What are the two factors that affect the frictional force between two surfaces?

After hitting the spring, the block is bounced back up the ramp. The maximum compression of the spring is Δx=0.03m, and the spri

Lyrx [107]

Answer:

v = 1.28 m/s

Explanation:

Given that,

Maximum compression of the spring, $\Delta x=0.03\ m$

Spring constant, k = 800 N/m

Mass of the block, m = 0.2 kg

To find,

The velocity of the block when it first reaches a height of 0.1 m above the ground on the ramp.

Solution,

When the block is bounced back up the ramp, the total energy of the system remains conserved. Let v is the velocity of the block such that,

Initial energy = Final energy

$\dfrac{1}{2}kx^2=mgh+\dfrac{1}{2}mv^2$

Substituting all the values in above equation,

$\dfrac{1}{2}\times 800\times 0.03^2=0.2\times 9.8\times 0.1+\dfrac{1}{2}\times 0.2\times v^2$

v = 1.28 m/s

Therefore the velocity of block when it first reaches a height of 0.1 m above the ground on the ramp is 1.28 m/s.

5 0

3 years ago

A disk has a radius of 30 cm and a mass of 0.3 kg and is turning at 3.0 rev/s. A trickle of sand falls onto the disk at a distan

Pani-rosa [81]

Answer:

The mass of the sand that will fall on the disk to decrease the is 0.3375 kg

Explanation:

Moment before = Moment after

$I \omega_i = I \omega_f +mr^2 \omega_f\\\\mr^2 \omega_f = I \omega_i - I \omega_f \\\\m = \frac{ I \omega_i - I \omega_f}{r^2 \omega_f }$

where;

I is moment of inertia = Mr² = 0.3 x (0.3)² = 0.027 kg.m²

substitute this in the above equation;

$m = \frac{ 0.027[3(2 \pi) - 2(2 \pi)]} {0.2^2 * 6\pi } = \frac{ 0.027[6 \pi - 4\pi]} {0.2^2 * 4\pi }\\\\m = 0.3375kg$

Therefore, the mass of the sand that will fall on the disk to decrease the is 0.3375 kg

7 0

3 years ago

The body weighing 2 kg moves through the horizontal surface and crosses the path x = 75 cm The coefficient of friction of the bo

Taya2010 [7]

The kinetic energy of the body in definitive position is 4.24 J.

Explanation:

As per the work energy theorem, the work done on any system or object to move it from one position to another is equal to the change in kinetic energy of the object. In this case, the body weighing 2 kg is moved over an horizontal surface for a distance of 75 cm. As there will be frictional force acting on the body while moving over the surface. This frictional force multiplied by the distance the object is moved will give the work done on the body.

Frictional force = Coeffficent of friction × Normal force.

As the weight of the body is 2 kg, the normal force acting on it will be mass multiplied with acceleration due to gravity.

Frictional force = - 0.8×9.8 × 2 =-15.68 N

So the work done will be the product of frictional force with the displacement of 75 cm or 0.75 m.

Work done = Frictional force × Displacement

Work done = -15.68×0.75 = -11.76 J.

So the work is done by the object.

If the kinetic energy of the body at starting is 16 J, then the kinetic energy of the body at definitive position will be obtained as below.

Work done = change in kinetic energy

-11.76 J = Final kinetic energy-16 J

Final Kinetic energy = - 11.76+16

Final kinetic energy = 4.24 J

Thus, the kinetic energy of the body in definitive position is 4.24 J.

3 0

3 years ago

Strong evidence for the existence of dark matter comes from observations of

adoni [48]

Answer:

NASA's Chandra X-ray Observatory and other telescopes such as the Hubble telescope.

Explanation:

Dark matter and normal matter have been wrenched apart by the tremendous collision of two large clusters of galaxies.

The above observations have provided the strongest evidence yet that most of the matter in the universe is dark.

5 0

3 years ago

Why do we say that a baseball falls toward Earth, and not Earth toward the baseball?

mezya [45]

Probably because the earth is stationary rotating on its axis whereas the baseball is the one doing all of the movement after being thrown

4 0

3 years ago