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3 years ago

8

Give at least 4 human activities which demonstrate the involvement of inertial frame of reference

1 answer:

Sergio039 [100]3 years ago

7 0

One clearcut example of an inertial reference frame is an isolated spaceship, far, far away from the Earth, the Sun, the Milky Way Galaxy, and all other massive objects. Fred places a blue ball into a claw at the left end of the ship, and red ball into a claw at the right end of the ship.

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An iron block of mass 10 kg rests on a wooden plane inclined at 30° to the horizontal. It is found

Kaylis [27]

I assume the 100 N force is a pulling force directed up the incline.

The net forces on the block acting parallel and perpendicular to the incline are

∑ F[para] = 100 N - F[friction] = 0

∑ F[perp] = F[normal] - mg cos(30°) = 0

The friction in this case is the maximum static friction - the block is held at rest by static friction, and a minimum 100 N force is required to get the block to start sliding up the incline.

Then

F[friction] = 100 N

F[normal] = mg cos(30°) = (10 kg) (9.8 m/s²) cos(30°) ≈ 84.9 N

If µ is the coefficient of static friction, then

F[friction] = µ F[normal]

⇒ µ = (100 N) / (84.9 N) ≈ 1.2

5 0

2 years ago

A bowling ball of mass 5.8 kg moves in a straight line at 4.34 m/s How fast must a Ping-Pong ball of mass 2.214 g move in a stra

lilavasa [31]

Answer: 11369.46 m/s

Explanation:

We have the following data:

$m_{1}=5.8 kg$ is the mass of the bowling ball

$V_{1}=4.34 m/s$ is the velocity of the bowling ball

$m_{2}=2.214 g \frac{1 kg}{1000 g}=0.002214 kg$ is the mass of the ping-pong ball

$V_{2}$ is the velocity of the ping-pong ball

Now, the momentum $p_{1}$ of the bowling ball is:

$p_{1}=m_{1}V_{1}$ (1)

$p_{1}=(5.8 kg)(4.34 m/s)$

$p_{1}=25.172 kg m/s$ (2)

And the momentum $p_{2}$ of the ping-pong ball is:

$p_{2}=m_{2}V_{2}$ (3)

If the momentum of the bowling ball is equal to the momentum of the ping-pong ball:

$p_{1}=p_{2}$ (4)

$m_{1}V_{1}=m_{2}V_{2}$ (5)

Isolating $V_{2}$ :

$V_{2}=\frac{m_{1}V_{1}}{m_{2}}$ (6)

$V_{2}=\frac{25.172 kg m/s}{0.002214 kg}$ (7)

Finally:

$V_{2}=11369.46 m/s$

6 0

2 years ago

As the bright sun shines upon the water, the water slowly disappears. The same sunlight gives energy to the surrounding plans to

mariarad [96]

first is physical .... sec chem

8 0

3 years ago

Perform the following calculations and give your answer with the correct number of significant figures

love history [14]

Answer:

see below

Explanation:

a. 0.1886 x 12 =2.2632

This has 2 sig figures so the answer can only have 2 sig figures

2.3

b. 2.995 - 0.16685 =2.82815

The most accurate in the problem is to thousands place so our answer can only be accurate to the thousands place

2.828

c. 910 x 0.18945=172.3995

The least number of significant figures is 3 so the answer can only have 3 significant figures

172

3 0

2 years ago

A man lifts a 25.9 kg bucket from a well and does 5.92 kJ of work.The acceleration of gravity is 9.8 m/s^2 . How deep is the wel

Jet001 [13]

Answer:

The well is 23.3 m

Explanation:

As the bucket is lifted out of the well, energy in the man is being transferred to the bucket as gravitational potential energy.

Work done against gravity = mass * height * acceleration due to to gravity

W = mgh

5 920 J = 25.9 kg * h * 9.8 m/s²

h = 23.3 m

4 0

3 years ago