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Marina CMI [18]

3 years ago

11

If a ball is given an acceleration of 3.0 meters/second^2 while being pushed with a force of 0.75 newtons, what is the mass of t

he ball?

2 answers:

Anit [1.1K]3 years ago

8 0

F=ma is one form of Newton's second law
m=F/a
m=(3/4)x1/3
m=9/4=2.25kg

yawa3891 [41]3 years ago

7 0

F = m.a

m = F/a

m = 0.75/3

m = 0.25 Kg OR 250 g

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What changes a ball velocity

Grace [21]

Answer/Explanation: Speed and direction can change with time. When you throw a ball into the air, it leaves your hand at a certain speed. As the ball rises, it slows down. Then, as the ball falls back toward the ground, it speeds up again. When the ball hits the ground, its direction of motion changes and it bounces back up into the air.

5 0

4 years ago

A 1300 kg car traveling at 35 mph rear-ends a 1000 kg car traveling 25 mph. Just after the collision (but before the driver’s sl

guajiro [1.7K]

Answer

given,

before collision

mass of car A = m_a = 1300 kg

velocity of car A = v_a = 35 mph

mass of car B = m_b= 1000 kg

velocity of car B = v_b = 25 mph

after collision

V_a = 30 mph

V_b = 31.5 mph

Initial momentum

$P_1 = m_av_a + m_b v_b$

$P_1 = 1300 \times 35+ 1000 \times 25$

$P_1 =70500 Kg.m/s$

final momentum

$P_2 = m_aV_a + m_b V_b$

$P_2 = 1300 \times 30+ 1000 \times 31.5$

$P_2 =70500 Kg.m/s$

here initial momentum is equal to the final momentum of the car.

hence, momentum is conserved in the collision.

6 0

3 years ago

The first artificial satellite to orbit the Earth was Sputnik I, launched October 4, 1957. The mass of Sputnik I was 83.5 kg, an

9966 [12]

Answer:

$-4.941*10^8J.$

Explanation:

To solve this exercise it is necessary to take into account the concepts related to gravitational potential energy, as well as the concept of perigee and apogee of a celestial body.

By conservation of energy we know that,

$\Delta U = \Delta_{perogee}-\Delta_{Apogee}$

Where,

$U= \frac{-GmM_e}{r}$

Replacing

$\Delta U = \frac{-GmM_e}{r_p}- \frac{-GmM_e}{r_a}$

$\Delta U = GmM_e (\frac{1}{r_A}-\frac{1}{r_p})$

Our values are given by,

$m = 85.5Kg$

$M_e = 5.97*10^{24}Kg$

$r_A = 7330Km$

$r_p = 6610Km$

$G = 6.67*10^{-11}Nm^2/Kg^2$

Replacing at the equation,

$\Delta U = (6.67*10^{-11})(85.5)(5.97*10^{24}) (\frac{1}{7330}-\frac{1}{6610})$

$\Delta U = -4.941*10^8J$

Therefore the Energy necessary for Sputnik I as it moved from apogee to perigee was $-4.941*10^8J.$

4 0

3 years ago

A 87 kg man has a total mechanical energy of 1780 J .If he is swinging downward and is currently 1.4 m above the ground, what is

borishaifa [10]

Answer:

6.4m/s

Explanation:

The total mechanical energy of the man is 1780J.

This mechanical energy is the energy due to the motion of the body and it is a form of kinetic energy.

Also, mass = 87kg

Kinetic energy = $\frac{1}{2}$ m v²

m is the mass

v is the velocity

1780 = $\frac{1}{2}$ x 87 x v²

v² = 40.9

v = 6.4m/s

4 0

3 years ago

If a 340 N girl sits on a seesaw and is lifted 1.4 m in 1.6 s, how much work was done by the child on the other side?

blondinia [14]

In physics, work is defined as the total energy when an object is moved to a certain displacement by the application of external force. It is calculated by the expression W = Fd. For this case, the displacement is apparently zero, then there is no work in the system above.

4 0

3 years ago