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kondaur [170]
4 years ago
7

projectile motion of a particle of mass M with charge Q is projected with an initial speed V in a driection opposite to a unifor

m electric fiedl of magnitude E.
Physics
1 answer:
SIZIF [17.4K]4 years ago
5 0

Answer:

Range, R = MV²/2QE

Explanation:

The question deals with the projectile motion of a particle mass M with charge Q, having an initial speed V in a direction opposite to that of a uniform electric field.

Since we are dealing with projectile motion in an electric field, the unknown variable here, would be the range, R of the projectile. We note that the electric field opposes the motion of the particle thereby reducing its kinetic energy. The particle stops when it loses  all its kinetic energy due to the work done on it in opposing its motion by the electric field. From work-kinetic energy principles, work done on charge by electric field = loss in kinetic energy of mass.

So, [tex]QER = MV²/2{/tex} where R is the distance (range) the mass moves before it stops

Therefore {tex}R = MV²/2QE{/tex}

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Explanation:

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6. The momentum of a 30.0 g bird with a speed of 12 m.s-1 is 0.36 kg.m.s-1. What will be its momentum 12s later if a constant .0
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Explanation:

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Consider a uniform horizontal wooden board that acts as a pedestrian bridge. The bridge has a mass of 300 kg and a length of 10
gayaneshka [121]

Answer:

F = 2123.33N

Explanation:

In order to calculate the torque applied by the left support, you take into account that the system is at equilibrium. Then, the resultant of the implied torques are zero.

\Sigma \tau=0

Next, you calculate the resultant of the torques around the right support, by taking into account that the torques are generated by the center of mass of the wooden, the person and the left support. Furthermore, you take into account that torques in a clockwise direction are negative and in counterclockwise are positive.

Then, you obtain the following formula:

-\tau_l+\tau_p+\tau_{cm}=0          (1)

τl: torque produced by the left support

τp: torque produced by the person

τcm: torque produced by the center of mass of the wooden

The torque is given by:

\tau=Fd           (2)

F: force applied

d: distance to the pivot of the torque, in this case, distance to the right support.

You replace the equation (2) into the equation (1) and take into account that the force applied by the person and the center of mass of the wood are the their weight:

-Fd_1+W_pd_2+W_{cm}d_3=0\\\\d_1=6.0m\\\\d_2=2.0m\\\\d_3=3.0m\\\\W_p=(200kg)(9.8m/s^2)=1960N\\\\W_{cm}=(300kg)(9.8m/s^2)=2940N

Where d1, d2 and d3 are distance to the right support.

You solve the equation for F and replace the values of the other parameters:

F=\frac{W_pd_2+W_d_3}{d_1}=\frac{(1960N)(2.0m)+(2940N)(3.0m)}{6.0m}\\\\F=2123.33N

The force applied by the left support is 2123.33 N

8 0
3 years ago
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