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2 years ago

Please answer my question I am having some trouble, please I would love for some help please and thank you.

1 answer:

8 0

the cliff doesnt matter think dropping a ball at head level around 6 feet and it has a force of 1 unit but you throw it 12 feet up and it falls with a force of 2 units aslong as the Coriolis effect isnt there so ball 1 has more force and it also has more air time as it goes up a certain distance falls back down from a greater height as it has been thrown straight up

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Why does time slow down the closer you get to a black hole, I know that time slows down the faster you move through space and th

mylen [45]

Answer:

Near the black hole, The gravitational pull is so strong and the black hole has a lot of stress energy. This causes the clock to tick slower than usual and due to this , Time slows down to a great extent.

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3 years ago

WHAT IS PIVOT POINT ?

geniusboy [140]

Pivots Points are price levels chartists can use to determine intraday support and resistance levels

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2 years ago

Read 2 more answers

Car A (mass 1100 kg) is stopped at a traffic light when it is rearended by car B(mass 1400 kg). Both cars then slide with locke

viva [34]

Answer:

Part a)

$v_a = 3.94 m/s$

Part b)

$v_b = 3.35 m/s$

Part C)

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

Explanation:

Part a)

As we know that car A moves by distance 6.1 m after collision under the frictional force

so the deceleration due to friction is given as

$a = -\frac{F_f}{m}$

$a = -\frac{\mu mg}{m}$

$a = - \mu g$

now we will have

$v_f^2 - v_i^2 = 2ad$

$0 - v_i^2 = 2(-\mu g)(6.1)$

$v_a = \sqrt{(2(0.13)(9.81)(6.1)}$

$v_a = 3.94 m/s$

Part b)

Similarly for car B the distance of stop is given as 4.4 m

so we will have

$v_b = \sqrt{2(0.13)(9.81)(4.4)}$

$v_b = 3.35 m/s$

Part C)

By momentum conservation we will have

$m_1v_{1i} = m_1v_{1f} + m_2v_{2f}$

$1400 v_b = 1100(3.94) + 1400(3.35)$

$v_b = 6.44 m/s$

Part d)

Due to large magnitude of friction between road and the car the momentum conservation may not be valid here as momentum conservation is valid only when external force on the system is zero.

3 0

3 years ago

Melting wax

MaRussiya [10]

It would have to be c because it is a chemical change. your welcome!!!!!

3 0

3 years ago

Read 2 more answers

A 86 kg human stands on the surface of Venus. The mass of Venus is 4.9 × 1024 kg and its radius is 6.1 × 106 m. Collapse questio

iris [78.8K]

Answer:

755.37 N

Explanation:

We are given that

Mass of venus= $m_1=4.9\times 10^{24}kg$

Radius= $r=6.1\times 10^6$ m

Mass of human= $m_2=86 kg$

We know that the gravitational force between two bodies

$F=G\frac{m_1m_2}{r^2}$

Where G=Gravitational constant= $6.67\times 10^{-11}Nm^2/kg^2$

Using the formula

The magnitude of the gravitational force exerted by Venus on the human= $F=\frac{6.67\times 10^{-11}\times 86\times 4.9\times 10^{24}}{(6.1\times 10^6)^2}$

The magnitude of the gravitational force exerted by Venus on the human=F= $755.37N$

7 0

2 years ago