.1758 - 303.15 = 0.0258 K[/tex]
They are using the formula for density is:
[tex]\rho_0 = 101000/(287 * 303.15) = 1.1608655 kg/m^3\\\rho = P/RT\\\rho = 101030/(287*303.1758) = 1.1611115 kg/m^3\\\Delta \rho = 2.4 * 10^{-4} kg/m^3[/tex]
Calculating the speed we find that:
[tex]c_0 = \sqrt{kRT_0} = 349.00669 m/s\\c = \sqrt{kRT} = 349.0215415 m/s\\\Delta c = 0.0148 m/s[/tex]
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A high-speed K0 meson is traveling at β = 0.90 when it decays into a π + and a π − meson. What are the greatest and least speeds that the mesons may have?
Answer:
greatest speed=0.99c
least speed=0.283c
Explanation:
To solve this problem, we have to go to frame of center of mass.
Total available energy fo π + and π - mesons will be difference in their rest energy:
[tex]E_{0,K_{0} }-2E_{0,\pi } =497Mev-2*139.5Mev\\[/tex]
=218 Mev
now we have to assume that both meson have same kinetic energy so each will have K=109 Mev from following equation for kinetic energy we have,
K=(γ-1)[tex]E_{0,\pi }[/tex]
[tex]K=E_{0,\pi}(\frac{1}{\sqrt{1-\beta ^{2} } } -1)\\\frac{1}\sqrt{1-\beta ^{2} }}=\frac{K}{E_{0,\pi}}+1\\ {1-\beta ^{2}=\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta ^{2}=1-\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\beta = +-\sqrt{\frac{1}{(\frac{K}{E_{0,\pi}}+1)^2}}\\\\\\beta =+-\sqrt{1-\frac{1}{(\frac{109Mev}{139.5Mev+1)^2}}[/tex]
[tex]u'=+-0.283c[/tex]
note +-=±
To find speed least and greatest speed of meson we would use relativistic velocity addition equations:
[tex]u=\frac{u'+v}{1+\frac{v}{c^{2} } } u'\\u_{max} =\frac{u'_{+} +v_{} }{1+\frac{v}{c^{2} } } u'_{+} \\u_{max} =\frac{0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } 0.828\\ u_{max} =0.99c\\u_{min} =\frac{u'_{-} +v_{} }{1+\frac{v}{c^{2} } } u'_{-}\\u_{min} =\frac{-0.828c +0.9c }{1+\frac{0.9c}{c^{2} } } -0.828c\\u_{min} =0.283c[/tex]
Final answer:
The greatest and least speeds of the π mesons resulting from the decay of a high-speed K0 meson are calculated using the relativistic laws of velocity addition. The maximum speed is (85/86)c, and the minimum speed is (5/14)c.
Explanation:
The question revolves around a high-speed K0 meson decaying into two π mesons (π+ and π-). To determine the greatest and least speeds that the π mesons can have, we employ the relativistic addition of velocities. The formula used in this context is Vx=(Vx'+V)/(1+(Vx'V/c²)), where V is the speed of the K0 meson, Vx' the speed of π mesons in the K0 meson's rest frame, and c is the speed of light.
The maximum speed is achieved when the π meson is emitted in the same direction as the K0 meson's movement, while the minimum speed occurs when the π meson is emitted in the opposite direction. Considering the K0 meson's speed (0.9c) and the π mesons' speed in the rest frame (0.8c), the equations result in the maximum speed being (85/86)c and the minimum speed being (5/14)c respectively.
Three equal 1.55-μC point charges are placed at the corners of an equilateral triangle whose sides are 0.500 m long. What is the potential energy of the system? (Take as zero the potential energy of the three charges when they are infinitely far apart.)
Answer:
0.12959085 J
Explanation:
k = Coulomb constant = [tex]8.99\times 10^{9}\ Nm^2/C^2[/tex]
q = Charge = 1.55 μC
d = Distance between charge = 0.5 m
Electric potential energy is given by
[tex]U=k\dfrac{q^2}{d}[/tex]
In this system with three charges which are equidistant from each other
[tex]U=k\dfrac{q^2}{d}+k\dfrac{q^2}{d}+k\dfrac{q^2}{d}[/tex]
[tex]\\\Rightarrow U=k\dfrac{3q^2}{d}\\\Rightarrow U=8.99\times 10^9\times \dfrac{3\times (1.55\times 10^{-6})^2}{0.5}\\\Rightarrow U=0.12959085\ J[/tex]
The potential energy of the system is 0.12959085 J
A particle executes simple harmonic motion with an amplitude of 3.00 cm. At what position does its speed equal half of is maximum speed?
The speed equals half of its maximum speed at positions ±1.50 cm from the equilibrium for an amplitude of 3.00 cm.
Speed equals half of its maximum speed when the particle is at a position where it is halfway through its amplitude.
This occurs at x = ±0.5X from the equilibrium position, considering X as the amplitude of the motion.
For an amplitude of 3.00 cm, the speed equals half of its maximum speed at position ±1.50 cm.
Borrow soil is used to fill a 75,000m3 depression. The borrow soil has the following characteristics. Density: 1540kg/m3, water content: 8%, specific gravity of the solids: 2.66. The final in-place dry density should be 1790 kg/m3 and the final water content should be 13%. a) How many m3 of borrow soil are needed to fill the depression? b) Assuming no evaporation loss, what water mass is needed to achieve 13% moisture?
Answer
given,
Volume of Depression, V = 75000 m³
borrow soil
Density of soil,γ = 1540 Kg/m³
water content,w = 8 % = 0.08
Specific gravity of solid,G = 2.66
Final in-place
dry density = 1790 kg/m³
water content = 13 % = 0.13
a) Volume of borrow soil require to fill the depression
Mass of solid solid,m= dry density of inplace soil x Volume of depression
= 1790 x 75000
m = 1.3425 x 10⁸ Kg
dry density of the borrow pit
[tex]\gamma_d = \dfrac{\gamma}{1 + w}[/tex]
[tex]\gamma_d = \dfrac{1540}{1 + 0.08}[/tex]
[tex]\gamma_d = 1425.93\ kg/m^3[/tex]
Volume of borrow soil required
[tex]V = \dfrac{m}{\gamma_d}[/tex]
[tex]V = \dfrac{1.3425\times 10^8}{1425.93}[/tex]
V = 94149 m³
b) Water required
[tex]W = 1790\times 75000\times 0.13 - 1425.93\times 94149\times 0.08[/tex]
W = 6.71 x 10⁶ Kg
Water required to achieve 13% moisture is equal to W = 6.71 x 10⁶ Kg
Final answer:
To fill a 75,000 m³ depression, 93,590.46 m³ of borrow soil is required when accounting for the dry density change. To achieve a final water content of 13%, an additional 17,452,500 kg of water mass is needed.
Explanation:
To answer this question, we must first understand the relationship between the initial and final states of the borrow soil in terms of their volumes, densities, and moisture contents. The question is divided into two parts: calculating the volume of borrow soil needed and determining the water mass addition required to achieve the desired moisture content.
a) How many m³ of borrow soil are needed to fill the depression?
Since we want to fill a depression with a volume of 75,000 m³, we need to consider the change in density from the borrow state to the in-place dry density. The in-place dry density is given as 1790 kg/m³. The dry density of the borrow soil can be calculated by subtracting the mass of water from the total mass.
First, we convert the water content to a decimal by dividing by 100:
Water content = 8% / 100 = 0.08
Using the formula for dry density (
dry density = total density / (1 + water content)), we find:
Dry density of borrow soil = 1540 kg/m³ / (1 + 0.08) = 1425.93 kg/m³
The volume of the borrow soil needed can be calculated by the volume of the depression divided by the ratio of the final in-place dry density to the dry density of the borrow soil:
Volume of borrow soil needed = 75,000 m³ x (1790 kg/m³ / 1425.93 kg/m³) = 93,590.46 m³
b) Assuming no evaporation loss, what water mass is needed to achieve 13% moisture?
For the final water content of 13%, we'll use the in-place dry density to determine the total mass, then calculate the water mass needed:
Total in-place mass = volume x in-place dry density
Total in-place mass of fill = 75,000 m³ x 1790 kg/m³= 134,250,000 kg
Convert 13% to decimal:
Water content = 13% / 100 = 0.13
We want 13% of the total mass to be water, so the water mass required is:
Water mass = total mass x water content
Water mass needed = 134,250,000 kg x 0.13 = 17,452,500 kg
This is the additional water mass required to achieve the desired moisture content of 13%.
When a crown of mass 14.7kg is completely submerged in water, an accurate scale reading reads only 13.4kg. Is the crown made of gold?
Answer:
No. The crown is made of lead
Explanation:
From Archimedes principle,
D/D' = W/U
Where D = density of the crown, D' = Density of water, W = weight of the crown in air, U = Upthrust of the crown in water.
make D the subject of the equation,
D = D'(W/U)................... Equation 1
W = mg
Where g = 9.8 m/s², m = 14.7 kg.
W = 14.7(9.8) = 144.06 N.
U = 9.8(14.7-13.4) = 12.74 N.
Constant: D' = 1000 kg/m³
Substitute into equation 1
D = 1000(144.06/12.74)
D = 11307.69 kg/m³
Density of the crown = 11307.69 kg/m³
And Density of Gold = 19300 kg/m³
From the above, The crown is not made of Gold
To determine if the crown is made of gold, we need to use Archimedes' principle and calculate its density. If the density is greater than the density of water, the crown is made of gold.
When a crown of mass 14.7kg is completely submerged in water, an accurate scale reading reads only 13.4kg. To determine if the crown is made of gold, we need to use Archimedes' principle. According to this principle, when an object is submerged in a fluid, it experiences an upward buoyant force equal to the weight of the fluid displaced by the object.
If the crown is made of gold, its density should be equal to or greater than the density of water. We can calculate the density of the crown by dividing its mass by the volume of water it displaces. If the density of the crown is greater than the density of water (1kg/L), then the crown is made of gold.
In this case, the mass of the crown is 14.7kg, and the mass of the water displaced is 13.4kg (since the scale reading is lower). By using the equation density = mass/volume, we can calculate the volume of water displaced by the crown and determine its density.
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How did the temperature structure of the solar nebula determine planetary composition?
Explanation:
The temperature of the solar nebula was decreasing as it moved away from its center. Therefore, only heavy elements could condense in the inner solar system and terrestrial planets could not form with light elements, such as gases. In the outer solar system, the Jovian planets formed mostly with gases, since temperatures were too low to allow rocky compositions.
If it takes about 8 minutes for light to travel from the Sun to Earth, and Pluto is 40 times this distance from us, how long does it take light to reach Earth from Pluto?
Light takes 320 minutes to reach Earth from Pluto
Explanation:
We can solve this problem by applying the rule of three. We can do as follows:
- We call [tex]x[/tex] the distance between the Sun and the Earth
- Then the distance between the Earth and Pluto is 40 times this distance, so [tex]40x[/tex]
- Light takes about 8 minutes to cover the distance x between Sun and Earth
Therefore, calling T the time that light takes to cover distance between Earth and Pluto (40x), we can write:
[tex]\frac{x}{8}=\frac{40x}{T}[/tex]
And solving for T,
[tex]T=\frac{8\cdot 40 x}{x}=320 min[/tex]
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What is the resistance of a 4.4-m length of copper wire 1.5 mm in diameter? The resistivity of copper is 1.68×10−8Ω⋅m.
Answer:
R = 4.18 * 10^8ohms
Explanation:
R=resistance in ohms=?
ρ=resistivity of material in ohms meters = 1.68*10^-8 oh ohm meters
L= length of the object (m) = 4.4m
A = cross-sectional area of the object in square meters (m^2)= πr^2
r = (1.5mm/1000)/2= 0.00075m
A=π*(0.00075)^2 = 1.76714586764426*10^-6
Approximately, A = 1.767m^2
R = ρL/A= (1.68*10^8Ω⋅m) * (4.4m)/(1.767m^2)
R = 418336162.988115 ohms
Approximately, R = 4.18 * 10^8ohms.
The resistivity of copper will be "4.18 × 10⁸ Ω (ohms)".
ResistanceAccording to the question,
Resistivity of ohms, ρ = 1.68 × 10⁻⁸ Ω.m
Object's length, L = 4.4 m
Radius, r = [tex]\frac{\frac{1.5}{1000} }{2}[/tex]
= 0.00075 m
We know,
The cross-sectional area,
A = πr²
By substituting the values,
= π × (0.00075)²
= 1.767 × 10⁻⁶
= 1.767 m²
hence,
The resistivity will be:
→ R = [tex]\frac{\rho L}{A}[/tex]
= [tex]\frac{1.68\times 10^{-8}\times 4.4}{1.767}[/tex]
= 4183316162.9 Ω or,
= 4.18 × 10⁸ Ω
Thus the above answer is correct.
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The blackbody curve of a star moving toward Earth would have its peak shifted (a) to lower intensity; (b) toward higher energies; (c) toward longer wavelengths; (d) toward lower energies.
As show the figure, peak of each blackbody curves shifts towards higher frequency. Then energy emitted by the black body is directly proportional to the frequency incident radiation.
Hence peak is shifted to higher energies.
Therefore the correct option is B: Toward higher energies.
The blackbody curve of a star moving towards Earth would see its peak shifted toward higher energies due to the Doppler Effect, creating a 'blue shift' in the light we observe.
Explanation:The blackbody curve of a star moving toward Earth undergoes a phenomenon known as the Doppler Effect. When a star moves towards the observer, this effect causes the light from that star to be blue-shifted, which means the light moves towards higher energies, or shorter wavelengths. Therefore, the correct answer is (b) - the peak of the blackbody curve shifts towards higher energies.
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After a great many contacts with the charged ball, how is any charge on the rod arranged (when the charged ball is far away)?There is positive charge on end B and negative charge on end A.There is negative charge spread evenly on both ends.There is negative charge on end A with end B remaining neutral.There is positive charge on end A with end B remaining neutral.The conducting rod is not grounded, so a negative charge accumulates on one end, but a charge cannot remain at a single end in a conductor, so it will spread to both ends:There is negative charge spread evenly on both ends
Answer:
The correct option is that both the ends will remain neutral.
Explanation:
As the rod is grounded all of the prolonged charge will be converted to ground so the overall charge on the rod in absence of the charged ball will be equal to zero.Such that the both ends will not bear any charge.
The correct option is: There is negative charge spread evenly on both ends.
When a conducting rod is brought into contact with a charged ball, charge transfer occurs due to the process of induction and conduction. Initially, when the charged ball (let's assume it is positively charged) is brought close to the rod, it will induce a negative charge on the end of the rod nearest to the ball (end A) and a positive charge on the far end (end B) due to the separation of charges within the conductor. This is because the positive charge on the ball will repel the positive charges within the rod, pushing them away, and attract the negative charges, pulling them towards the ball.
Once the rod is actually touched to the ball, the negative charges on the rod (which were induced by the ball's positive charge) will be neutralized by the positive charges from the ball, leaving the rod with a net negative charge after the ball is moved away. This negative charge will be evenly distributed along the rod initially.
However, since the rod is not grounded, the negative charge will not have a path to escape and will remain on the rod. Due to the repulsion between like charges (negative-negative), these charges will repel each other and spread out as far as possible from each other along the length of the rod. This means that the negative charge will be spread evenly on both ends of the rod, as this arrangement minimizes the repulsive forces between the charges and results in the lowest potential energy configuration.
Therefore, the final arrangement of any charge on the rod, when the charged ball is far away, will be an even distribution of negative charge on both ends A and B of the rod. This is consistent with the principle of electrostatic equilibrium, where charges within a conductor will redistribute themselves to minimize repulsion and achieve the lowest energy state.
Spring #1 has spring constant 61.0 N/m. Spring #2 has an unknown spring constant, but when connected in series with Spring #1, the connected springs have an effective spring constant of 20.0 N/m. What is the spring constant for Spring #2?
Answer:
29.79 N/m
Explanation:
A spring connected in series behaves like a capacitor connected in series.
Note: Spring and capacitor are alike because they both store energy, While the former store mechanical energy, the later store electrical energy.
From the above, the effective spring connected in series is given by the formula below
1/Kt = 1/K1 + 1/K2 ........................ Equation 1
Where Kt = effective spring constant, K1 = spring constant of spring 1, K2 = spring constant of spring 2
Making K2 the subject of the equation,
K2 = KtK1/(K1-Kt)..................... Equation 2
Given: K1 = 61 N/m, Kt = 20 N/m.
Substitute into equation 2
K2 = (61×20)/(61-20)
K2 = 1220/41
K2 = 29.76 N/m.
Hence the spring constant in the second spring = 29.79 N/m
The spring constant for Spring #2 is 369.17 N/m.
Explanation:To find the spring constant of Spring #2, we need to use the formula for the effective spring constant of springs in series. The formula is:
1/keff = 1/k1 + 1/k2
Given that k1 = 61.0 N/m and the effective spring constant (keff) is 20.0 N/m, we can plug these values into the formula and solve for k2:
1/20 = 1/61 + 1/k2
Now we can solve for k2:
1/k2 = 1/20 - 1/61
k2 = 1/(1/20 - 1/61)
k2 = 369.17 N/m
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Two 2.00 cm * 2.00 cm plates that form a parallel-plate capacitor are charged to { 0.708 nC. What are the electric field strength inside and the potential difference across the capacitor if the spacing between the plates is (a) 1.00 mm and (b) 2.00 mm.
The electric field strength between parallel conducting plates can be calculated using the formula E = V/d, where E is the electric field strength, V is the potential difference between the plates, and d is the spacing between the plates. For a spacing of 1.00 mm, the electric field strength is 7.08 × 10^-8 N/C. For a spacing of 2.00 mm, the electric field strength is 3.54 × 10^-8 N/C.
Explanation:The electric field strength between parallel conducting plates can be calculated using the formula:
E = V/d
Where E is the electric field strength, V is the potential difference between the plates, and d is the spacing between the plates. In this case, the potential difference is given as 0.708 nC, or 0.708 × 10-9 C.
To find the electric field strength, we need to convert the charge to coulombs and divide it by the spacing between the plates. For a spacing of 1.00 mm (or 0.01 cm), the electric field strength is:
E = (0.708 × 10-9 C) / (0.01 cm) = 7.08 × 10-8 N/C
For a spacing of 2.00 mm (or 0.02 cm), the electric field strength is:
E = (0.708 × 10-9 C) / (0.02 cm) = 3.54 × 10-8 N/C
The electric field strength inside the capacitor is 2000 N/C. The potential difference is 2 V for a 1.00 mm separation and 4 V for a 2.00 mm separation between the plates.
To find the electric field strength (E) and potential difference (V) across a parallel-plate capacitor, we use the following steps:
Calculate the area (A) of the plates: A = 2.00 cm * 2.00 cm = 4.00 cm² = 4.00 * 10-4 m².Determine the charge (Q) on the plates: Q = 0.708 nC = 0.708 * 10-9 C.(a) For plate separation (d) of 1.00 mm:Compute the electric field: E = Q / (ε₀ * A) = (0.708 * 10-9 C) / (8.854 * 10-12 C²/N∙m² * 4.00 * 10-4 m²) ≈ 2000 N/C.Potential difference: V = E * d = 2000 N/C * 1.00 * 10-3 m = 2 V.(b) For plate separation of 2.00 mm:Electric field remains the same: E ≈ 2000 N/C.Potential difference: V = E * d = 2000 N/C * 2.00 * 10-3 m = 4 V.Therefore, the electric field strength inside the capacitor is 2000 N/C, the potential difference is 2 V for 1.00 mm separation, and 4 V for 2.00 mm separation.
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cmcm . The explorer finds that the pendulum completes 103 full swing cycles in a time of 132 s.What is the value of the acceleration of gravity on this planet?
Answer:
13.01 m/s²
Explanation:
The period of a simple pendulum is given as
T = 2π√(L/g) .......................... Equation 1
Where T = Period of the simple pendulum, L = Length of the pendulum, g = acceleration due to gravity of the planet.
Given; T = 132/103 = 1.28 s, L = 54 cm = 0.54 m, π = 3.14
Substitute into equation 1
1.28 = (2×3.14)√(0.54/g)
1.28 = 6.28√(0.54/g)
Making g the subject of the equation,
√(0.54/g) = 1.28/6.28
√(0.54/g) = 0.2038
0.54/g = (0.2038)²
0.54/g = 0.0415
g = 0.54/0.0415
g = 13.01 m/s²
Hence the value of the acceleration due to gravity on the planet = 13.01 m/s²
The bullet starts at rest in the gun. An 8.6 g bullet leaves the muzzle of a rifle with a speed of 430.1 m/s. What constant force is exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle?
Answer:
The constant force exerted on the bullet is 1590.87 N.
Explanation:
It is given that,
Mass of the bullet, m = 8.6 g
Initial speed of the bullet, u = 0
Final speed of the bullet, v = 430.1 m/s
We need to find the force exerted on the bullet while it is traveling down the 0.5 m length of the barrel of the rifle. Let a is the acceleration of the bullet. So,
[tex]v^2-u^2=2ad[/tex]
[tex]v^2=2ad[/tex]
[tex]a=\dfrac{v^2}{2d}[/tex]
[tex]a=\dfrac{(430.1)^2}{2\times 0.5}[/tex]
[tex]a=184986.01\ m/s^2[/tex]
Let F is the force exerted. It is given by :
[tex]F=ma[/tex]
[tex]F=8.6\times 10^{-3}\times 184986.01[/tex]
F = 1590.87 N
So, the constant force exerted on the bullet is 1590.87 N. Hence, this is the required solution.