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3 years ago

The roller coaster from problem #1 then tops a second hill at 15.0 m/s, how high is the second hill?

Physics

1 answer:

navik [9.2K]3 years ago

4 0

Answer:

68.5

Explanation:

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Rate at which wave travels or propagates

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Answer:

v =fλ v = f λ where v

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3 years ago

When the rubber sheet is pulled down, the volume of the bell jar is ____________ and the pressure in the bell jar is ___________

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2 years ago

A roller skater of 47kg moving with a velocity of 12 m/s to the east picks up a bag of 6.0 kg. What is the final velocity of the

11Alexandr11 [23.1K]

Answer:

v_f = 10.85 m/s

Explanation:

We will apply the law of conservation of momentum here:

$m_{1}v_{1i} + m_{2}v_{2i} = m_{1}v_{1f}+m_{2}v_{2f}\\$

where,

m₁ = mass of roller skater = 47 kg

m₂ = mass of bag = 6 kg

v_1i = initial speed of roller skater = 12 m/s

v_2i = initial speed of the bag = 0 m/s

v_1f = final speed of the roller skater = ?

v_2f = final speed of the bag = ?

Both the bag and the skater will have same speed at the end because kater is carrying the bag:

v_1f = v_2f = v_f

Therefore, the equation will become:

$(47\ kg)(12\ m/s)+(6\ kg)(0\ m/s)=(47\ kg)(v_{f})+(5\ kg)(v_{f})\\564\ N.s = (47\ kg+5\ kg)(v_{f})\\v_{f} = \frac{564\ N.s}{52\ kg}\\$

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2 years ago

Which statement describes an intensive property of matter

Fantom [35]

Answer:

where are the statements so i can see which one is correct

Explanation:

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3 years ago

In an experiment, a rectangular block with height h is allowed to float in two separate liquids. In the first liquid, which is w

Amiraneli [1.4K]

Answer:

The relative density of the second liquid is 7.

Explanation:

From archimede's principle we know that the force that a liquid exerts on a object equals to the weight of the liquid that the object displaces.

Let us assume that the volume of the object is 'V'

Thus for the liquid in which the block is completely submerged

The buoyant force should be equal to weight of liquid

Mathematically

$F_{buoyant}=Weight\\\\\rho _{1}\times V\times g=m\times g\\\\\therefore \rho _{1}=\frac{m}{V}...............(i)$

Thus for the liquid in which the block is 1/7 submerged

The buoyant force should be equal to weight of liquid

Mathematically

$F'_{buoyant}=Weight\\\\\rho _{2}\times \frac{V}{7}\times g=m\times g\\\\\therefore \rho _{2}=\frac{7m}{V}.................(ii)$

Comparing equation 'i' and 'ii' we see that

$\rho_{2}=7\times \rho _{1}$

Since the first liquid is water thus $\rho _{1}=1gm/cm^3$

Thus the relative density of the second liquid is 7.

6 0

3 years ago

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