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Savatey [412]
3 years ago
6

In your experiment, you measure a total deflection of 4.12 cm when an electric field of 1.10×103V/m is established between the p

lates (with no magnetic field present). When you add the magnetic field as described in Part C, to what value do you have to adjust its magnitude B0 to observe no deflection? Assume that the plates are 6.00 cm long and that the distance between them and the screen is 12.0 cm. Express your answer numerically in tesla.
Physics
1 answer:
Bond [772]3 years ago
5 0

Answer:

B_0 = 1.69 \times 10^{-4}\ T

Explanation:

given,

total deflection = 4.12 cm

Electric field = 1.1 ×10³ V/m

plate length = 6 cm

distance between them = 12 cm

using formula

v_0 = \sqrt{\dfrac{q\epsilon_0d}{ym}(\dfrac{d}{2}+L)}

q = 1.6 × 10⁻¹⁹ C

m = 9.11 x 10⁻³¹ kg

d = 0.06 m

L = 0.12 m

v_0 = \sqrt{\dfrac{1.6 \times 10^{-19}\times 1.1 \times 10^{3}\times 0.06}{0.0412\times 9.11 \times 10^{-31} }(\dfrac{0.06}{2}+0.12)}

v_0 = 6496355.63 m/s

v_0 = \dfrac{E}{B_0}

B_0 = \dfrac{E}{v_0}

B_0 = \dfrac{1.1\times 10^{3}}{6496355.63}

B_0 = 1.69 \times 10^{-4}\ T

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8090 [49]

The work done by tension force of 14N applied on the laptop by a rope as it moves 2.0 mm up the slope is 0.028 J

W = F d cos θ

W = Work done

F = Force

d = Displacement

θ = Angle between force and displacement vector

F = 14 N

d = 2 mm = 0.002 m

θ = 0

W = 14 * 0.002 * 1

W = 0.028 J

Work done is the change in energy of an object. So if an object moves a certain distance, work is done on the object. If the force and displacement are perpendicular to each other there is no work done on the object.

Therefore, the work done by tension on the laptop is 0.028 J

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8 0
1 year ago
Which of the following examples illustrates static friction?
vivado [14]

Answer:

A box sits stationary  on a ramp

Explanation:

Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.

Static force of friction is calculated as follows:

F= μη

F is static force of friction.

μ is the coefficient of static friction.

η is the normal force.

6 0
3 years ago
A proton is released from rest at the positive plate of a parallelplatecapacitor. It crosses the capacitor and reaches the negat
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Answer:

2.1406 ×10^6 m/sec

Explanation:

we know that energy is always conserved

so from the law of energy conservation

qV=\frac{1}{2}mv^2

here V is the potential difference  

we know that mass of proton = 1.67×10^{-27} kg

we have given speed =50000m/sec

so potential difference V=\frac{\frac{1}{2}\times 1.67\times 10^{-27}50000^2}{1.6\times 10^{-19}}=13.045

now mass of electron =9.11×10^{-31}

so for electron

\frac{1}{2}\times 9.11\times 10^{-31}v^2=1.6\times 10^{-19}\times 13.045=2.1406\times 10^6 m/sec

so the velocity of electron will be 2.1406×10^6 m/sec

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3 years ago
if a star 100 light years from earth is beginning to expand into a giant star how long will it take for astronomers to observe t
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Answer:

100years later

Explanation:

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7 0
3 years ago
What is the deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h? (such deceleration caused on
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The deceleration of the rocket sled if it comes to rest in 1.1 s from a speed of 1000 km/h is  252.52\ m/s^2.

The acceleration in opposite direction is known as the deceleration. Basically the deceleration is negative value of the acceleration since the negative sign depicts its opposite in direction.

The given data:

time, t = 1.1 s

initial speed, u = 1000 km/h = \frac{2500}{9}\ m/s

final speed, v = 0 m/s

So we will be using the equation of motion, that is,

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6 0
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