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krek1111 [17]
3 years ago
5

Which of the following is the best example of uniform circular motion? A. A child riding on a constantly spinning merry-go-round

B. A car on a road curving back and forth as it passes through hills C. A pendulum bob swinging back and forth in a curved path D. A baseball that is hit up and out of the park for a home run
Physics
1 answer:
Ulleksa [173]3 years ago
5 0

A. because as the merry-go-round spins the child accelerates towards the center of the merry-go-round at a uniform rate.

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TheE-field strength is 50,000 N/C inside a parallel plate capacitor with 2.0 mmspacing. A proton is released from rest at the po
Ghella [55]

Answer:

The speed is 138852.4 \frac{m}{s}

Explanation:

Electric field magnitude (E) and electric force magnitude (F) are related by the equation

F=Eq

with q the charge of the proton (1.61\times10^{-19} C). Because between parallel plates electric field is almost constant, electric force is constant too:

F=(50000)(1.61\times10^{-19})=8.05\times10^{-15}N

because electric force is constant, then by Newton's second law acceleration (a) is constant too, it is:

a=\frac{F}{m}

with m the mass of the proton (1.67\times10^{-27} kg):

a=\frac{8.05\times10^{-15}}{1.67\times10^{-27}}= 4.82\times10^{12}\frac{m}{s^{2}}

Now with this constant acceleration we can use the kinematic equation

v^{2}=v_{0}^{2}+2ad

with v the final speed, v_i the initial velocity that is zero (because proton starts at rest) and d is the distance between the plates, so:

v=\sqrt{2ad}=\sqrt{2(4.82\times10^{12}\frac{m}{s^{2}})(2.0\times10^{-3}m)}

v=138852.4 \frac{m}{s}

5 0
3 years ago
पत्र- अपने क्षेत्र की सड़कों की बुरी दशा की जानकारी देते हुए और उन्हें ठीक कराने की प्रार्थना करते हुए नगर निगम अधिकारी को पत्र
Ksivusya [100]

Answer:

पत्र- अपने क्षेत्र की सड़कों की बुरी दशा की जानकारी देते हुए और उन्हें ठीक कराने की प्रार्थना करते हुए नगर निगम अधिकारी को पत्र लिखिए |

5 0
2 years ago
Look at the equation shown below. What does the letter Q represent?
Westkost [7]

Answer:

Explanation:

I=CURRENT

Q=CHARGE

t=TIME

6 0
3 years ago
Read 2 more answers
An astronaut is standing on the surface of a planetary satellite that has a radius of 1.74 × 10^6 m and a mass of 7.35 × 10^22 k
ExtremeBDS [4]

Answer:

2.87 km/s

Explanation:

radius of planet, R = 1.74 x 10^6 m

Mass of planet, M = 7.35 x 10^22 kg

height, h = 2.55 x 10^6 m

G = 6.67 x 106-11 Nm^2/kg^2

Use teh formula for acceleration due to gravity

g=\frac{GM}{R^{2}}

g=\frac{6.67\times 10^{-11}\times 7.35\times 10^{22}}{1.74^{2}\times 10^{12}}

g = 1.62 m/s^2

initial velocity, u = ?, h = 2.55 x 10^6 m , final velocity, v = 0

Use third equation of motion

v^{2}=u^{2}-2gh

0 = v² - 2 x 1.62 x 2.55 x 10^6

v² = 8262000

v = 2874.37 m/s

v = 2.87 km/s

Thus, the initial speed should be 2.87 km/s.

6 0
3 years ago
A solenoidal coil with 26 turns of wire is wound tightly around another coil with 350 turns. The inner solenoid is 20.0 cm long
noname [10]

Answer:

Part a)

\phi = 2.76 \times 10^{-7} T m^2

Part B)

M = 5.52 \times 10^{-5} H

Part C)

EMF = 0.1 V/s

Explanation:

Part a)

Magnetic field due to a long ideal solenoid is given by

B = \mu_0 n i

n = number of turns per unit length

n = \frac{N}{L}

n = \frac{350}{0.20}

n = 1750 turn/m

now we know that magnetic field due to solenoid is

B = (4\pi \times 10^{-7})(1750)(0.100)

B = 2.2 \times 10^{-4} T

Now magnetic flux due to this magnetic field is given by

\phi = B.A

\phi = (2.2 \times 10^{-4})(\pi r^2)

\phi = (2.2 \times 10^{-4})(\pi(0.02)^2)

\phi = 2.76 \times 10^{-7} T m^2

Part B)

Now for mutual inductance we know that

\phi_{total} = M i

\phi_{total} = N\phi

\phi_{total} = 20(2.76 \times 10^{-4})

\phi_{total} = 5.52 \times 10^{-6}

now we have

M = \frac{5.52 \times 10^{-6}}{0.100}

M = 5.52 \times 10^{-5} H

Part C)

As we know that induced EMF is given as

EMF = M \frac{di}{dt}

EMF = 5.52 \times 10^{-5} (1800)

EMF = 0.1 V/s

3 0
3 years ago
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