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3 years ago

11

HELP ASAPP PLEASEE

2 answers:

Goshia [24]3 years ago

7 0

To calculate the Potential Difference between the plates in an Uniform Electric Field we have the formula:
$u = e \times d$
U = Potential difference = ?
E = Electric Field = 4.1 x 10 N/C
D = Distance = 0.01m = 1 x 10^-2 m

$u = (4.1 \times 10) \times (1 \times 10^{ - 2}) \\ u = 41 \times 10^{ - 2} \: volts \: or \: \: 0.41 \: volts$

olga55 [171]3 years ago

6 0

Answer: It can be moved 0.01 m toward the negative plate.

Explanation: I just took the test.

!!!!!!!

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4 years ago

A boy sleds down a hill and onto a frictionless ice- covered lake at 10.0 m/s. In the middle of the lake is a 1000-kg boulder. W

mina [271]

Answer:

The speed of the sled is 9.2 m/s

The speed of the boulder is 0.82 m/s

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As per the question:

Mass of the boulder, $m_{B} = 1000\ kg$

Mass of the sled, $m_{S} = 2.50\ kg$

Mass of the boy, $m_{b} = 40\ kg$

Initial Velocity, v = 10.0 m/s

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To calculate the speed of both the sled and the boulder after the occurrence of the collision:

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Initial velocity of the boulder, $v_{B} = 0\ m/s$

Since, the collision is elastic, both the energy and momentum rem,ain conserved.

Now,

Using the conservation of momentum:

$mv + m_{B}v_{B} = mv' + m_{B}v'_{B}$

where

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$v'_{B}$ = final velocity of the boulder

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Now,

Using conservation of energy:

$\frac{1}{2}mv^{2} + \frac{1}{2}m_{B}v_{B}^{2} = \frac{1}{2}mv'^{2} + \frac{1}{2}m_{B}v'_{B}^{2}$

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$42.50v'^{2} + 1000v'_{B}^{2} = 4250$ (2)

Now, from eqn (1) and (2):

$v' = \frac{m - m_{B}}{m + m_{B}}\times v$

$v' = \frac{42.50 - 1000}{42.5 + 1000}\times 10 = - 9.2\ m/s$

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$v'_{B} = \frac{2m}{m + m_{B}}\times v$

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3 years ago

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mars1129 [50]

Answer:

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mrs_skeptik [129]

Answer:

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