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anygoal [31]
3 years ago
8

At the beginning of a unit on forces, Ms. Alton is leading a class discussion asking her students

Physics
1 answer:
VMariaS [17]3 years ago
6 0

Answer:

(iv), (v), (vi) would be incorrect.

Explanation:

(iv) Force isn't transferred from one colliding object to another, but momentum can be.

(v) An object doesn't stop immediately a force stops acting on it. Think of a thrown ball.

(vi) For an object not to move, it means that the net force on the object is zero, and not necessarily that there are no forces acting on the object. For example, an object could be pushed on one side, and be pushed on the other side with an equal force in the opposite direction. The forces would cancel each other and the net force would be zero.

The rest should be correct.

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How long does it take a vehicle to reach a velocity of 32 m/s if it accelerates from rest at a rate of 4.2 m/s^2?
Marta_Voda [28]

Answer:

7.62

Explanation:

because you have to divide 32/4.2

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3 years ago
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Artyom0805 [142]
It's the angle made by the incident ray when it's perpendicular to the surface. (Perpendicular lines are the lines that form a graph or like a 90-degree angle)
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3 years ago
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Which of the following examples illustrates static friction?
vivado [14]

Answer:

A box sits stationary  on a ramp

Explanation:

Static friction is a force which keeps an object at rest as it is in the case of the box. It has to be overcome for the object to be set into motion.

Static force of friction is calculated as follows:

F= μη

F is static force of friction.

μ is the coefficient of static friction.

η is the normal force.

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4 years ago
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TiliK225 [7]

Explanation:

The answer is:

A squirrel runs up the trunk of a tree.

3 0
3 years ago
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A wad of clay of mass m1 = 0.49 kg with an initial horizontal velocity v1 = 1.89 m/s hits and adheres to the massless rigid bar
notka56 [123]

Answer:

<h2>The angular velocity just after collision is given as</h2><h2>\omega = 0.23 rad/s</h2><h2>At the time of collision the hinge point will exert net external force on it so linear momentum is not conserved</h2>

Explanation:

As per given figure we know that there is no external torque about hinge point on the system of given mass

So here we will have

L_i = L_f

now we can say

m_1v_1\frac{L}{2} = (m_2L^2 + m_1(\frac{L}{2})^2)\omega

so we will have

0.49(1.89)(0.45) = (2.13(0.90)^2 + 0.49(0.45)^2)\omega

\omega = 0.23 rad/s

Linear momentum of the system is not conserved because at the time of collision the hinge point will exert net external force on the system of mass

So we can use angular momentum conservation about the hinge point

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3 years ago
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