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1 year ago

What is the weight of an astronaut with a mass of 75 kg on the moon? The gravitational field strength on the moon is 1.6 Nkg-1.

Physics

1 answer:

bearhunter [10]1 year ago

7 0

okay this is kinda easy

What is the gravitational field strength on the moon?

The Moon has a gravitational field strength of 1.6 N/kg.

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A particularly scary roller coaster contains a loop-the-loop in which the car and rider are completely upside down. If the radiu

Pepsi [2]

Answer:

$v = 10.89\ m/s$

Explanation:

given,

radius of loop = 12.1 m

to find the minimum speed transverse by the rider to not to fall out upside down

centripetal force = $\dfrac{mv^2}{r}$

gravitational force = m g

computing both the equation]

$mg = \dfrac{mv^2}{r}$

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$v = \sqrt{118.58}$

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2 years ago

Why is the curve between 1950 and 1980 relatively flat and centered around zero degrees difference from the baseline? (Hint: how

zimovet [89]

Look at the title of the graph, in small print under it.

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2 years ago

The severity of a fall depends on your speed when you strike the ground. All factors but the acceleration from gravity being the

Diano4ka-milaya [45]

Answer:

The object could fall from six times the original height and still be safe

Explanation:

Free Falling

When an object is released from rest in free air (no friction), the motion is completely dependant on the acceleration of gravity g.

If we drop an object of mass m near the Earth surface from a height h, it has initial mechanical energy of

$U=m.g.h$

When the object strikes the ground, all the mechanical energy (only potential energy) becomes into kinetic energy

$\displaystyle K=\frac{1}{2}m.v^2$

Where v is the speed just before hitting the ground

If we know the speed v is safe for the integrity of the object, then we can know the height it was dropped from

$\displaystyle m.g.h=\frac{1}{2}m.v^2$

Solving for h

$\displaystyle h=\frac{m.v^2}{2mg}=\frac{v^2}{2g}$

If the drop had occurred in the Moon, then

$\displaystyle h_M=\frac{v_M^2}{2g_M}$

Where hM, vM and gM are the corresponding parameters on the Moon. We know v is the safe hitting speed and the gravitational acceleration on the Moon is g_M=1/6 g

$\displaystyle h_M=\frac{v^2}{2\frac{1}{6}g}$