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Vlada [557]
3 years ago
15

A large crate of mass m is placed on the back of a truck but not tied down. As the truck accelerates forward with an acceleratio

n a, the crate remains at rest relative to the truck. What force causes the crate to accelerate forward?
a) normal force
b) force of gravity
c) force of friction between the crate and the floor of the truck
d) the "ma" force
e) none of these
Physics
2 answers:
Margaret [11]3 years ago
5 0

Answer: Option C

Explanation: Let's analyze each possible force:

A)normal force, it pushes in the direction in which the box has contact with some surface, in this case, is in the bottom of the truck, so this force pushes upwards.

B) Force of gravity always pushes downwards, so it can not accelerate the box in the same direction in which the truck is moving.

c) The force of friction points in the opposite direction in which the box would move, if the truck moves to the right, the box will move to the left, so the friction pushes also to the right, which is the same direction in which the truck is moving.

D) there is nothing called the "ma" force.

Now, the correct option would be C; the force of friction (the statical one, because we know that the box does not move)

Andrew [12]3 years ago
4 0
"Force of friction between the crate and the floor of the truck" is the one force among the choices given in the question that <span>causes the crate to accelerate forward. The correct option among all the options that are given in the question is the third option or option "c". I hope the answer helps you.</span>
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Do you divide mass by volume to get density
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3 years ago
A cannon, positioned on a hill, shoots a cannonball horizontally at 23 m/s. The cannonball hits the stone wall 1.96 m below the
irina [24]

Answer: 14. 49 m

Explanation:

We can solve this problem with the following equations:

x=V_{o} cos \theta t (1)

y-y_{o}=V_{o} sin \theta t-\frac{1}{2}gt^{2} (2)

Where:

x is the horizontal distance between the cannon and the ball

V_{o}=23 m/s is the cannonball initial velocity

\theta=0\° since the cannonball was shoot horizontally

t is the time

y=0 is the final height of the cannonball

y_{o}=1.96 m is the initial height of the cannonball

g=9.8 m/s^{2} is the acceleration due gravity

Isolating t from (2):

t=\sqrt{-\frac{2(y-y_{o})}{g}} (3)

t=\sqrt{-\frac{2(0 m-1.96 m)}{9.8 m/s^{2}}} (4)

t=0.63 s (5)

Substituting (5) in (1):

x=(23 m/s) cos(0\°) 0.63 s (6)

Finally:

x=14.49 m

5 0
3 years ago
(a) The electric potential due to a point charge is given by V = kq⁄r where q is the charge, r is the distance from q and k = 8.
LiRa [457]

Answer:

a)  [volts] = [N m / C],

b) The lines or surface that has the same potential are called equipotential

c) the equipotential lines must also be perpendicular to the electric field lines

Explanation:

a) find the units of the volt

the electric potential energy is

             V = k q / r

             V = [N m² / C²] C / m

              V = [N m / C]

The electric potential is defined as

             V = E .s

             V = [N / C] [m]

             V = [N m / C] = [volt]

we see that in the two expressions the same result is obtained therefore the volt is

            [volts] = [N m / C]

b) The lines or surface that has the same potential are called equipotential surfaces, the great utility of these lines or surfaces is that a face can be displaced on it without doing work.

c) The electric potential is defined as the gradient of the electric field

             v =  - \frac{dE}{dx} i^

therefore the equipotential lines must also be perpendicular to the electric field lines

4 0
3 years ago
A magnetically soft material is placed in a strong magnetic field. What is the most likely outcome?
Svet_ta [14]

D-It will become a temporary magnet because the domains will easily realign.

9 0
2 years ago
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