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RSB [31]
2 years ago
12

el of a Shs when places in water. The same piece of day ficas ftis made into the stage of a boat. Why does the day ficat when ma

de into the shape of a boat? Once ses dese when sterec re me seace of a beet. O mesec staze disclasg a greater volume of water than the ball shape. The bevant force is greater on the day when it has a spherical shape.​
Physics
1 answer:
SCORPION-xisa [38]2 years ago
8 0
If my truck loses a ski, and the sky is yellow, then how many ounces of milk will I need to make a house?
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A 1.0 kg ball falls from rest a distance of 19.6 m.
Nimfa-mama [501]

Answer:

192.08J

19.6m/s

Explanation:

Since there will be no potential energy when the ball is on the ground, the change in potential energy is equal to the potential energy at the start when the ball is 19.6m above the ground.

PE=mgh

=(1)(9.8)(19.6)

=192.08J

v²=u²+2as, where v is the final velocity, u is initial velocity, a is acceleration and s is distance. Initial velocity is 0 since it starts at rest.

v²=u²+2as

v²=0²+2(9.8)(19.6)

v=√384.16

=19.6m/s

6 0
2 years ago
Two students race up a flight of steps together. "Student A" reaches the top of the steps 2 seconds ahead of
oksano4ka [1.4K]
Um student a because they were there a few seconds ahead
3 0
3 years ago
When in orbit, astronauts experience weightlessness what is this caused by?
erica [24]

no gravity or dark matter

4 0
3 years ago
Read 2 more answers
True or False: Amplitude can increase or decrease wavelength
Marat540 [252]

Answer:

A. False, frequency can increase or decrease wavelength.

For example: a high frequency would mean there are shorter wavelengths that occur in a period. Meanwhile, a low frequency would indicate that the wavelengths are longer and in longer periods.

4 0
3 years ago
A spherical asteroid of average density would have a mass of 8.7×1013kg if its radius were 2.0 km. 1. If you and your spacesuit
Law Incorporation [45]

1. 0.16 N

The weight of a man on the surface of asteroid is equal to the gravitational force exerted on the man:

F=G\frac{Mm}{r^2}

where

G is the gravitational constant

M=8.7\cdot 10^{13}kg is the mass of the asteroid

m = 100 kg is the mass of the man

r = 2.0 km = 2000 m is the distance of the man from the centre of the asteroid

Substituting, we find

F=(6.67\cdot 10^{-11}m^3 kg^{-1} s^{-2})\frac{(8.7\cdot 10^{13} kg)(110 kg)}{(2000 m)^2}=0.16 N

2. 1.7 m/s

In order to stay in orbit just above the surface of the asteroid (so, at a distance r=2000 m from its centre), the gravitational force must be equal to the centripetal force

m\frac{v^2}{r}=G\frac{Mm}{r^2}

where v is the minimum speed required to stay in orbit.

Re-arranging the equation and solving for v, we find:

v=\sqrt{\frac{GM}{r}}=\sqrt{\frac{(6.67\cdot 10^{-11} m^3 kg^{-1} s^{-2})(8.7\cdot 10^{13} kg)}{2000 m}}=1.7 m/s

3 0
3 years ago
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