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3 years ago

How many days are in a year on mars

Physics

1 answer:

Triss [41]3 years ago

7 0

687 days
hope this helps
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A force of 5 N is applied to the end of a lever that has a length

Klio2033 [76]

Answer:

The answer is 10Nm

Explanation: I ended up just messing around with the numbers, I multiplied 5 and 2 got 10 as my answer and it was right.

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3 years ago

What is tge scientific teem of the ratio between power out to power in (physics)

Art [367]

Answer:

In physics, power is the amount of energy transferred or converted per unit time. In the ... Power (physics) ... Angular acceleration / displacement / frequency / velocity. show. Scientists ... Hence the formula is valid for any general situation. ... because they define the maximum performance of a device in terms of velocity ratios

Explanation:

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3 years ago

What is it called when sediment is dropped and comes to rest?

rosijanka [135]

It's called deposition.

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3 years ago

An aircraft with a mass of 10,000 kg starts from rest at sea level and takes off, then flies to a cruising speed of 620 km/h and

Natasha_Volkova [10]

Answer:

The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

Explanation:

Given that,

Mass of aircraft = 10000 kg

Speed = 620 km/h = 172.22 m/s

Altitude = 10 km = 1000 m

We calculate the change in potential energy

$\Delta P.E=mg(h_{2}-h_{1})$

$\Delta P.E=10000\times9.8\times(10000-0)$

$\Delta P.E=10000\times9.8\times10000$

$\Delta P.E=980000000\ J$

$\Delta P.E=980\ MJ$

For g = 10 m/s²,

The change in potential energy will be 1000 MJ.

We calculate the change in kinetic energy

$\Delta K.E=\dfrac{1}{2}m(v_{2}^2-v_{1}^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2-0^2)$

$\Delta K.E=\dfrac{1}{2}\times10000\times(172.22^2)$

$\Delta K.E=148298642\ J$

$\Delta K.E=148.3\ MJ$

For g = 10 m/s²,

The change in kinetic energy will be 150 MJ.

Hence, The change in potential energy and kinetic energy are 980 MJ and 148.3 MJ.

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3 years ago

a large heavy truck and a baby carriage roll down a hill. Neglecting friction, at the bottom of the hill, the baby carriage will

Lisa [10]

Answer:

please give me brainlist and follow

Explanation:

At the bottom of the hill, the baby carriage will likely have less momentum Therefore, option D is correct. Solution: ... Therefore, at the bottom of the hill, the heavy truck will have more momentum and baby carriage will have less momentum.

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3 years ago