Answer:
A. 3.9 V B. 1.9 fA
Explanation:
Part A
What emf is induced in the ring as the field changes?
Express your answer to two significant figures and include the appropriate units.
The induced emf ε = ΔΦ/Δt where ΔΦ = change in magnetic flux = ΔABcosθ where A = area of coil and B = magnetic field strength, θ = angle between A and B = 0 (since the axis of the ring is parallel )Δt = change in time
ε = ΔΦ/Δt
ε = ΔABcos0°/Δt
ε = AΔB/Δt
A = πd²/4 where d = diameter of ring = 2.0 cm = 2.0 × 10⁻² m, A = π(2.0 × 10⁻² m)²/4 = π4.0 × 10⁻⁴ m²/4 = 3.142 × 10⁻⁴ m², ΔB = change in magnetic field strength = B₁ - B₀ where B₁ = final magnetic field strength = 2.5 T and B₀ = initial magnetic field strength = 0 T. ΔB = B₁ - B₀ = 2.5 T -0 T = 2.5 T and Δt = 200 μs = 200 × 10⁻⁶ s.
So, ε = AΔB/Δt
ε = 3.142 × 10⁻⁴ m² × 2.5 T/200 × 10⁻⁶ s
ε = 7.854 × 10⁻⁴ m²-T/2 × 10⁻⁴ s
ε = 3.926 V
ε ≅ 3.9 V
Part B
If the band is gold with a cross-section area of 4.0 mm2, what is the induced current? Assume the band is of jeweler's gold and its resistivity is 13.2 x 1010 Ω*m.
Express your answer to two significant figures and include the appropriate units.
Since current, i = ε/R where ε = induced emf = 3.926 V and R = resistance of band = ρl/A where ρ = resistivity of band = 13.2 × 10¹⁰ Ωm, l = length of band = πd where d = diameter of band = 2.0 cm = 2.0 × 10⁻² m. So, l = π2.0 × 10⁻² m = 6.283 × 10⁻² m and A = cross-sectional area of band = 4.0 mm² = 4.0 × 10⁻⁶ m².
So, i = ε/R
= ε/ρl/A
= εA/ρl
= 3.926 V × 4.0 × 10⁻⁶ m²/(13.2 × 10¹⁰ Ωm × 6.283 × 10⁻² m)
= 15.704 × 10⁻⁶ V-m²/(82.9356 × 10⁸ Ωm²
= 0.1894 × 10⁻¹⁴ A
= 1.894 × 10⁻¹⁵ A
≅ 1.9 fA
In medieval times it was believed that projectiles were pushed through the air until they reached their impetus.
a. True
b. Fals
Answer:
false
Explanation:
Identical satellites X and Y of mass m are in circular orbits around a planet of mass M. The radius of the planet is R. Satellite X has an orbital radius of 3R, and satellite Y has an orbital radius of 4R. The kinetic energy of satellite X is Kx . Satellite X is moved to the same orbit as satellite Y by a force doing work on the satellite. In terms of Kx , the work done on satellite X by the force is
Answer:
The work down on satellite X by the force in terms of Kx is [tex]\dfrac{-K_x}{4}[/tex].
Explanation:
The work done is given as in terms of
[tex]W=\Delta TE[/tex]
Where ΔTE is the change in total energy.
This is given as
[tex]W=\Delta TE\\W=TE_y-TE_x\\W=\dfrac{-GMm}{2(4R)}-\dfrac{-GMm}{2(3R)}\\W=\dfrac{-GMm}{8R}+\dfrac{GMm}{6R}\\W=\dfrac{-6GMm+8GMm}{48R}\\W=\dfrac{2GMm}{48R}\\W=\dfrac{GMm}{24R}[/tex]
Rearranging it in terms of K_x gives
[tex]W=\dfrac{GMm}{24R}\\W=\dfrac{GMm}{-4\times -6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{6R}\\W=\dfrac{1}{-4}\dfrac{-GMm}{2(3R)}\\W=\dfrac{1}{-4}K_x\\W=\dfrac{-K_x}{4}[/tex]
A circular cylinder has a diameter of 3.0 cm and a mass of 25 g. It floats in water with its long axis perpendicular to the water's surface. It is pushed down into the water by a small distance and released; it then bobs up and down. Part A What is the oscillation frequency
Answer:
f = 5.3 Hz
Explanation:
To solve this problem, let's find the equation that describes the process, using Newton's second law
∑ F = ma
where the acceleration is
a = [tex]\frac{d^2 y}{dt^2 }[/tex]
B- W = m \frac{d^2 y}{dt^2 }
To solve this problem we create a change in the reference system, we place the zero at the equilibrium point
B = W
In this frame of reference, the variable y' when it is oscillating is positive and negative, therefore Newton's equation remains
B’= m [tex]\frac{d^2 y'}{dt^2 }[/tex]
the thrust is given by the Archimedes relation
B = ρ_liquid g V_liquid
the volume is
V = π r² y'
we substitute
- ρ_liquid g π r² y’ = m \frac{d^2 y'}{dt^2 }
[tex]\frac{d^2 y'}{dt^2} + \rho_liquid \ g \ \pi r^2/m ) y' \ =0[/tex]
this differential equation has a solution of type
y = A cos (wt + Ф)
where
w² = ρ_liquid g π r² /m
angular velocity and frequency are related
w = 2π f
we substitute
4π² f² = ρ_liquid g π r² / m
f = [tex]\frac{1}{2\pi } \ \sqrt{ \frac{ \rho_{liquid} \ \pi r^2 \ g}{m } }[/tex]
calculate
f = [tex]\frac{1}{2 \pi } \sqrt{ \frac{ 1000 \ \pi \ 0.03^2 \ 9.8 }{0.025} }[/tex]
f = 5.3 Hz
You and a friend are playing with a Coke can that you froze so it's solid to demonstrate some ideas of Rotational Physics. First, though, you want to calculate the Rotational Kinetic Energy of the can as it rolls down a sidewalk without slipping. This means it has both linear kinetic energy and rotational kinetic energy. [The freezing only matters because if there is liquid inside, the calculation for the Moment of inertia becomes more complicated]. A Coke can can be modeled as a solid cylinder rotating about its axis through the center of the cylinder. This can has a mass of 0.33 kg and a radius of 3.20 cm. You'll need to look up the equation for the Moment of Inertia in your textbook. It is rotating with a linear velocity of 6.00 meters / second in the counter-clockwise (or positive) direction. You can use this to determine the angular velocity of the can (since it is rolling without slipping). What is the Total Kinetic Energy of the Coke can
Answer:
K_{total} = 8.91 J
Explanation:
In this exercise you are asked to find the kinetic energy of the can of coca-cola
K_total = K_ {Translation} + K_ {rotation}
the translational kinetic energy is
K_ {translation} = ½ m v²
the kinetic energy of rotation is
K_ {rotation} = ½ I w²
The moment of inertia of a cylinder is
I = ½ m r²
we substitute
K_ {total} = ½ m v² + ½ (½ m r²) w²
angular and linear velocity are related
v = w r
we substitute
K_ {total} = ½ m v² + ¼ m r² v² / r²
K_ {total} = m v² (½ + ¼)
K_ {total} = ¾ m v²
let's calculate
K_ {total} = ¾ 0.33 6.00²
K_{total} = 8.91 J
what kind of charge does an object have if it has extra positive charges
A skater spins with an angular speed of 5.9 rad/s with her arms outstretched. She lowers her arms, decreasing her moment of inertia by a factor of 1.7. Ignoring friction on the skates, determine the ratio of her final kinetic energy to her initial kinetic energy.
Answer:
the ratio of her final kinetic energy to her initial kinetic energy is 1.7.
Explanation:
Given;
initial angular speed, ω₁ = 5.9 rad/s
let her initial moment of inertia = I₁
her final moment of inertia [tex]I_2 = \frac{I_1}{1.7}[/tex]
Apply the principle of conservation of angular momentum to determine the final angular speed of the girl;
[tex]\omega_1I_1 = \omega_f I_2\\\\\omega_f = \frac{\omega _1 I_1}{I_2} \\\\\omega_f = \frac{5.9 \times I_1}{I_1/1.7} \\\\\omega = 5.9 \times 1.7 \\\\\omega_f = 10.03 \ rad/s[/tex]
The initial rotational kinetic energy is given as;
[tex]K.E_I = \frac{1}{2}I_1 \omega_I ^2[/tex]
The final rotational kinetic energy is given as;
[tex]K.E_f = \frac{1}{2}I_2 \omega_f ^2[/tex]
The ratio of her final kinetic energy to her initial kinetic energy is given as;
[tex]\frac{K.E_f}{K.E_I}= \frac{\frac{1}{2}I_2 \omega_f^2 }{\frac{1}{2} I_1\omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{I_2 \omega_f^2}{ I_1\omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{I_1/1.7 \times \omega_f^2}{ I_1 \times \omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{ \omega_f^2}{ 1.7 \omega _1^2} \\\\\frac{K.E_f}{K.E_I}= \frac{ (10.03)^2}{ 1.7(5.9)^2} = \frac{17}{10} = 1.7[/tex]
Therefore, the ratio of her final kinetic energy to her initial kinetic energy is 1.7.
A car is travelling at 27m/s and decelerates at a=5m/s2 for a distance of 10m. Calculate its final velocity. (Hint does deceleration imply that the acceleration is positive or negative?)[
Answer:
use the formula to calculate acceleration and you'll get the answers
PLEASE HELP
Which of the following are examples of gravity in action? Select all that apply.
A. an earthquake
B. a planet orbiting the sun
C. a ball flying through the air
D. precipitation falling to Earth
When 16.35 moles of SI reacts with 11.26 moles of N2, how many moles of SI3N4 are formed
Answer:
5.45 moles
Explanation:
The chemical balanced equation of this reaction is;
3Si + 2N2 → Si3N4
From the balanced equation, we can see that that 3 moles of Si reacts with 2 moles of N2 to produce 1 mole of Si3N4.
Thus imies that the molar ration of Si to N2 is 3:2.
Now, we are told that 16.35 moles of Si reacts with 11.26 moles of N2.
16.35
Thus, using the ratio 3:2, we can say that moles of 16.35 miles of Si will react completely with (16.35 × ⅔) = 10.9 moles of N2.
Remaining N2 = 11.26 - 10.9 = 0.36 will be the excess.
From our balanced chemical equation, we saw that;
3 moles of Si produced 1 mole of Si3N4.
Thus; 16.35 moles of Si will produce;
no. of moles of Si3N4 produced = (1 × 16.35)/(3.0) = 5.45 moles
Extra CreditA particle is directed along the axis of the instrument in the gure. Aparallel plate capacitor sets up an electric eld E, which is orientedperpendicular to a uniform magnetic eld B. If the plates are separated byd= 2:0 mm and the value of the magnetic eld isB= 0:60T. Calculatethe potential di erence, between the capacitor plates, required to allow aparticle
This question is incomplete, the complete question is;
A particle is directed along the axis of the instrument in the figure below. A parallel plate capacitor sets up an electric field E, which is oriented perpendicular to a uniform magnetic field B. If the plates are separated by d = 2.0 mm and the value of the magnetic field is B = 0.60T.
Calculate the potential difference, between the capacitor plates, required to allow a particle with speed v = 5.0 × 10⁵ m/s to pass straight through without deflection.
Hint : ΔV = Ed
Answer:
the required potential difference, between the capacitor plates is 600 V
Explanation:
Given the data in the question;
B = 0.60 T
d = 2.0 mm = 0.002 m
v = 5.0 × 10⁵ m/s.
since particle pass straight through without deflection.
F[tex]_{net[/tex] = 0
so, F[tex]_E[/tex] = F[tex]_B[/tex]
qE = qvB
divide both sides by q
E = vB
we substitute
E = (5.0 × 10⁵) × 0.6
E = 300000 N/C
given that; potential difference ΔV = Ed
we substitute
ΔV = 300000 × 0.002
ΔV = 600 V
Therefore, the required potential difference, between the capacitor plates is 600 V
4. Speedy leaves the ground with an initial vertical velocity of 53 m/s and a horizontal velocity of 42 m/s.
How much time does he spend in the air?
How far (horizontally) does he travel during this time?
5. The Angry Bird is fired at an angle of 35 above the horizontal at a speed of 72 m/s.
Draw the initial velocity vector
Determine the initial horizontal velocity
Determine the initial vertical velocity
How much time does it spend in the air?
What horizontal distance does it go?
physics grade9 teacher guide
Answer:
huh
Explanation:
what does loudness of a sound depend on?
Answer:
Amplitude
Explanation:
The loudness of a sound depends on the amplitude of vibration producing the sound
Increasing the telescope diameter beyond the value found in part (a) will increase the light-gathering power of the telescope, allowing more distant and dimmer astronomical objects to be studied, but it will not improve the resolution. In what ways are the Keck telescopes (each of 10-m diameter) atop Mauna Kea in Hawaii superior to the Hale Telescope (5-m diameter) on Palomar Mountain in California
Answer:
Ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
Explanation:
If increasing the Diameter of a Telescope beyond a given value will increase the ability of the telescope to capture more light and also capture astronomical objects located in a very distant position without improving resolution.
Hence the superiority of Keck telescope atop Mauna Kea over Hale Telescope atop Palomar mountain in California is the ability of the Keck telescope to capture more distant object despite been atop Mauna kea that Hale Telescope may not capture even if it is atop Palomar mountain in California
Plutonium-238 has a half life of 87.7 years. What percentage of a 5 kilogram (kg) sample remains after 50 years?
Answer:
i dont know but i should know try g o o g l e
Explanation:
g A high-speed flywheel in a motor is spinning at 500 rpm when a power failure suddenly occurs. The flywheel has mass 39.0kg and diameter 78.0cm. The power is off for 34.0s, and during this time the flywheel slows due to friction in its axle bearings. During the time the power is off, the flywheel makes 170 complete revolutions.At what rate is the flywheel spinning when the power comes back on?
Answer:
[tex]10.54\ \text{rad/s}[/tex]
Explanation:
[tex]\omega_i[/tex] = Initial angular velocity = 500 rpm = [tex]500\times \dfrac{2\pi}{60}\ \text{rad/s}[/tex]
[tex]\omega_f[/tex] = Final angular velocity
t = Time = 34 s
[tex]\theta[/tex] = Angular displacement = 170 revs = [tex]170\times 2\pi\ \text{rad}[/tex]
[tex]\alpha[/tex] = Angulr acceleration
From the kinematic equations of angular motion we have
[tex]\theta=\omega_it+\dfrac{1}{2}\alpha t^2\\\Rightarrow \alpha=\dfrac{\theta-\omega_it}{\dfrac{1}{2}t^2}\\\Rightarrow \alpha=\dfrac{170\times 2\pi-500\times \dfrac{2\pi}{60}\times 34}{\dfrac{1}{2}\times 34^2}\\\Rightarrow \alpha=-1.23\ \text{rad/s}^2[/tex]
[tex]\omega_f=\omega_i+\alpha t\\\Rightarrow \omega_f=500\times \dfrac{2\pi}{60}+(-1.23)\times 34\\\Rightarrow \omega_f=10.54\ \text{rad/s}[/tex]
The rate at which the wheel is spinning when the power comes back on is [tex]10.54\ \text{rad/s}[/tex].
If F = force, which equation illustrates the Law of Conservation of Momentum?
A) F1 = F2
B) F1 = - F2
C) - F1 = -F 2
D) F1 + - F2 = F3
Answer:
b
Explanation:
f1=-f2 that could be thank u
A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surface of the water. It takes a time of 2.00 s for the boat to travel from its highest point to its lowest, a total distance of 0.600 m . The fisherman sees that the wave crests are spaced a horizontal distance of 6.40 m apart.
Required:
a. How fast are the waves traveling?
b. What is the amplitude of each wave?
c. If the total vertical distance traveled by the boat were 0.30 m but the other data remained the same, how would the answers to parts (a) and (b) be affected?
Answer:
a. Speed = 1.6 m/s
b. Amplitude = 0.3 m
c. Speed = 1.6 m/s
Amplitude = 0.15 m
Explanation:
a.
The frequency of the wave must be equal to the reciprocal of the time taken by the boat to move from the highest point to the highest point again. This time will be twice the value of the time taken to travel from the highest point to the lowest point:
frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz
The wavelength of the wave is the distance between consecutive crests of wave. Therefore,
Wavelength = 6.4 m
Now, the speed of the wave is given as:
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
Speed = 1.6 m/s
b.
Amplitude is the distance between the mean position of the wave and the extreme position. Hence, it will be half the distance between the highest and lowest point:
Amplitude = (0.5)(0.6 m)
Amplitude = 0.3 m
c.
frequency = [tex]\frac{1}{2(2\ s)}[/tex] = 0.25 Hz
Speed = (Frequency)(Wavelength)
Speed = (0.25 Hz)(6.4 m)
Speed = 1.6 m/s
Amplitude = (0.5)(0.3 m)
Amplitude = 0.15 m
Saved Which of the following is NOT an important function of facial display? Multinio Choic
A. emotion
B. attractiveness
c. Primacy
d. identity
Answer:
C
Explanation:
Primacy means being first or important so thats not an important facial display as the others.
In a physics lab experiment for the determination of moment of inertia, a team weighs an object and finds a mass of 4.07 kg. They then hang the object on a pivot located 0.155 m from the object's center of mass and set it swinging at a small amplitude. As two of the team members carefully count 113 cycles of oscillation, the third member measures a duration of 247 s. What is the moment of inertia of the object with respect to its center of mass about an axis parallel to the pivot axis
Answer:
I = 0.65 kgm²
Explanation:
Since the mass is an inertial pendulum, we use the formula for the period, T of an inertial pendulum.
T = 2π√(I/mgh) where I = moment of inertia of object about pivot point, m = mass of object5 = 4.07 kg, g = acceleration due to gravity = 9.8 m/s² and h = distance of center of mass of object from pivot point = 0.155 m.
Given that the team measures 113 cycles of oscillation in 247 s, the period, T = time of oscillations/total number of oscillations = 247 s/113 oscillations = 2.186 s/oscillation
So, T = 2.186 s
We now find I by making it subject of the formula in the equation for T.
So,
T = 2π√(I/mgh)
dividing both sides by 2π, we have
T/2π = √(I/mgh)
squaring both sides, we have
(T/2π)² = [√(I/mgh)]²
T²/4π² = I/mgh
multiplying both sides by mgh, we have
T²mgh/4π² = I
I = T²mgh/4π²
substituting the values of the variables into the equation, we have
I = T²mgh/4π²
I = (2.186 s)² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²
I = 4.778 s² × 4.07 kg × 9.8 m/s² × 0.155 m/4π²
I = 29.539 kgm²/4π²
I = 0.748 kgm²
Now I = I' + mh² (parallel axis theorem) where I' = moment of inertia of object about its center of mass, m = mass of object = 4.07 kg and h = distance of center of mass object from pivot point.
So, I' = I - mh²
Substituting the values of the variables into the equation, we have
I' = I - mh²
I' = 0.748 kgm² - 4.07 kg × (0.155 m)²
I' = 0.748 kgm² - 4.07 kg × 0.02403 m²
I' = 0.748 kgm² - 0.098 kgm²
I = 0.65 kgm²
A 1.10 kg block is attached to a spring with spring constant 17 N/m. While the block is sitting at rest, a student hits it with a hammer and almost instantaneously gives it a speed of 46 cm/s.
A) What is the amplitude of the subsequent oscillations?
B) What is the block's speed at the point where x = 0.25 A?
Answer:
Explanation:
The kinetic energy of block will be converted into potential energy of spring .
If A be the amplitude of oscillations
1 /2 k A² = 1/2 m v²
17 A² = 1.1 x .46²
A² = .0137
A= 11.7 cm
B )
when x = .25 A = .25 x 11.7 = 2.9 cm
potential energy = 1/2 k x²
= .5 x 17 x ( .029 )² = .00715 J
kinetic energy = 1/2 m v²
1/2 m v² + .00715 = .5 1.1 x .46²
1/2 m v² + .00715 = .1164
1/2 m v² = .10925
.5 x 1.1 x v²= .10925
v² = .1986
v = .4456 m /s
= 44.56 cm /s
A truck is traveling on a level road. The driver suddenly applies the brakes, causing the truck to decelerate by an amount g/2. This causes a box in the rear of the truck to slide forward. If the coefficient of sliding friction between the box and the truckbed is 2/5, find the acceleration of the box relative to the truck and relative to the road.
Answer:
Truck [tex]\dfrac{g}{10}[/tex]
Road [tex]-\dfrac{g}{10}[/tex]
Explanation:
[tex]a_1[/tex] = Acceleration of truck = [tex]-\dfrac{g}{2}[/tex]
[tex]\mu[/tex] = Coefficient of friction = [tex]\dfrac{2}{5}[/tex]
Frictional force is given by
[tex]f=-\mu mg\\\Rightarrow f=-\dfrac{2}{5}mg\\\Rightarrow ma_2=-\dfrac{2}{5}mg\\\Rightarrow a_2=-\dfrac{2}{5}g[/tex]
Net acceleration is given by
[tex]a=a_2-a_1\\\Rightarrow a=-\dfrac{2}{5}g+\dfrac{g}{2}\\\Rightarrow a=\dfrac{g}{10}[/tex]
The acceleration of the box relative to the truck is [tex]\dfrac{g}{10}[/tex] and [tex]-\dfrac{g}{10}[/tex] relative to the road.
Urgent!!!!! A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes. The data are shown in the table. Estimate the average temperature of the air in the room at 20 min. explain your answer.
The average temperature of the air in the room at 20 min is 23°C.
What is temperature?Temperature is the degree of hotness or coldness of the object.
A student heated 235 g of water in a beaker until the water reached 100°C. The student removed the beaker from the heat and placed the beaker on a counter in a 23°C room. The student recorded the temperature of the water every 4 minutes for 20 minutes.
The surrounding is vast, its temperature does not get affected by small amount of water. So, the temperature of air remains constant.
Thus, the average temperature of the air in the room at 20 min is 23°C.
Learn more about temperature.
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1.What is the Kinetic energy of a 3 kg object moving at 4 m/s?
Plz help I’ll give points
Answer:
24 J
Explanation:
[tex]K = \frac{1}{2} mv^{2} = \frac{1}{2} (3kg)(4m/s)^{2} = 24 J[/tex]
stored energy is _________ ___________
kinetic energy
energy in motion
potential energy
Answer:
Potential energy
Explanation:
Potential energy is stored energy
Two point charges, initially 3 cm apart, are moved to a distance of 1 cm apart. By what factor does the resulting electric force between them change?
A. 3
B. 1/9
C. 1/3
D. 9
How does the presence of a nucleus provide a method of basic cell
classification? *
Answer:
The nucleus-containing cells are called eukaryotic cells. Eukaryote means having membrane-bound organelles.
Explanation:
I hope this is what you were looking for?!
Hope this helps!
Have a great day!
-Hailey!
The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.
What is a nucleus?
The nucleus is the controlling center of the body. It performs all the major activities in the cell. The cell which possesses a nucleus is called a eukaryote. The cell which does not have a nucleus is called prokaryote.
A prokaryotic cell is a straightforward, one-celled (unicellular) organism that is devoid of a nucleus or any other organelle that is membrane-bound. The nucleoid, a dark area in the center of the cell, is where prokaryotic DNA is located.
Therefore, The cells which have a nucleus are called Eukaryotic cells. The cells which do not have a nucleus are called prokaryotic cells.
To learn more about cell, refer to the link:
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A 2:2 kg toy train is con ned to roll along a straight, frictionless track parallel to the x-axis. The train starts at the origin moving at a speed of 1:6m=s in the +x direction, and continues until it reaches a position 7:5m down the track from where it started. During its journey, it experiences a force pointing in the same direction as the vector 0:6 +0:8 , with magnitude initially 2:8N and decreasing linearly with its x-position to 0N when the train has finished its journey.
Required:
a. Calculate the work done by this force over the entire journey of the train.
b. Find the speed of the train at the end of its journey.
Answer:
a) 10.51 J
b) 3.48 m/s
Explanation:
Given data :
mass of train ( M ) = 2.2 kg
Given initial velocity ( u ) = 1.6 m/s
a) calculating work done by the force over the journey of the train
F = mx + b ------ ( 1 )
m = slope = ( Δ f / Δ x ) = 2.8 / -7.5 = - 0.373 N/m
x = distance travelled on the x axis by the train = 7.5 m
F = force experienced by the train = 2.8 N
x = 0
∴ b = 2.8
hence equation 1 can be written as
F = ( -0.373) x + 2.8 ----- ( 2 )
hence to determine the work done by the force
W = [tex]\int\limits^7_0 { ( -0.373) x + 2.8 )} \, dx[/tex] Note: the limits are actually 7.5 and 0
∴ W ( work done ) = -10.49 + 21 = 10.51 J
b) calculate the speed of the train at the end of its journey
we will apply the work energy theorem
W = 1/2 m*v^2 - 1/2 m*u^2
∴ V^2 = 2 / M ( W + 1/2 M*u^2 ) ( input values into equation )
V^2 = 12.11
hence V = 3.48 m/s
A wheel has eight spokes and a radius of 30 cm. It is mounted on a fixed axle and is spinning at 2.5 rev/s. You want to shoot a 24- cm arrow parallel to this axle and through the wheel without hitting any of the spokes Assume that the arrow and the spokes hitting any of the spokes. Assume that the arrow and the spokes are very thin. (a) What minimum speed must the arrow have to pass through (a) What minimum speed must the arrow have to pass through without contact
Answer:
4.8 m/s
Explanation:
Given: angular velocity of wheel ω = 2.5 rev/sec
radius r = 30 cm
length of arrow = 24 cm
For arrow to pass through spinning ring it has to pass between any two spokes of the wheel.
angle between two spokes = π/4
time taken by a spook to reach the position of adjacent spoke t =θ/ω
= π/4/(2.5×2π) = 1/20 sec
for the arrow to pass through the spokes of the wheel it should take time t <1/20 sec to pass through the wheel
a) therefore, minimum speed = (24/100)/(1/20) = 4.8 m/s
Think of a hydropower dam . How is electrical energy produced from potential and kinetic energy ?
hydroelectric dam converts the potential energy stored in a water reservoir behind a dam to mechanical energy—mechanical energy is also known as kinetic energy. ... The generator converts the turbine's mechanical energy into electricity.
Hope this helps!
Answer:
Potential energy and kinetic energy are constituents of mechanical energy.
When a turbine is switched on, it rotates with mechanical energy.
Since a motor runs the turbine, it converts this mechanical energy to electrical energy.
