All categories

3 years ago

What are two factors affecting friction? How do they affect friction? ch

Physics

1 answer:

AlexFokin [52]3 years ago

5 0

The roughness of the surface, the mass of the object, and the area of contact.

You might be interested in

A 730-N man stands in the middle of a frozen pond of radius 5.0 m. He is unable to get to the other side because of a lack of fr

STatiana [176]

Answer:62.11 s

Explanation:

Given

weight of man =730 N

mass of man $m_1=\frac{730}{9.8}=74.48 kg$

radius of pound r=5 m

mass of book $m_2=1.2 kg$

velocity of book $v_2=5 m/s$

let $v_1$ be the velocity of man

conserving momentum

$0=m_1v_1+m_2v_2$

$0=74.48\times v_1+1.2\times 5$

$v_1=\frac{6}{74.48}=0.0805 m/s$

time taken by man to reach south end

$t=\frac{5}{0.0805}=62.11 s$

5 0

3 years ago

When the amplitudes of waves are equal, which frequency waves have the most energy?

Molodets [167]

Answer:

higher frequency waves will have more energy

Explanation:

5 0

2 years ago

A quantity that is fully described by magnitude alone is a ___________ quantity. A quantity that is fully described by both magn

deff fn [24]

Answer:

Scalar quantity

Vector quantity

Explanation:

A scalar quantity is a quantity that is fully described by magnitude alone. Examples include; mass, temperature etc

A vector quantity is described by both magnitude and direction. E.g force, weight etc

7 0

2 years ago

One end of a thin rod is attached to a pivot, about which it can rotate without friction. Air resistance is absent. The rod has

Scrat [10]

Answer:

5.24 m/s

Explanation:

So for the rod to be able to rise upward to the straight up position, the kinetic energy caused by linear speed v0 must be just enough to convert into the potential energy.

Since the rod is uniform in mass, we can treat the body as 1 point, at its center of mass, or geometric center, aka 0.35 / 2 = 0.175 m from the pivot.

For the rod to swing from bottom to top, the center must have moved a distance of h = 0.175 * 2 = 0.35 m, vertically speaking.

Since we neglect friction and air resistance, according to the law of energy conservation then:

$E_k = E_p$

$mv^2/2 = mgh$

Where v is the speed at the center of mass, g = 9.81 m/s2 is the gravitational acceleration, and m is the mass. We can divide both sides by m

$v^2 = 2gh = 2*9.81*0.35 = 6.867$

$v = \sqrt{6.867} = 2.62 m/s$

As this is only the speed at the center of mass, the speed at the bottom end would be different, to calculate this, we need to find the common angular speed:

$\omega = v / r = 2.62 / 0.175 = 14.97 rad/s$