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nataly862011 [7]
3 years ago
6

An electron is released from rest at a distance of 0.300 m from a large insulating sheet of charge that has uniform surface char

ge density 4.90×10^−12 C/m^2C/m^2.
a. How much work is done on the electron by the electric field of the sheet as the electron moves from its initial position to a point 0.050 m from the sheet?
b. What is the speed of the electron when it is 0.050 m from the sheet?
Physics
1 answer:
jarptica [38.1K]3 years ago
8 0
Yes that is the correct answer
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Two bodies of mass m₁ & m₂ are moving with the same velocity 'v' K.E. will be greater for??
boyakko [2]

Answer:

the one with a higher mass

Explanation:

The body with more mass will have the greater kinetic energy of the two.

Kinetic energy is the energy due to the motion of body. It is mathematically expressed as:

           K.E  = \frac{1}{2}  m v²

m is the mass

v is the velocity

 Since the velocity of the two bodies are the same, and mass is directly proportional to kinetic energy, the body with more mass will have a higher kinetic energy.

 So between mass m1 and mass m2, the one with a greater mass will have a higher kinetic energy

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2 years ago
Friction is a ____________ force<br> a. Artificial<br> b. Natural<br> c. Pessimistic<br> d. Negative
Daniel [21]

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2 years ago
Read 2 more answers
The displacement of a wave traveling in the positive x-direction is y(x, t)|= (3.5 cm)cos(2.7x − 92t), where x is in m and t is
zubka84 [21]

Answer

Given,

y(x, t) = (3.5 cm) cos(2.7 x − 92 t)

comparing the given equation with general equation

y(x,t) = A cos(k x - ω t)

 A = 3.5 cm  , k = 2.7 rad/m    , ω = 92 rad/s

we know,

a) ω =2πf

   f = 92/ 2π

   f = 14.64 Hz

b) Wavelength of the wave

 we now, k = 2π/λ

      2π/λ = 2.7

      λ = 2 π/2.7

      λ = 2.33 m

c) Speed of wave

     v = ν λ

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5 0
3 years ago
Please help me with this physics prooblem
zaharov [31]

Take the missile's starting position to be the origin. Assuming the angles given are taken to be counterclockwise from the positive horizontal axis, the missile has position vector with components

x=v_0\cos20.0^\circ t+\dfrac12a_xt^2

y=v_0\sin20.0^\circ t+\dfrac12a_yt^2

The missile's final position after 9.20 s has to be a vector whose distance from the origin is 19,500 m and situated 32.0 deg relative the positive horizontal axis. This means the final position should have components

x_{9.20\,\mathrm s}=(19,500\,\mathrm m)\cos32.0^\circ

y_{9.20\,\mathrm s}=(19,500\,\mathrm m)\sin32.0^\circ

So we have enough information to solve for the components of the acceleration vector, a_x and a_y:

x_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\cos20.0^\circ(9.20\,\mathrm s)+\dfrac12a_x(9.20\,\mathrm s)^2\implies a_x=21.0\,\dfrac{\mathrm m}{\mathrm s^2}

y_{9.20\,\mathrm s}=\left(1810\,\dfrac{\mathrm m}{\mathrm s}\right)\sin20.0^\circ(9.20\,\mathrm s)+\dfrac12a_y(9.20\,\mathrm s)^2\implies a_y=110\,\dfrac{\mathrm m}{\mathrm s^2}

The acceleration vector then has direction \theta where

\tan\theta=\dfrac{a_y}{a_x}\implies\theta=79.2^\circ

5 0
3 years ago
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