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Alchen [17]
2 years ago
13

A 2-kg wood block is pulled by a string across a rough horizontal floor. The string exerts a tension force of 30 N on the block

at an angle of 20º above the horizontal. The block moves at constant speed. If the block is pulled for a distance of 3.0 m, how much work is done by the tension force?
Physics
1 answer:
Luden [163]2 years ago
7 0

Answer:

Work done, W = 84.57 Joules

Explanation:

It is given that,

Mass of the wooden block, m = 2 kg

Tension force acting on the string, F = 30 N

Angle made by the block with the horizontal, \theta=20^{\circ}

Distance covered by the block, d = 3 m

Let W is the work done by the tension force. It can be calculated as :

W=F\ cos\theta\times d

W=30\times cos(20)\times 3

W = 84.57 Joules

So, the work done by the tension force is 84.57 Joules. Hence, this is the required solution.

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A pet-store supply truck moves at 25.0 m/s north along a highway. inside, a dog moves at 1.75 m/s at an angle of 35.0° east of
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<u>Answer:</u>

 Velocity of the dog relative to the road = 26.04 m/s 3.15⁰ north of east.

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  Speed of truck = 25 m/s north = 25 j m/s

  Speed of dog = 1.75 m/s at an angle of 35.0° east of north = (1.75 cos 35 i + 1.75 sin 35 j)m/s

                          = (1.43 i + 1.00 j) m/s

    Velocity of the dog relative to the road = 25 j + 1.43 i + 1.00 j = 1.43 i + 26.00 j

    Magnitude of velocity = 26.04 m/s

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At a certain instant after jumping from the airplane A, a skydiver B is in the position shown and has reached a terminal (consta
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A 10-turn coil of wire having a diameter of 1.0 cm and a resistance of 0.50 Ω is in a 1.0 mT magnetic field, with the coil orien
n200080 [17]

Answer:

The voltage across the capacitor is 1.57 V.

Explanation:

Given that,

Number of turns = 10

Diameter = 1.0 cm

Resistance = 0.50 Ω

Capacitor = 1.0μ F

Magnetic field = 1.0 mT

We need to calculate the flux

Using formula of flux

\phi=NBA

Put the value into the formula

\phi=10\times1.0\times10^{-3}\times\pi\times(0.5\times10^{-2})^2

\phi=7.85\times10^{-7}\ Tm^2

We need to calculate the induced emf

Using formula of induced emf

\epsilon=\dfrac{d\phi}{dt}

Put the value into the formula

\epsilon=\dfrac{7.85\times10^{-7}}{dt}

Put the value of emf from ohm's law

\epsilon =IR

IR=\dfrac{7.85\times10^{-7}}{dt}

Idt=\dfrac{7.85\times10^{-7}}{R}

Idt=\dfrac{7.85\times10^{-7}}{0.50}

Idt=0.00000157=1.57\times10^{-6}\ C

We know that,

Idt=dq

dq=1.57\times10^{-6}\ C

We need to calculate the voltage across the capacitor

Using formula of charge

dq=C dV

dV=\dfrac{dq}{C}

Put the value into the formula

dV=\dfrac{1.57\times10^{-6}}{1.0\times10^{-6}}

dV=1.57\ V

Hence, The voltage across the capacitor is 1.57 V.

5 0
2 years ago
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