3 years ago

according to newton's first law of motion what is the reason for a ball throwing up in the airfall back to earth

1 answer:

5 0

The force of earth's gravitational field is always directed downwards (towards the center of the earth. When the ball is thrown up, it is going against the earth's gravitational field and so, the earth's gravitational force pulls it back down, accelerating it downwards.

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Can anyone help me in this question?

Anna71 [15]

Answer:

I guess the answer is charging by friction

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3 years ago

A cow is walking at a velocity of 4 m/s and speeds up to 6 m/s in 5 seconds. Calculate the acceleration of the cow. Round your a

siniylev [52]

Answer:

0.4 m/s²

Explanation:

average acceleration = (final velocity - initial velocity)/ time elapsed

average acceleration = (6 - 4)/5

average acceleration = 2/5

average acceleration = 0.4 meters/seconds squared or 0.4m/s²

3 0

2 years ago

The mass of a sports car is 1000 kg. The shape of the car is such that the aerodynamic drag coefficient is 0.260 and the frontal

alisha [4.7K]

Answer:

The initial acceleration is 0,221 m/s^2.

Explanation:

$Fcar = Fairfriction\\m*a = 0.5*A*c*density*v^2\\a = 0.5*A*c*density*v^2 / 1000 kg = 0.5*2.10m^2*0.260*1.295kg/m^3*(25 m/s)^2 / 1000 kg = 0,221 m/s^2$

8 0

3 years ago

Read 2 more answers

Darius' boat sails into the harbor with a speed of 80m/s. After 20 seconds, Darius' boat has come to a stop at the dock. What is

Novosadov [1.4K]

Answer:

The boat's acceleration is 4 m/s²

Explanation:

This question seeks to test the knowledge of acceleration and how to calculate acceleration when the speed is provided. Hence, the formula for acceleration would be used here.

Acceleration (m/s²) = speed (in m/s) ÷ time (in seconds)

Acceleration = 80 m/s ÷ 20 secs

Acceleration = 4 m/s² or 4 ms⁻²

The boat's acceleration is 4 m/s²

4 0

2 years ago

A 1200 kg car traveling east at 4.5 m/s crashes into the side of a 2100 kg truck that is not moving. During the collision, the v

Rasek [7]

Answer:

Explanation:

This is a simple Law of Momentum Conservation problem of the inelastic type. The equation for this is

$[m_1v_1+m_2v_2]_b=[(m_1+m_2)v]_a$ Filling in:

$[1200(4.5)+2100(0)]=[(1200+2100)v]$ which simplifies to

5400 + 0 = 3300v

so v = 1.6 m/s to the east, choice B

6 0

2 years ago