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torisob [31]
3 years ago
13

What is the net force on a truck if the force of friction is 31 N and the force of the engine is 79 N?

Physics
1 answer:
NeTakaya3 years ago
6 0

Answer:  C) 48 N forward

===================================================

Explanation:

Subtract the two forces, since they are working opposite one another.

79-31 = 48

The net force is 48 newtons forward. The truck is moving forward since the engine's force (79 N) is larger than the force of friction (31 N).

The force of friction cannot exceed the engine's force because friction only slows down an object, and does not push the object backward. In other words, if the object was stationary, then the ground isn't pushing the object backwards assuming you're on a level flat ground.

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Answer:

0.075 T

Explanation:

When a current-carrying wire is immersed in a region with magnetic field, the wire experiences a force, given by

F=ILB sin \theta

where

I is the current in the wire

L is the length of the wire

B is the strength of the magnetic field

\theta is the angle between the direction of I and B

In this problem we have:

L = 0.65 m is the length of the wire

I = 8.2 A is the current in the wire

F = 0.40 N is the force experienced by the wire

\theta=90^{\circ} since the current is at right angle with the magnetic field

Solving the formula for B, we find the strength of the magnetic field:

B=\frac{F}{IL sin \theta}=\frac{0.40}{(8.2)(0.65)(sin 90^{\circ})}=0.075 T

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Select the correct answer. Which of the following causes air pollution inside the house? A. Smoking cigarettes B. Growing housep
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2 years ago
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A rock, initially at rest with respect to Earth and located an infinite distance away is released and accelerates toward Earth.
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the rock speed is increased

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The spring in the muzzle of a child's spring gun has a spring constant of 730 N/m. To shoot a ball from the gun, first the sprin
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Answer:

a. V=11.84 m/s

b.x=0.052m

Explanation:

a).

Given

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v_f^2=v_i^2+2*g*h

v_i^2=2*g*h=2*9.8m/s^2*1.9m

v_i=\sqrt{2*9.8m/s^2*1.9m}=\sqrt{37.24 m^2/s^2}

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V=\frac{v_i}{sin(31)}=\frac{6.1m/s}{sin(31)}

V=11.84 m/s

b).

K_k=\frac{1}{2}*K*x^2

No friction on the ball so:

x^2=\frac{2*K_k}{K}

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