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Vanyuwa [196]
3 years ago
11

A 150-N box is being pulled horizontally in a wagon accelerating uniformly at 3.00 m/s2. The box does not move relative to the w

agon, the coefficient of static friction between the box and the wagon's surface is 0.600, and the coefficient of kinetic friction is 0.400. The friction force on this box is closest to
A) 450 N.
B) 90.0 N.
C) 60.0 N.
D) 45.9 N
Physics
1 answer:
Zepler [3.9K]3 years ago
8 0

Answer:

Frictional force, F = 45.9 N

Explanation:

It is given that,

Weight of the box, W = 150 N

Acceleration, a=3\ m/s^2

The coefficient of static friction between the box and the wagon's surface is 0.6 and the coefficient of kinetic friction is 0.4.  

It is mentioned that the box does not move relative to the wagon. The force of friction is equal to the applied force. Let a is the acceleration. So,

m=\dfrac{W}{g}

m=\dfrac{150}{9.8}

m=15.3\ kg

Frictional force is given by :

F=ma

F=15.3\times 3

F = 45.9 N

So, the friction force on this box is closest to 45.9 N. Hence, this is the required solution.

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Answer: (d)

Explanation:

Given

Mass of the first ram m_1=49\ kg

The velocity of this ram is v_1=-7\ m/s

Mass of the second ram m_2=52\ kg

The velocity of this ram v_2=9\ m/s

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Conserving the momentum

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Momentum after the collision will be

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The peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

<h3>Relationship between electric and magnetic field</h3>

The relationship between electric and magnetic field at a given peak electric field is given as;

c = (E₀) / (B₀)

where;

  • c is speed of light
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B₀ = 9.67 x 10⁻¹⁰ T

Thus, the peak magnetic field of the electromagnetic wave in the red part of the visible spectrum is 9.67 x 10⁻¹⁰ T.

Learn more about peak magnetic field here: brainly.com/question/24487261

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Answer:

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