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3 years ago

A 50.0-g ball traveling at 25.0 m/s bounces off a brick wall and

Physics

1 answer:

liubo4ka [24]3 years ago

4 0

Answer:

13,400 m/s²

Explanation:

Average acceleration is the change in velocity over time:

a = Δv / t

a = (22.0 m/s − (-25.0 m/s)) / 0.00350 s

a = 13,400 m/s²

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Explain why the boiling point of a liquid varies with atmospheric pressure

OLEGan [10]

The boiling point of water, or any liquid, varies according to the surrounding atmospheric pressure. A liquid boils, or begins turning to vapor, when its internal vapor pressure equals the atmospheric pressure.

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3 years ago

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Answer:

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3 years ago

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DochEvi [55]

Answer:

Explanation:

A novae in astronomy means an explosion in the white dwarf star which had tapped enough gas from a companion star,hence it releases an incredible amount of energy which is Over a million times brighter than it normal stars.

A super novae on the other hand is a cosmic explosion that can be a billion times brighter than the normal.

From this one can see that a perculiar similarity between a novae and super novae is that both generate huge explosion and bright Ness, and a major difference is super novae release huge amount of brightness and energy more than the novae

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3 years ago

A quarter is flipped from a height of 1.45 m above the ground. How much time will it take to reach the ground if the person flip

Artemon [7]

It will take the quarter 0.151 seconds to reach the ground.

<u>Given the following data:</u>

Height = 1.45 meters

Initial velocity = 10.32 m/s

We know that acceleration due to gravity (a) for an object is equal to 9.8 meter per seconds square.

To find how much time it will take the quarter to reach the ground, we would use the second equation of motion.

Mathematically, the second equation of motion is given by the formula;

$S = ut + \frac{1}{2} at^2$

Where:

S is the height or distance covered.

u is the initial velocity.

a is the acceleration.

t is the time measured in seconds.

Substituting the values into the formula, we have;

$1.45 = 10.32(t) + \frac{1}{2} (9.8)t^2\\\\1.45 = 10.32t + 4.9t^2\\\\4.9t^2 + 10.32t - 1.45 = 0$

The standard form of a quadratic equation is:

$ax^2 + bx + c = 0$

a = 4.9, b = 10.32 and c = 1.45

We would solve the above quadratic equation by using the quadratic equation formula;

$x = \frac{-b\; \pm \;\sqrt{b^2 - 4ac}}{2a}$

Substituting the values, we have;

$t = \frac{-10.32\; \pm \;\sqrt{10.32^2\; - \;4(4.9)(1.45)}}{2(4.9)}\\\\t = \frac{-10.32\; \pm \;\sqrt{106.5024\; - \;28.42}}{9.8}\\\\t = \frac{-10.32\; \pm \;\sqrt{78.0824}}{9.8}\\\\t = \frac{-10.32\; \pm \;8.84}{9.8}\\\\t = \frac{-10.32\; + \;8.84}{9.8}\\\\t = \frac{1.48}{9.8}$

<em>Time, t = 0.151 seconds.</em>

Therefore, it will take the quarter 0.151 seconds to reach the ground.

Read more: brainly.com/question/8898885

3 0

3 years ago

what is the difference between kinetic energy and random motion of a cylider of oxygen carried by a car and one standing on a pl

pochemuha

Answer:

See explanation below

Explanation:

If we are talking about the kinetic energy of the cylinder of oxygen:

The kinetic energy possessed by any object is given by

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

In this case, we have one cylinder carried by a car and one standing on a platform: this means that the speed of the cylinder carried by the car will be different from zero (and so also its kinetic energy will be different from zer), while the speed of the cylinder standing on the platform will be zero (and so its kinetic energy also zero). Therefore, the kinetic energy of the cylinder carried by the car will be larger than that standing on a platform.

Instead, if we are talking about the kinetic energy due to the random motion of the molecules of oxygen inside the cylinder:

The kinetic energy of the molecules in a gas is directly proportional to the absolute temperature of the gas:

$K=\frac{3}{2}kT$

where k is called Boltzmann constant and T is the absolute temperature of the gas. Therefore, we see that K does not depend on whether the gas is in motion or not, but only on its temperature - therefore, in this case there is no difference between the kinetic energy of the cylinder carried by the car and that standing on the platform (assuming they are at the same temperature)

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3 years ago