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3 years ago

Which car is harder to stop and why?

Physics

2 answers:

kondor19780726 [428]3 years ago

8 0

Answer:

The momentum of bath cars is 40000 Ns which make the difficulty to stop each car in aspect of fprce is the same.

Explanation:

Momentum (P) =mass(m) × velocity (v)

For car A,

P = m × v = 1000 × 40 = 40000 Ns

For car B,

P = m × v = 4000 × 10 = 40000 Ns

Force (F) = Momentum change(ΔΡ)/ time taken(t)

F = ΔΡ/t

When stopping the car the momentum changes from 40000 Ns to 0

So momentum change in both cars is the same. So to stop the two cars in a given time (t) you need the same force, which means you will feel same difficulty.

Bond [772]3 years ago

8 0

Answer:

The third option

Explanation:

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3 years ago

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Answer:

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Explanation:

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3 years ago

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3 years ago

A wooden object (conically shaped) has a diameter of 8cm and height of 14cm. It floats in oil with 6cm of its height above oil l

baherus [9]

Answer:

(a) The density of the object is 316/343 × the density of the oil

(b) The fraction of oil displaced after immersing the object is 0.461 of the oil volume

Explanation:

(a) The volume, V of a cone of height, h and base diameter, D = 2×r is given by the following equation;

$V = \dfrac{\pi r^{2} h}{3}$

The volume of the object is therefore;

$\dfrac{\pi \times 4^{2} \times 14}{3} = 74\tfrac{2}{3}\pi \, cm^3$

Where 6 cm is above the oil level we have;

$\dfrac{\pi \times \left (6 \times \dfrac{4}{14} \right )^{2} \times 6}{3} = 5\tfrac{43}{49}\pi \, cm^3$ above the oil level

Therefore, volume of the oil displaced = $68\tfrac{116}{147}\pi$ cm³ = 216.11 cm³

The density of the object is thus;

$\dfrac{68\tfrac{116}{147}\pi}{ 74\tfrac{2}{3}\pi} \times Density \ of \ the \ oil = \dfrac{316}{343} \right ) \times Density \ of \ the \ oil$

The density of the object = 316/343 × the density of the oil.

(b) The volume of the oil = 2 × Volume of the object = $2 \times 74\tfrac{2}{3}\pi \, cm^3 = 149\tfrac{1}{3}\pi \, cm^3$

The fraction of the volume displaced, x, after immersing the object is given as follows;

$x = \dfrac{68\tfrac{116}{147}\pi}{ 149\tfrac{1}{3}\pi} = \dfrac{158}{343} = 0.461$

The fraction of oil displaced after immersing the object = 0.461 of the volume of the oil

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3 years ago