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3 years ago

Following an inelastic collision, what can kinetic energy be converted into

Physics

2 answers:

Leona [35]3 years ago

8 0

Kinetic energy can be converted into Vibrational energy.

kobusy [5.1K]3 years ago

6 0

Vibrational energy, which as we know, causes heat!

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What is the acceleration of an object with a mass of

aev [14]

Heya!!

For calculate aceleration, lets applicate second law of Newton:

$\boxed{F=ma}$

<u>Δ Being Δ</u>

F = Force = 183 N

m = Mass = 367 kg

a = Aceleration = ?

⇒ Let's replace according the formula and clear "a":

$\boxed{a=183 \ N / 367\ kg}$

⇒ Resolving

$\boxed{ a = 0.49 \ m/s^{2}}$

Result:

The aceleration is <u>0,49 meters per second squared (m/s²)</u>

Good Luck!!

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3 years ago

Which is not standing in the way of astronomers getting a good view of distant stars? A. some stars are too far away for our tel

Naya [18.7K]

Answer: D. There is a lot of light pollution on earth

Explanation: The light pollution on Earth has nothing to do with the stars in the sky

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3 years ago

A device that measures potential is a(n) circuit.

Darina [25.2K]

Answer:

voltmeter

Explanation:

A voltmeter is the instrument used to measure a potential difference between two points in an electric circuit.

8 0

3 years ago

How much time does it take for an eagle flying at a speed of 50 kilometers per hour to travel a distance of 2000 kilometers?

gtnhenbr [62]

Answer:40 hour

Explanation:

8 0

3 years ago

Suppose you push a hockey puck of mass m across frictionless ice for a time \Delta t, starting from rest, giving thepuck speed v

bazaltina [42]

Answer:

1. $t_2 = 2t_1$

2. $t_2 = t_1\sqrt{2}$

Explanation:

1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.

In Newton's 2nd law: F = ma

where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

$a_1 = 2a_2$

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

$t = v / a_1$

Now that acceleration is halved:

$t = \frac{v}{2a_2}$

$\frac{v}{a_2} = 2t$

You would need to push for twice amount of time $t_2 = 2t_1$

2. The distance traveled by the puck is as the following equation:

$d = at^2$

So if the acceleration is halved while maintaining the same d:

$\frac{d_1}{d_2} = \frac{a_1t_1^2/2}{a_2t_2^2/2}$

As $d_1 = d_2$ , then $d_1/d_2 = 1$ . Also $a_1 = 2a_2$

$1 = \frac{2a_2t_1^2}{a_2t_2^2}$

$t_2^2 = 2t_1^2$

$t_2 = t_1\sqrt{2}\approx 1.14t_1$

So t increased by 1.14

7 0

3 years ago