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rewona [7]
3 years ago
7

the gravitational force between two masses at distance of 2.5×10^6 metre is 250 Newton . what should be distance between them to

reduce the gravitational force by half?​
Physics
2 answers:
Aleks04 [339]3 years ago
7 0

Answer:

<u>distance</u><u> </u><u>between</u><u> </u><u>should</u><u> </u><u>be</u><u> </u><u>1</u><u>.</u><u>2</u><u>5</u><u> </u><u>×</u><u> </u><u>1</u><u>0</u><u>^</u><u>6</u><u> </u><u>m</u>

Explanation:

Force is proportional to separation distance.

250 \: newtons = 2.5 \times  {10}^{6}  \: metres \\ ( \frac{1}{2}  \times 250) \: newtons = d \\  \\ d =  \frac{125 \times 2.5 \times  {10}^{6} }{250}  \\ d = 1.25 \times  {10}^{6}  \: metres

solmaris [256]3 years ago
3 0

Answer:

distance between should be 1.25×10^6.

Explanation:

force Is proportional to separation .

mrk above answer branilest .

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A 4-kg toy car with a speed of 5 m/s collides head-on with a stationary 1-kg car. After the collision, the cars are locked toget
mihalych1998 [28]

Kinetic energy lost in collision is 10 J.

<u>Explanation:</u>

Given,

Mass, m_{1} = 4 kg

Speed, v_{1} = 5 m/s

m_{2} = 1 kg

v_{2} = 0

Speed after collision = 4 m/s

Kinetic energy lost, K×E = ?

During collision, momentum is conserved.

Before collision, the kinetic energy is

\frac{1}{2} m1 (v1)^2 + \frac{1}{2} m2(v2)^2

By plugging in the values we get,

KE = \frac{1}{2} * 4 * (5)^2 + \frac{1}{2} * 1 * (0)^2\\\\KE = \frac{1}{2} * 4 * 25 + 0\\\\

K×E = 50 J

Therefore, kinetic energy before collision is 50 J

Kinetic energy after collision:

KE = \frac{1}{2} (4 + 1) * (4)^2 + KE(lost)

KE = 40J + KE(lost)

Since,

Initial Kinetic energy = Final kinetic energy

50 J = 40 J + K×E(lost)

K×E(lost) = 50 J - 40 J

K×E(lost) = 10 J

Therefore, kinetic energy lost in collision is 10 J.

4 0
3 years ago
The electric field of a sinusoidal electromagnetic wave obeys the equation E = (375V /m) cos[(1.99× 107rad/m)x + (5.97 × 1015rad
kenny6666 [7]

Answer:

a)  v = 2,9992 10⁸ m / s , b)  Eo = 375 V / m ,  B = 1.25 10⁻⁶ T,

c)     λ = 3,157 10⁻⁷ m,   f = 9.50 10¹⁴ Hz ,  T = 1.05 10⁻¹⁵ s , UV

Explanation:

In this problem they give us the equation of the traveling wave

        E = 375 cos [1.99 10⁷ x + 5.97 10¹⁵ t]

a) what the wave velocity

all waves must meet

        v = λ f

In this case, because of an electromagnetic wave, the speed must be the speed of light.

        k = 2π / λ

        λ = 2π / k

        λ = 2π / 1.99 10⁷

        λ = 3,157 10⁻⁷ m

        w = 2π f

        f = w / 2 π

        f = 5.97 10¹⁵ / 2π

        f = 9.50 10¹⁴ Hz

the wave speed is

        v = 3,157 10⁻⁷   9.50 10¹⁴

        v = 2,9992 10⁸ m / s

b) The electric field is

           Eo = 375 V / m

to find the magnetic field we use

           E / B = c

           B = E / c

            B = 375 / 2,9992 10⁸

            B = 1.25 10⁻⁶ T

c) The period is

           T = 1 / f

            T = 1 / 9.50 10¹⁴

            T = 1.05 10⁻¹⁵ s

the wavelength value is

          λ = 3,157 10-7 m (109 nm / 1m) = 315.7 nm

this wavelength corresponds to the ultraviolet

5 0
3 years ago
needhelpp101 Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential ene
Vlad1618 [11]

If you remember the formula for potential energy,
then this question is a piece-o-cake.

  <em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>

-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ? 

Compared to gravity on Earth, it's only  16.5 percent as much on the Moon. 
So naturally, from the formula, you'd expect the Potential Energy to be less
on the Moon.

4 0
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A light wave traveling through medium 1 strikes a boundary with medium 2 at at a 45 degree angle. the light then enters the seco
Alona [7]

Here light ray strikes to interface at an angle of 45 degree and then refracts into other medium such that it will bend towards boundary.

So here the angle of incidence will be less than the angle of refraction as light moves towards the boundary after refraction which mean it will bend away from the normal

here it can be said that medium 2 will be rarer then medium 1

So here the possible options are

1. Water  

Air

2. Diamond  

Air

So in above two options medium 1 is denser and medium 2 is rarer

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3 years ago
An astronaut takes an object to the moon where there is less gravity. Explain how the mass and weight of the object on the moon
alexandr1967 [171]
The mass of the object will remain the same rather it's on the moon or on the Earth and even in other places. But the weight will change on the moon, so its weight will be different from the one it had on Earth
3 0
3 years ago
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