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Whitepunk [10]
3 years ago
6

Object A attracts object B with a gravitational force of 10 newtons from a given distance. If the distance between the two objec

ts is doubled, what is the changed force of attraction between them?
A. 2.5 newtons
B. 5 newtons
C. 20 newtons
D. 100 newtons
Physics
2 answers:
balandron [24]3 years ago
7 0

By definition, the gravitational force is given by:

F = G(\frac{m1m2}{r^2})

Where,

G: gravitational constant

m1: mass of object 1

m2: mass of object 2

r: distance between objects:

We know that the force is equal to 10N:

G(\frac{m1m2}{r^2}) = 10

If the distance is double we have:

F = G(\frac{m1m2}{(2r)^2}) = G(\frac{m1m2}{4r^2}) = \frac{1}{4}G(\frac{m1m2}{r^2})

\frac{1}{4}(10) = 2.5

Therefore, the new force is:

2.5 newtons

Answer:

the changed force of attraction between them is:

A. 2.5 newtons

levacccp [35]3 years ago
5 0
2.5
(Because formula has r squared, or 2 times 2 or 4)
So 10:4=2.5
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svlad2 [7]

Answer:

v=20m/S

p=-37.5kPa

Explanation:

Hello! This exercise should be resolved in the next two steps

1. Using the continuity equation that indicates that the flow entering the nozzle must be the same as the output, remember that the flow equation consists in multiplying the area by the speed

Q=VA

for he exitt

Q=flow=5m^3/s

A=area=0.25m^2

V=Speed

solving for V

V=\frac{Q}{A} \\V=\frac{5}{0.25} =20m/s

velocity at the exit=20m/s

for entry

V=\frac{5}{1} =5m/s

2.

To find the pressure we use the Bernoulli equation that states that the flow energy is conserved.

\frac{P1}{\alpha } +\frac{v1^2}{2g} =\frac{P2} {\alpha } +\frac{v2^2}{2g}

where

P=presure

α=9.810KN/m^3 specific weight for water

V=speed

g=gravity

solving for P1

(\frac{p1}{\alpha } +\frac{V1^2-V2^2}{2g})\alpha  =p2\\(\frac{150}{9.81 } +\frac{5^2-20^2}{2(9.81)})9.81  =p2\\P2=-37.5kPa

the pressure at exit is -37.5kPa

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3 years ago
The Earth’s internal __________ source provides the energy for our dynamic planet, providing it with the driving force for on-go
monitta
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Answer:

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Explanation:

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One watt is A. equal to one kilogram per second B. the power that can be delivered by an average horse C. the same as one-foot p
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Answer:

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Explanation:

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If there is a huge boulder on your lawn and you want to determine its density, but it is
Olin [163]

There are different options here but all of them work by approximating and assuming.

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ii) that the bottom surface of the boulder is known.

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All the above methods are estimating methods.

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Density of water

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This is what I think after correction and allthe best!

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