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3 years ago

Why is gravity so hard to overcome when we try to go up?

Physics

1 answer:

kupik [55]3 years ago

8 0

Answer:

Because you need a lot of force I mean a lot to overcome it like a jet engine .

What comes up must come down

Explanation:

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A certain bridge is 4,224 feet long. What constant rate, in miles per hour, must be maintained in order to walk across the bridg

Archy [21]

Answer:

4miles/hour

Explanation:

the solution for this question requires that the quantities are converted to the appropriate units as required by the question.

Rate in miles per hour = distance in miles / time in hour

to convert 12 minutes to hours; recall that 60 minutes make 1 hour

12 minutes to hour = 12/60 = 0.2hr

to convert 4224 feet to miles; recall 5280 feet is equivalent to 1 mile

4224 feet to miles = 4224/5280 = 0.8 miles

∴ rate = 0.8 / 0.2

rate = 4 miles per hour

the constant rate in miles per hour = 4 miles/hour

4 0

3 years ago

Express force in terms of base units

MariettaO [177]

<h2><u>Answer:</u></h2>

<u></u>

<u>Force = kgm/s² ⟶ Newton</u>

<h2><u>Explanation:</u></h2>

<u></u>

According to the formula we've learnt,

Force = mass × acceleration.

To find force in terms of base units, we must first identify the base SI units of mass & acceleration.

Base SI unit of mass = kg (kilogram)

Base SI unit of acceleration = m/s² (meter per second squared)

So, the base unit of force is,

Force = mass × acceleration.

Force = kg × m/s²

<u>Force = kgm/s² ⟶ Newton</u>

________________

Hope it helps!

$\mathfrak{Lucazz}$

4 0

3 years ago

What cell organelle is necessary for cellular respiration

podryga [215]

It is the cytoplasm.

4 0

3 years ago

Read 2 more answers

Please help me out as soon as you can. Really need help.

olga nikolaevna [1]

The answer would be 6 because 2.0x3= 6

(newton’s 2nd law)

mark me brainliest

4 0

3 years ago

The two masses in the Atwood's machine shown in the figure are initially at rest at the same height. After they are released, th

Inga [223]

According to the description given in the photo, the attached figure represents the problem graphically for the Atwood machine.

To solve this problem we must apply the concept related to the conservation of energy theorem.

PART A ) For energy conservation the initial kinetic and potential energy will be the same as the final kinetic and potential energy, so

$E_i = E_f$

$0 = \frac{1}{2} (m_1+m_2)v_f^2-m_2gh+m_1gh$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

PART B) Replacing the values given as,

$h= 1.7m\\m_1 = 3.5kg\\m_2 = 4.3kg \\g = 9.8m/s^2 \\$

$v_f = \sqrt{2gh(\frac{m_2-m_1}{m_1+m_2})}$

$v_f = \sqrt{2(9.8)(1.7)(\frac{4.3-3.5}{3.5+4.3})}$

$v_f = 1.8486m/s$

Therefore the speed of the masses would be 1.8486m/s

6 0

4 years ago