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elena-s [515]
3 years ago
11

Which fundamental force has a small range and is always an attractive force?

Physics
1 answer:
erica [24]3 years ago
5 0

Answer:

gravitational force is a fundamental force and also , it does have a small range and is always an attractive force.

Explanation:

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2. The gravitational field intensity is measured in: a. N c. kg d. Nm2/kg? b. N/kg​
alexdok [17]

Answer:

N/kg

Explanation:

newton/kilogram is the answer.please mark me brainliest

4 0
3 years ago
In an attempt to reduce the extraordinarily long travel times for voyaging to distant stars, some people have suggested travelin
Vika [28.1K]

Answer:

Explanation:

Let the required velocity of rocket be v .

We shall use the formula of time dilation to find the velocity of rocket .

t = \frac{t'}{\sqrt{1-\frac{v^2}{c^2} } }

t = 430

t' = 38

430=\frac{38}{\sqrt{1-\frac{v^2}{c^2} } }

\sqrt{1-\frac{v^2}{c^2} } }=\frac{38}{430}

1-\frac{v^2}{c^2} = .0078

\frac{v^2}{c^2} =.9922

\frac{v}{c} = .996

v  = .996 x 3 x 10⁸ m /s

= 2.988 x 10⁸ m /s

B )

Kinetic energy of rocket

= 1/2 m v²

= .5 x 20000 x (2.988 x 10⁸ )²

= 8.9 x 10²⁰ J .

C )

This energy is 8.9 times the energy requirement of United states in the year 2005 .

7 0
3 years ago
If Maria lifts a mass of 10kg into a table of 1.25m high. How much gpe has the mass gained
SCORPION-xisa [38]

Answer:

gravitational potential energy of 10 kg mass is 122.625 J

Explanation:

Gravitational Potential energy gain is given as

U = mgh

here we know that

m = 10 kg

h = 1.25 m

now from above formula we have

U = (10)(1.25)(9.81)

U = 122.625 J

So gravitational potential energy of 10 kg mass is 122.625 J

3 0
3 years ago
The closely packed cones in the fovea of the eye have a diameter of about 2 μm. For the eye to discern two images on the fovea
Marina CMI [18]

Answer:

The distance between the object is l=0.0056\  cm

Explanation:

The free body diagram of this setup is on the first uploaded image

From the question

   The diameter of closely packed cones in the fovea of the eye is  =  2 \mu m

     The distance of separation by one cone(not excited ) is d = 4\mu m = 4*10^{-4}cm

     The distance between the two point-like object  is  l

     The diameter of the eye is D = 2 cm

     The distance of the two point-like object from the near point of the eye is A = 28 cm

 From the diagram we see that the light from the two point-like object form a triangle of similar base l and d  and height D and A

So for a triangle with similar base we have that

                \frac{l}{A} =\frac{d}{D}

                \frac{l}{28} = \frac{4*10^{-4}}{2}

making l the subject we have

            l = \frac{28 *4*10^{-4}}{2}

              l=0.0056\  cm

               

         

       

   

5 0
3 years ago
Read 2 more answers
Pulling a rubber band back and then letting it fly across the room is an example of
Cloud [144]

Answer:

b

Explanation:

because a elastic band uses elastic energy

7 0
3 years ago
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