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2 years ago

5

A small glass bead charged to 8.0 nC is in the plane that bisects a thin, uniformly charged, 10-cm long glass rod and is 4.0 cm

from the rod's center. The bead is repelled from the rod with a force of 940 μN.
Required:
What is the total charge on the rod?

1 answer:

Assoli18 [71]2 years ago

8 0

Answer:

71nC is the total charge of the rod

Explanation:

See attached file

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A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then

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C) $F_{f2}=207.58N$

Explanation:

This question is incomplete. The full question was:

<em>A skateboarder with mass ms = 54 kg is standing at the top of a ramp which is hy = 3.3 m above the ground. The skateboarder then jumps on his skateboard and descends down the ramp. His speed at the bottom of the ramp is vf = 6.2 m/s. </em>

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<em>Part (b) The ramp makes an angle θ with the ground, where θ = 30°. Write an expression for the magnitude of the friction force, fr, between the ramp and the skateboarder. </em>

<em>Part (c) When the skateboarder reaches the bottom of the ramp, he continues moving with the speed vf onto a flat surface covered with grass. The friction between the grass and the skateboarder brings him to a complete stop after 5.00 m. Calculate the magnitude of the friction force, Fgrass in newtons, between the skateboarder and the grass.</em>

For part A), we make a balance of energy to calculate the work done by the friction force:

$W_{ff}=\Delta E$

$W_{ff}=1/2*m*vf^2-m*g*hy$

$W_{ff}=-744.12J$

For part B), we use our previous value for the work:

$W_{ff}=-F_f*(hy/sin\theta)$ Solving for friction force:

$F_f=-W_{ff}*sin\theta /hy$

$F_f=112.75N$

For part C), we first calculate the acceleration by kinematics and then calculate the module of friction force by dynamics:

$Vf^2=Vo^2+2*a*d$

Solving for a:

$a=-3.844m/s^2$

Now, by dynamics:

$|F_f|=|m*a|$

$|F_f|=207.58N$

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