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Serhud [2]
3 years ago
11

" alt="V = \frac{4}{3} \pi .r3" align="absmiddle" class="latex-formula">
Physics
1 answer:
pogonyaev3 years ago
6 0
The formula is the volume of a *sphere* ... l hope it helped:)
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To maintain the same amount of torque due to a mass on a balance as the mass is increased, how should the position of the mass c
olasank [31]

Answer:

the mass should be bring closer to the point about which we are finding torque

Explanation:

τ = Σr × F = rmg

where m is the mass, g is acceleration due to gravity, and r is the distance

Torque is directly proportional to -

1.mass, m , of object

2. distance, r, of the mass from the point about which we are finding the torque.

So if we increase or decrease them then the torque will also increase or decrease.

So if we increase the mass the torque will increase but since we have to maintain same torque therefore we have to decrease the distance of mass from the point about which we are finding torque.

Therefore the mass should be bring closer to the point about which we are finding torque.

8 0
3 years ago
Water stored in a reservoir is used to generate electricity as it flows down and into a river. The river flows to the ocean, whe
Thepotemich [5.8K]

where is there started interested
7 0
2 years ago
Traveling into space from Earth, which of these would he be able to reach first?
mr Goodwill [35]

You didn't Provide choices to choose from. Please fix question and I'd love to be able to help.

5 0
2 years ago
Read 2 more answers
An astronaut on a distant planet wants to determine its acceleration due to gravity. the astronaut throws a rock straight up wit
Tamiku [17]
Define
u = 16 m/s, the vertical launch velocity
g = acceleration due to gravity, measured positive downward
s = vertical distance traveled
t  = 21.2 s, total time of travel.

The vertical motion obeys the equation
s = ut - (1/2)gt²

When the rock is at ground level, s = 0.
Therefore
(16 m/s)(21.2 s) - 0.5*(g m/s²)*(21.2 s)² = 0
339.2 - 224.72g = 0
g = 1.5094 m/s²

Answer:
The acceleration due to gravity is 1.509 m/s² measured positive downward.


7 0
3 years ago
A student weighing 700 N climbs at constant speed to the top of an 8 m vertical rope in 10 s. The average power expended by the
nasty-shy [4]

Answer:

The average power the student expended to overcome gravity is 560W, Watts is a units of work it is Joules /time or kg*m^{2}/ s^{3}

Explanation:

Weight = 700N

F= m*g*h \\m*g= 700N\\F= 700N*h\\F=700N*8m\\F= 5600 N*m \\F=5600 J

The power is the work in (Joules) or (N*m) in a determinate time (s) to get Watts (W) units for work

Work= \frac{5600 J}{10s} = \frac{5600 \frac{kg*m^{2} }{s^{2} } }{10 s}  \\Work= 560\frac{kg*m^{2} }{s^{3} } \\Work =560 W

3 0
2 years ago
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