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Alina [70]
3 years ago
13

A proton is released from rest in a uniform electric field. After the proton has traveled a distance of 10.0 cm, its speed is 1.

4 x 10^6 m/s in the positive x direction.
Required:
a. Find the magnitude and direction of the electric field.
b. Find the speed of the proton
Physics
1 answer:
GREYUIT [131]3 years ago
6 0

Answer:

Explanation:

distance travelled s = 10 cm

speed v = 1.4 x 10⁶ m /s

v² = u² + 2as

u = 0

v² = 2as

( 1.4 x 10⁶ )² = 2 x a x .10

a = 9.8 x 10¹² m /s²

force on proton = mass x acceleration

= 1.67 x 10⁻²⁷ x 9.8 x 10¹²

= 16.366 x 10⁻¹⁵ N .

If magnitude of electric field be E

force on proton

= E x charge on proton

= E x 1.6 x 10⁻¹⁹

E x 1.6 x 10⁻¹⁹ = 16.366 x 10⁻¹⁵

E = 10.22 x 10⁴ N/C

The direction of electric field will be positive x - direction .

b )

Speed of proton = 1.4 x 10⁶  m /s .

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Answer:

1) as far as I remember

Let's take 20 as vf (final velocity) and 11 as (initial velocity) and 4 as time

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So 20-11/4

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To impress his friends while riding on a carnival
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B. decreases while his angular speed remains  unchanged.

Explanation:

His angular speed will always be the same as the wheel's angular speed, which remains constant as it's in uniform motion. As for linear speed, which is defined as the product of angular speed and distance r to the center of rotation, and his distance to center is decreasing, his linear speed must be decreasing as well.

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What is the smallest time interval in which a 5.7 T magnetic field can be turned on or off if the induced emf around the patient
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The smallest time interval in which the magnetic field can be turned on or off to induced the emf is 47.5 s.

<h3>Emf induced in the coil</h3>

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3 0
2 years ago
A block of ice(m = 14.0 kg) with an attached rope is at rest on a frictionless surface. You pull the block with a horizontal for
nadezda [96]

Answer:

a) The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) The final speed of the block of ice is 9.8 meters per second.

Explanation:

a) We need to calculate the weight, normal force from the ground to the block and the pull force. By 3rd Newton's Law we know that normal force is the reaction of the weight of the block of ice on a horizontal.

The weight of the block (W), measured in newtons, is:

W = m\cdot g (1)

Where:

m - Mass of the block of ice, measured in kilograms.

g  - Gravitational acceleration, measured in meters per square second.

If we know that m = 14\,kg and g = 9.807\,\frac{m}{s^{2}}, the magnitudes of the weight and normal force of the block of ice are, respectively:

N = W = (14\,kg)\cdot \left(9.807\,\frac{m}{s^{2}} \right)

N = W = 137.298\,N

And the pull force is:

F_{pull} = 98\,N

The weight and the normal force of the block has a magnitude of 137.298 newtons and the pull force exerted on the block has a magnitude of 98 newtons.

b) Since the block of ice is on a frictionless surface and pull force is parallel to the direction of motion and uniform in time, we can apply the Impact Theorem, which states that:

m\cdot v_{o} +\Sigma F \cdot \Delta t = m\cdot v_{f} (2)

Where:

v_{o}, v_{f} - Initial and final speeds of the block, measured in meters per second.

\Sigma F - Horizontal net force, measured in newtons.

\Delta t - Impact time, measured in seconds.

Now we clear the final speed in (2):

v_{f} = v_{o}+\frac{\Sigma F\cdot \Delta t}{m}

If we know that v_{o} = 0\,\frac{m}{s}, m = 14\,kg, \Sigma F = 98\,N and \Delta t = 1.40\,s, then final speed of the ice block is:

v_{f} = 0\,\frac{m}{s}+\frac{(98\,N)\cdot (1.40\,s)}{14\,kg}

v_{f} = 9.8\,\frac{m}{s}

The final speed of the block of ice is 9.8 meters per second.

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