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3 years ago

A 125 W motor accelerates a block along a level, frictionless surface at an average speed of 5.0

Physics

1 answer:

Lelu [443]3 years ago

3 0

Answer:

25N

Explanation:

Given parameters:

Power of the motor = 125W

Average speed = 5m/s

Unknown:

Force supplied to the motor = ?

Solution:

Work done is the force applied to move a body through a particular distance;

Work done = Force x distance

Also,

Work done = Power x time

So;

Force x distance = Power x time

since force is the unknown;

Force = $\frac{Power x time}{distance}$

Speed = $\frac{distance }{time}$

Force = $\frac{Power}{speed}$

Now solve;

Force = $\frac{125}{5}$ = 25N

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If a charge travels through a magnetic field, it experiences a magnetic force and its velocity is perpendicular to the direction

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Answer:

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Explanation:

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3 years ago

The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

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Neporo4naja [7]

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Metals are able to conduct electric currents due to the free movement of negatively charged particles i.e., electrons, across the conductor.

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Serggg [28]

Explanation:

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