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quester [9]
2 years ago
6

As shown in the figure, a given force is applied to a rod in several different ways. In which case is the torque about the pivot

P due to this force the smallest?

Physics
1 answer:
LekaFEV [45]2 years ago
3 0

Answer:

OPTION 5

Explanation:

As the force has no moment arm, it creates no torque about point P.

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A ball moving with an initial velocity of 5 m/s comes to rest after 2s. What was the ball's acceleration?
Inga [223]

Answer:

-2.5m/s²

Explanation:

The acceleration of a body is giving by the rate of change of the body's velocity. It is given by

a = Δv / t        ----------------(i)

Where;

a = acceleration (measured in m/s²)

Δv = change in velocity = final velocity - initial velocity   (measure in m/s)

t = time taken for the change (measured in seconds(s))

From the question;

i. initial velocity = 5m/s

final velocity = 0 [since the body (ball) comes to rest]

Δv = 0 - 5 = -5m/s

ii. time taken = t = 2s

<em>Substitute these values into equation (i) as follows;</em>

a = (-5m/s) / (2s)

a = -2.5m/s²

Therefore, the acceleration of the ball is -2.5m/s²

NB: The negative sign shows that the ball was actually decelerating.

6 0
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An experimental apparatus has two parallel horizontal metal rails separated by 1.0 m. A 3.0 Ω resistor is connected from the lef
Blizzard [7]

Answer:

The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

Explanation:

Given that,

Distance = 1.0 m

Resistance = 3.0 Ω

Speed = 35 m/s

Angle = 53°

Magnetic field B=5.0\times10^{-5}\ T

(a). We need to calculate the induced emf

Using formula of emf

E = Blv\sin\theta

Where, B = magnetic field

l = length

v = velocity

Put the value into the formula

E=5.0\times10^{-5}\times1.0\times35\sin53^{\circ}

E=1.398\times10^{-3}\ V

We need to calculate the induced current

E =IR

I=\dfrac{E}{R}

Put the value into the formula

I=\dfrac{1.398\times10^{-3}}{3.0}

I=0.5\ mA

(b). We need to calculate the power dissipated through the resistor

Using formula of power

P=I^2 R

Put the value into the formula

P=(0.5\times10^{-3})^2\times3.0

P=7.5\times10^{-7}\ Watt

Hence, The induced current and the power dissipated through the resistor are 0.5 mA and 7.5\times10^{-7}\ Watt.

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Answer:

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