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3 years ago

15

What is the velocity of an object with a mass of 4 kg and a momentum of 24 kg.m/s

1 answer:

Eva8 [605]3 years ago

5 0

Answer:

6m/s

Explanation:

momentum = mass × change in velocity

∆p =m(v)

24 = 4(v)

V =>24/4 = 6m/s

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which of the following instruments measures air pressure? group of answer choices barometer hygrometer radiometer thermometer

Zepler [3.9K]

The tool used to gauge atmospheric pressure is a barometer. Concerning the device used to gauge air pressure, we have been questioned.

Mercury is used in barometers. A reservoir of mercury is compressed by atmospheric pressure, and depending on the force of the air, the mercury is propelled up into the tube to a specific height. A straightforward barometer consists of a long glass tube that is filled with mercury and is closed at one end and open at the other. As most cases, the air pressure is expressed in mm Hg.
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4 0

1 year ago

a 905 - g meteor impacts the earth at a speed of 1623 m/s. if all of its energy is entirely converted to heat in the meteor, wha

nlexa [21]

Answer: 2859.78 k

Explanation: By using the law of conservation of energy, the kinetic energy of the meteor equals the heat energy.

Kinetic energy = 1/2mv^2

Heat energy = mcΔθ

Where m = mass of meteor , v = velocity of meteor = 1623 m/s

c = specific heat capacity of meteor (iron) = 460.548 j/kg/k

Δθ = change in temperature of meteor = ?

From law ofconservation of energy, we have that

1/2mv^2 = mcΔθ

By cancelling "m" on both sides, we have that

v^2/2 = cΔθ

v^2 = 2cΔθ

(1623)^2 = 2× 460.548 × Δθ

2634129 = 921.096 × Δθ

Δθ = 2634129 / 921.096

Δθ = 2859.78 k

6 0

3 years ago

A remote-controlled plane has a mass of 0.386 kg and flies in a circle with a radius of 75 m. The centripetal force on the plane

kolezko [41]

Answer:

28 m/s

Explanation:

correct

6 0

4 years ago

Why is it easier to change meters to centimeters than it is changing yards to feet?

storchak [24]

Because metric units use the deca system, 1km = 1000m = 100 000cm etc...

6 0

3 years ago

A small block of mass m = 0.032 kg can slide along the frictionless loop-the-loop, with loop radius R = 12 cm. The block is rele

Lunna [17]

Answer:

Part a)

$W = 0.15 J$

Part b)

$W = 0.11 J$

Part c)

$U = 0.19 J$

Part d)

$U = 0.038 J$

Part e)

$U = 0.075 J$

Part f)

It is independent of the speed of the object so all part answers will remain the same

Explanation:

Part a)

As we know that Point P is at height 5R while point Q is at height R

so the work done by gravity from P to Q is given as

$W = mg(5R - R)$

$W = 0.032(9.8)(4)(0.12)$

$W = 0.15 J$

Part b)

When it reaches to the top of the loop then its final height from ground is

h = 2R

so work done from P to Q is given as

$W = mg(5R - 2R)$

$W = 3mgR$

$W = 0.11 J$

Part c)

Potential energy at P point is given as

$U = mgH$

$U = 0.032(9.8)(5)(0.12)$

$U = 0.19 J$

Part d)

Potential energy at Q point is given as

$U = mgH$

$U = 0.032(9.8)(0.12)$

$U = 0.038 J$

Part e)

Potential energy at top point is given as

$U = mgH$

$U = 0.032(9.8)(2)(0.12)$

$U = 0.075 J$

Part f)

Since all the answer from part a) to part e) depends only upon the position of the object.

So here we can say that it is independent of the speed of the object so all part answers will remain the same

8 0

4 years ago