All categories

3 years ago

Is the momentum of a 0.1 kg Mass moving with a velocity of 5 miles per second West

Physics

2 answers:

Vsevolod [243]3 years ago

8 0

Momentum = mass times velocity

0.1 kg x 5 miles per second West

= 804.7 kg-meters/sec West .

dem82 [27]3 years ago

6 0

Momentum equals Mass x Velocity
Mass equals 0.1kg
Velocity equals 5m/s

So the momentum has to = 0.1 x 5 = 0.5kgm/s

I hope this helped

You might be interested in

Which best describes the runners

qaws [65]

Answer:

The answer is

Explanation:

Daniela had a 5 meter head start, and Leonard caught up to her at 25 meters.

brainiest would be cool

3 0

3 years ago

An astronaut has a mass of 50.0 kg on earth. what is her mass on the moon, where gravity is 1 6 that on earth?

never [62]

I believe the correct gravity on the moon is 1/6 of Earth. Take note there is a difference between 1 6 and 1/6.

HOWEVER, we should realize that the trick here is that the question asks about the MASS of the astronaut and not his weight. Mass is an inherent property of an object, it is unaffected by external factors such as gravity. What will change as the astronaut moves from Earth to the moon is his weight, which has the formula: weight = mass times gravity.

Therefore if he has a mass of 50 kg on Earth, then he will also have a mass of 50 kg on moon.

6 0

4 years ago

The lower the angle of the slope, ________ the acceleration along the ramp, therefore, the speed at the bottom of a slope will b

pogonyaev

Answer:

Lower

gsintheta (gsinθ)

Explanation:

The sum of forces resolved parallel to the inclined plane is given by;

F - mgsinθ = 0

ma - mgsinθ = 0

ma = mgsinθ

a = gsinθ

Acceleration is proportional to angle of inclination, thus the lower the angle of the slope, lower the acceleration along the ramp.

therefore, the speed at the bottom of a slope will be lower, (velocity is directly proportional to acceleration) and, consequently, the control will be better.

The acceleration along the ramp, is gsintheta (gsinθ)

3 0

4 years ago

A 3.91 kg cart is moving at 5.7 m/s when it collides with a 4 kg cart which was at rest. They collide and stick together.

Nesterboy [21]

Answer:

The velocity after the collision is 2.82 m/s

Explanation:

Law Of Conservation Of Linear Momentum

It states the total momentum of a system of bodies is conserved unless an external force is applied to it. The formula for the momentum of a body with mass m and speed v is

P=mv.

If we have a system of two bodies, then the total momentum is the sum of the individual momentums:

$P=m_1v_1+m_2v_2$

If a collision occurs and the velocities change to v', the final momentum is:

$P'=m_1v'_1+m_2v'_2$

Since the total momentum is conserved, then:

P = P'

Or, equivalently:

$m_1v_1+m_2v_2=m_1v'_1+m_2v'_2$

If both masses stick together after the collision at a common speed v', then:

$m_1v_1+m_2v_2=(m_1+m_2)v'$

The common velocity after this situation is:

$\displaystyle v'=\frac{m_1v_1+m_2v_2}{m_1+m_2}$

There is an m1=3.91 kg car moving at v1=5.7 m/s that collides with an m2=4 kg cart that was at rest v2=0.

After the collision, both cars stick together. Let's compute the common speed after that:

$\displaystyle v'=\frac{3.91*5.7+4*0}{3.91+4}$

$\displaystyle v'=\frac{22.287}{7.91}$

$\boxed{v' = 2.82\ m/s}$

The velocity after the collision is 2.82 m/s

6 0

3 years ago

A swift blow with the hand can break a pine board. As the hand hits the board, the kinetic energy of the hand is transformed int

Stells [14]

Answer:

A. The hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand = 1025 N

Explanation:

A.To determine the speed the hand must move with to break the board, the force workdone in breaking the board is found first.

Workdone = force × distance

Minimum force required = 870 N;

Distance moved by board/Deflection in order to break = 1.4 cm = 0.014 M

Workd done = 870 N × 0.014 m = 12.18 Nm or 12.18 J

This work done = Kinetic energy of the hand

Kinetic energy = mv²/2 ; where m is mass and v is velocity

Mass of hand = 0.50 Kg, velocity = ?, K.E. = 12.18 J

v² = 2 KE/m

v = √2KE/m

v = √(2 × 12.18/0.50)

v = 6.98 m/s

Therefore, the hand must move with a velocity of 6.98 m/s to break the board.

B. Average force on the hand

This can be determined using the equation of motion, v² = u² + 2as to find acceleration, since force = mass × acceleration

From the equation of motion, a = v² - u²/2s

At rest, v = 0, u = 6.98, s = 1.2 cm = 0.012 m

a = 0² - 6.98²/ 2 × 0.012

a = -2030 m/s²

Force = 2030 m/s² × 0. 50 kg = 1015 N

Therefore, Average force on the hand = 1025 N

3 0

3 years ago