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2 years ago

A man pushed the box with a force of 20 N. The mass of the box doubled, but he still pushed with a force of 20 N.

Physics

1 answer:

Nuetrik [128]2 years ago

7 0

Answer:

Option (c) Acceleration is 1/2x the original acceleration

Explanation:

To know the the correct answer to the question, we shall determine the acceleration of the box in each case.

Case 1:

Mass (m) = m

Force (F) = 20 N

Acceleration 1 (a₁) =?

F = ma₁

20 = m × a₁

Divide both side by m

a₁ = 20 / m

Case 2:

Mass (m) = 2m

Force (F) = 20 N

Acceleration 2 (a₂) =?

F = ma₂

20 = 2m × a₂

Divide both side by 2m

a₂ = 20 / 2m

a₂ = 10 / m

Finally, we shall determine the acceleration of the box after the mass was doubled. This can be obtained as illustrated below:

Acceleration 1 (a₁) = 20 / m

Acceleration 2 (a₂) = 10 / m

a₂ : a₁ = 10 / m : 20 / m

a₂ / a₁ = 10 / m ÷ 20 / m

a₂ / a₁ = (10 / m) × (m / 20)

a₂ / a₁ = ½

Cross multiply

a₂ = ½a₁

Thus, the acceleration of the box after the mass was doubled is ½ times the original acceleration.

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Squids and octopuses propel themselves by expelling water. They do this by keeping water in a cavity and then suddenly contracti

Alex

Answer:

A) The speed of the water must be 8.30 m/s.

B) Total kinetic energy created by this maneuver is 70.12 Joules.

Explanation:

A) Mass of squid with water = 6.50 kg

Mass of water in squid cavuty = 1.55 kg

Mass of squid = $m_1=6.50 kg- 1.55 kg=4.95 kg$

Velocity achieved by squid = $v_1=2.60 m/s$

Momentum gained by squid = $P=m_1v_1$

Mass of water = $m_2=1.55 kg$

Velocity by which water was released by squid = $v_2$

Momentum gained by water but in opposite direction = $P'=m_2v_2$

P = P'

$m_1v_1=m_2v_2$

$v_2=\frac{m_1v_1}{m_2}=\frac{4.95 kg\times 2.60 m/s}{1.55 kg}=8.30 m/s$

B) Kinetic energy does the squid create by this maneuver:

Kinetic energy of squid = K.E = $\frac{1}{2}m_1v_1^{2}$

Kinetic energy of water = K.E' = $\frac{1}{2}m_2v_2^{2}$

Total kinetic energy created by this maneuver:

$K.E+K.E'=\frac{1}{2}m_1v_1^{2}+\frac{1}{2}m_2v_2^{2}$

$=\frac{1}{2}\times 4.95 kg\times (2.60 m/s)^2+\frac{1}{2}\times 1.55 kg\times (8.30 m/s)^2=70.12 Joules$

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3 years ago

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2 years ago

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A woman of mass 50 kg is swimming with a velocity of 1.6 m/s. If she stops stroking and glides to a stop in the water, what is t

Snezhnost [94]

Answer:

Impulse of force = -80 Ns

Explanation:

Given the following data;

Mass = 50kg

Initial velocity = 1.6m/s

Since she glides to a stop, her final velocity equals to zero (0).

Now, we would find the change in velocity.

$Change \; in \; velocity = final \; velocity - initial \; velocity$

Substituting into the equation above;

Change in velocity = 0 - 1.6 = 1.6m/s

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Substituting into the equation, we have;

$Impulse \; of \; force = 50 * -1.6$

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