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3 years ago

What is the momentum of a 15 kg tire rolling down a hill at 3 m/s? 5 kg • m/s 18 kg • m/s 45 kg • m/s 60 kg • m/s

Physics

2 answers:

maw [93]3 years ago

7 0

p=mv

?=15kg * 3m/s

45p

galben [10]3 years ago

4 0

Answer:

45 kg • m/s

Explanation:

just answered question

You might be interested in

At the top of the hill, the

WARRIOR [948]

Answer:

The answer to your question is below

Explanation:

To explain what happens with the ball we must remember the Law of Conservation of Energy.

This law states that the energy can be neither created nor destroyed only converted from one form of energy to another.

Then,

At the top of the hill, the potential energy is maximum and the kinetic energy equals to zero.

When the ball starts to roll down the potential energy will be lower and the kinetic energy will have a low value.

At the middle of the hill, both energies have the same values.

At the end of the hill, the potential energy will be equal to zero and the kinetic energy will be maximum.

8 0

3 years ago

Assuming the incline to be frictionless and the zero of gravitational potential energy to be at the elevation of the horizontal

Leya [2.2K]

The energy in the system is given by the law of conservation of energy.

The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h.

Reason:

The given parameters are;

Surface of the inclined plane = Frictionless

Potential energy at the horizontal line = Zero of gravitational potential

By the law of Conservation of Energy, we have;

The spring will be compressed by a distance, x

The energy stored in the compressed spring, K.E. = (1/2)·k·x²

Energy in the block when the block comes to rest at a height, h₁, will be, P.E. = m·g·h₁

Therefore, by conservation of energy, we have;

The initial potential of the block = The stored energy in the compressed string + The gravitational potential energy of the block when it has compressed.

Therefore, the correct option is; The energy stored in the spring plus the gravitational potential energy of the block when it has fully compressed the spring will be equal to m·g·h

Learn more here;

brainly.com/question/17713698

4 0

2 years ago

2. A jack exerts a vertical force of 4.5 X 103

skad [1K]

Correct Question:-

A jack exerts a vertical force of 4.5 × 10³

newtons to raise a car 0.25 meter. How much

work is done by the jack?

$\\ \\$

Given :-

$\star \sf \small force = 4.5 \times {10}^{3} \: newton$

$\star \sf \small distance = 0.25 \: meter$

$\\ \\$

To find:-

$\sf \star \: work = \: ?$

$\\ \\$

Solution:-

we know :-

$\bf \dag \boxed{ \rm work = force \times distance}$

$\\ \\$

So:-

$\dashrightarrow \sf work = force \times distance$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times 0.5 \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{0 \cancel.5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = (4.5 \times 1 {0}^{3} ) \times \cancel \frac{5}{10} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{4\cancel.5}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10} \times 1 {0}^{3} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{3 - 1} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{10 {}^{0} } \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45}{1} \times 1 {0}^{2} \times \dfrac{1}{2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times \cancel{10}}{ \cancel2} \\$

$\\ \\$

$\dashrightarrow \sf work = \dfrac{45 \times 10 \times 5}{ 1} \\$

$\\ \\$

$\dashrightarrow \sf work =225 \times 10$

$\\ \\$

$\dashrightarrow \bf work =\red{2250\: joule}$

5 0

2 years ago

A 30.0-kg child sits on one end of a long uniform beam having a mass of 20.0 kg, and a 40.0-kg child sits on the other end. The

qaws [65]

let the length of the beam be "L"

from the diagram

AD = length of beam = L

AC = CD = AD/2 = L/2

BC = AC - AB = (L/2) - 1.10

BD = AD - AB = L - 1.10

m = mass of beam = 20 kg

m₁ = mass of child on left end = 30 kg

m₂ = mass of child on right end = 40 kg

using equilibrium of torque about B

(m₁ g) (AB) = (mg) (BC) + (m₂ g) (BD)

30 (1.10) = (20) ((L/2) - 1.10) + (40) (L - 1.10)

L = 1.98 m

4 0

3 years ago

Name the substance in which starch becomes dark blue in colour.

matrenka [14]

Any substance that contains starch turns blue-black in presence of iodine solution.

6 0

2 years ago