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marishachu [46]
4 years ago
10

A particle that carries a net charge of -41.8 μc is held in a region of constant, uniform electric field. the electric field vec

tor is oriented 55.2° clockwise from the vertical axis, as shown. if the magnitude of the electric field is 4.82 n/c, how much work is done by the electric field as the particle is made to move a distance of d = 0.556 m straight up?
Physics
1 answer:
miss Akunina [59]4 years ago
3 0
The total work done by the electric field on the charge is given by the scalar product between the electric force acting on the charge and the displacement of the charge:
W=F d cos \theta
where the force is F=qE, d=0.556 and \theta=55.2^{\circ}. Using the value of q and E given by the problem, we find
W=qEdcos\theta = 6.39\cdot10^{-5}J
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A horizontal 810-N merry-go-round of radius 1.60 m is started from rest by a constant horizontal force of 55 N applied tangentia
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Answer:

576 joules

Explanation:

From the question we are given the following:

weight = 810 N

radius (r) = 1.6 m

horizontal force (F) = 55 N

time (t) = 4 s

acceleration due to gravity (g) = 9.8 m/s^{2}

K.E = 0.5 x MI x ω^{2}

where MI is the moment of inertia and ω is the angular velocity

MI = 0.5 x m x r^2

mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg

MI = 0.5 x 82.65 x 1.6^{2}

MI = 105.8 kg.m^{2}

angular velocity (ω) = a x t

angular acceleration (a) = torque ÷ MI

where torque = F x r = 55 x 1.6 = 88 N.m

a= 88 ÷ 105.8 = 0.83 rad /s^{2}

therefore

angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s

K.E = 0.5 x MI x ω^{2}

K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules

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3 years ago
A billiard ball is moving in the x-direction at 30.0 cm/s and strikes another billiard ball moving in the y-direction at 40.0 cm
____ [38]

Answer:

53.13 °

Explanation:

In order to do this, we just need to apply the following:

tanα = Dy/Dx

Where:

Vy: speed of the ball in the y axis.

Vx: speed of the ball in the x axis.

At this point we do not need the speed of the first ball after the collision because in that moment is already heading in the direction that we are looking for. Therefore, we just need to use the innitial data to calculate the direction which the first ball will go.

According to this, then:

tanα = (40/30)

tanα = 1.3333

α = tan⁻¹(1.3333)

<h2>α = 53.13°</h2>

This means that the final direction of the first ball is 53.13° and in the x axis because the starting momentum of this ball in the x axis has not dissapeared.

Hope this helps

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