Ask question

Ask question

All categories

2 years ago

13

Friction is a (1)______

1 answer:

spin [16.1K]2 years ago

7 0

Answer:

friction is motion heat that collect hot vibration

You might be interested in

Please any one need help on this

Romashka [77]

-- The wavelength and the amplitude were described in my answer to your previous question.

-- A "compression" is a place where the wave is <em>compressed</em>. It's the darker section of the wave in the picture, where the wavelength is temporarily shorter, so several waves are all bunched up (compressed) in a small time.

-- A "rarefaction" is exactly the opposite of a "compression". It's a place where the wave gets more "<em>rare</em>" ... the wavelength temporarily gets longer, so that several waves get stretched out, and there are fewer of them in some period of time. The arrow in the picture points to a rarefaction.

3 0

2 years ago

The circumference of the earth is 40,000 km. The distance from the

Finger [1]

Answer:

The distance from the North Pole to the equator is $1*10^{7}$ m.

Explanation:

Circumference of Earth = 40,000 km ......................(1)

Distance from the North Pole to the Equator is = 1/4th of the Circumference of Earth ...................... (2)

Let Distance from the North Pole to the Equator be d ,

the equation formed will be ,

d = 1/4 * Circumference of Earth ........(3)......... ( from equation 1 )

put the value of Circumference of Earth in equation (3),

d = 1/4 * 40,000 km

d = 10,000 km

converting km to m ,

d = 10,000 * $10^{3}$ m

d = 1 * $10^{7}$ m

The distance from the North Pole to the equator is $1*10^{7}$ m.

7 0

2 years ago

If an object force of 50 N is used to move an object a distance of 20 m, what distance must the object be moved if the input for

steposvetlana [31]

Answer:

$\ d_{out} = 100 \ m.$

Explanation:

Given data:

$F_{in} = 50 \ \rm N$

$F_{out} = 10 \ \rm N$

$d_{in} = 20 \ m$

Let the distance traveled by the object in the second case be $d_{out}.$

In the given problem, work done by the forces are same in both the cases.

Thus,

$W_{in} = W_{out}$

$F_{in}.d_{in} = F_{out}.d_{out}$

$\Rightarrow \ d_{out} = \frac{F_{in}.d_{in}}{F_{out}}$

$\ d_{out} = \frac{50 \times 20}{10}$

$\ d_{out} = 100 \ m.$

5 0

2 years ago

Q. A car is traveling

Norma-Jean [14]

Answer:

I don't even know can anybody help this person

6 0

3 years ago

If a 54 kg sprinter can accelerate from a standing start to a speed of 10 m/s in 3 s, what average power is generated?

lora16 [44]

Answer:

Power, P = 600 watts

Explanation:

It is given that,

Mass of sprinter, m = 54 kg

Speed, v = 10 m/s

Time taken, t = 3 s

We need to find the average power generated. The work done divided by time taken is called power generated by the sprinter i.e.

$P=\dfrac{W}{t}$

Work done is equal to the change in kinetic energy of the sprinter.

$P=\dfrac{\dfrac{1}{2}mv^2}{t}$

$P=\dfrac{\dfrac{1}{2}\times 54\ kg\times (10\ m/s)^2}{3\ s}$

P = 900 watts

So, the average power generated by the sprinter is 900 watts. Hence, this is the required solution.

3 0

2 years ago