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a_sh-v [17]
2 years ago
5

A 47-kg packing crate is pulled with constant speed across a rough floor with a rope that is at an angle of 37 degrees above the

horizontal. If the tension in the rope is 142 N, how much work is done on the crate to move it 6.1m?
Physics
1 answer:
levacccp [35]2 years ago
8 0

Tension in the rope due to applied force will be given as

F = 142 N

angle of applied force with horizontal is 37 degree

displacement along the floor = 6.1 m

so here we can use the formula of work done

W = F d cos\theta

now we can plug in all values above

W = 142 * 6.1 * cos37

W = 691.8 J

So here work done to pull is given by 691.8 J


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ziro4ka [17]

Answer:

net force

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net force is the total amount of force exerted on an object.

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The magnitude of the voltage induced in a conductor moving through a stationary magnetic field depends on the _______ and the __
Virty [35]

The correct answer is: Option (D) length, speed

Explanation:

According to Faraday's Law of Induction:

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As you can see that ξ (Emf/voltage induction) is directly proportional to the length and the speed of the conductor. Therefore, the correct answer will be Option (D) Length, Speed

3 0
3 years ago
Read 2 more answers
A.
Brut [27]

Answer:

\boxed{\sf Work \ done = 4 \ J}

Given:

Force = 8 N

Distance covered by the body = 50 cm = 0.5 m

Explanation:

Work Done = Force × Distance covered by the body

= 8 × 0.5

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6 0
2 years ago
A charge of -8.5 µC is traveling at a speed of 9.0 106 m/s in a region of space where there is a magnetic field. The angle betwe
Sindrei [870]

Answer:

The magnitude of the magnetic field is 9.3\times 10^{-5}\ T.

Explanation:

Given that,

Charge, q=-8.5\ \mu C=-8.5\times 10^{-6}\ C

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The angle between the velocity of the charge and the field is 56°.

The magnitude of force, F=5.9\times 10^{-3}\ N

We need to find the magnitude of the magnetic field. When a charged particle moves in the magnetic field, the magnetic force is experienced by it. The force is given by :

F=qvB\ \sin\theta

B is the magnetic field.

B=\dfrac{F}{qv\ \sin\theta}\\\\B=\dfrac{5.9\times 10^{-3}}{8.5\times 10^{-6}\times 9\times 10^6\ \sin(56)}\\\\B=9.3\times 10^{-5}\ T

So, the magnitude of the magnetic field is 9.3\times 10^{-5}\ T. Hence, this is the required solution.

4 0
3 years ago
One of the factors that determines the ? of a capacitor is the frequency measured in hertz.
ycow [4]

Answer:

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The reactance of the capacitor tells somehow the "resistance" of the capacitor to the passage of current through it. In fact:

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