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3 years ago

How could you increase the amount of kenitic energy

Physics

1 answer:

Fiesta28 [93]3 years ago

8 0

Answer:

As You know kinetic Energy is equal (mv^2)/2

You can increase it by increasing the mass(keeping the velocity)

Or You can add velocity

Explanation:

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What is the transition from a gas to a liquid?

postnew [5]

The transition from gas to liquid is called condensation. An example would be water droplets forming on an ice cold glass placed in room temperature.

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3 years ago

How is the mass of a body measured ? Write in brief.

skad [1K]

Answer:

Mass is the quantity of matter contained in a body. The mass of an object is measured in kilogram (kg).

Explanation:

3 0

3 years ago

Read 2 more answers

If the kinetic energy of a given mass is to be doubled, its speed must be multiplied by

Tomtit [17]

Answer:

Either B or D. The answer itself is 2.

Explanation:

The equation for the kinetic energy would be 1/2*mv^2.

When m is doubled, we can plug in 1 and 2 to compare our answers.

Plugging in 1 for mass would give us the answer 1/2*v^2.

Plugging in 2 for mass would give us v^2. This means that the velocity was multiplied by 2, meaning that the answer is it is multiplied by 2.

I am not sure which answer is correct since there seems to be two answer choices with 2 in it, but the answer is either B or D (I will call it ABCD because I do not want to cause confusion by saying 2 multiple times).

3 0

3 years ago

An astronaut, of total mass 86.0 kg including her suit, stands on a spherical satellite of mass 360 kg, both at rest relative a

Afina-wow [57]

Answer: 0.56 m/s

Explanation:

Hi, to answer this question we have to apply the formula of the conservation of momentum.

m1 v1 = m2 v2 (because the system is stationary at the beginning)

Where:

m1 = mass of the astronaut

v1= velocity of the astronaut

m2= mass of the satellite

v2= velocity of the satellite

Replacing with the values given and solving:

86 kg (2.35m/s) = 360 kg v2

202.1 kgm/s=360kg v2

202.1kgm/s /360kg =v2

v2 = 0.56 m/s

Feel free to ask for more if needed or if you did not understand something.

5 0

3 years ago

A merry-go-round of radius R, shown in the figure, is rotating at constant angular speed. The friction in its bearings is so sma

mel-nik [20]

The angular speed of the merry-go-round reduced more as the sandbag is

placed further from the axis than increasing the mass of the sandbag.

The rank from largest to smallest angular speed is presented as follows;

[m = 10 kg, r = 0.25·R]

${}$ ⇩

[m = 20 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R]

${}$ ⇩

[m = 10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R]

${}$ ⇩

[m = 10 kg, r = 1.0·R]

Reasons:

The given combination in the question as obtained from a similar question online are;

1: m = 20 kg, r = 0.25·R

2: m = 10 kg, r = 1.0·R

3: m = 10 kg, r = 0.25·R

4: m = 15 kg, r = 0.75·R

5: m = 10 kg, r = 0.5·R

6: m = 40 kg, r = 0.25·R

According to the principle of conservation of angular momentum, we have;

$I_i \cdot \omega _i = I_f \cdot \omega _f$

The moment of inertia of the merry-go-round, $I_m$ = 0.5·M·R²

Moment of inertia of the sandbag = m·r²

Therefore;

0.5·M·R²· $\omega _i$ = (0.5·M·R² + m·r²)· $\omega _f$

Given that 0.5·M·R²· $\omega _i$ is constant, as the value of m·r² increases, the value of $\omega _f$ decreases.

The values of m·r² for each combination are;

Combination 1: m = 20 kg, r = 0.25·R; m·r² = 1.25·R²

Combination 2: m = 10 kg, r = 1.0·R; m·r² = 10·R²

Combination 3: m = 10 kg, r = 0.25·R; m·r² = 0.625·R²

Combination 4: m = 15 kg, r = 0.75·R; m·r² = 8.4375·R²

Combination 5: m = 10 kg, r = 0.5·R; m·r² = 2.5·R²

Combination 6: m = 40 kg, r = 0.25·R; m·r² = 2.5·R²

Therefore, the rank from largest to smallest angular speed is as follows;

Combination 3 > Combination 1 > Combination 5 = Combination 6 >

Combination 2

Which gives;

[m = 10 kg, r = 0.25·R] > [m = 20 kg, r = 0.25·R] > [m = 10 kg, r = 0.5·R] > [m =

10 kg, r = 0.5·R] = [m = 40 kg, r = 0.25·R] > [m = 10 kg, r = 1.0·R].

Learn more here:

brainly.com/question/15188750

6 0

2 years ago