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yulyashka [42]
3 years ago
11

PLEASE HELP ASAP!! WILL GIVE BRAINLIEST

Physics
1 answer:
nignag [31]3 years ago
8 0
I think it may be c
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If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much w
aleksklad [387]

Answer:3W

If it takes an amount of work W to move two q point charges from infinity to a distance d apart from each other, then how much work should it take to move three q point charges from infinity to a distance d apart from each other?

A) 2W

B) 3W

C) 4W

D) 6W

Explanation: calculating work done,W, in moving two positive q point charges from infinity to a valued distance d from each other  is

W = k(+q)(+q)/ d

k is couloumb's constant

work done in moving 3 equal positive charges from infinity to a finite distance is given by

W₂=W₄=W₆=k(+q)(+q)/ d

Total work done, W' =k(+q)(+q)/ d + k(+q)(+q)/ d + k(+q)(+q)/ d

= W + W + W = 3W

7 0
3 years ago
If all else stays the same, which would cause an increase in the gravitational force on a space shuttle?
Alenkasestr [34]

Answer: Decreasing the distance of the space shuttle from Earth .

Explanation:

According to expression of gravitational force:

F=\frac{G\times m_1\times m_2}{r^2}

G = gravitational constant

m_1, m_2 = masses of two objects

r = Distance between the two objects.

F = Gravitational force

From the above expression we can say that gravitational force is inversely proportional to squared of the distance between the two masses.

F\propto \frac{1}{r^2}

So, in order to increase the gravitational force on space shuttle distance between the space space shuttle  must be decreased.

Hence, the correct answer 'decreasing the distance of the space shuttle from Earth '.

8 0
3 years ago
Read 2 more answers
Please help! you are amazing! brainliest!
ELEN [110]
I think it’s a nebula i hope it’s right
8 0
3 years ago
Read 2 more answers
A 2.7-kg block is released from rest and allowed to slide down a frictionless surface and into a spring. The far end of the spri
exis [7]

a) The speed of the block at a height of 0.25 m is 2.38 m/s

b) The compression of the spring is 0.25 m

c) The final height of the block is 0.54 m

Explanation:

a)

We can solve the problem by using the law of conservation of energy. In fact, the total mechanical energy (sum of kinetic+gravitational potential energy) must be conserved in absence of friction. So we can write:

U_i +K_i = U_f + K_f

where

U_i is the initial potential energy, at the top

K_i is the initial kinetic energy, at the top

U_f is the final potential energy, at halfway

K_f is the final kinetic energy, at halfway

The equation can be rewritten as

mgh_i + \frac{1}{2}mu^2 = mgh_f + \frac{1}{2}mv^2

where:

m = 2.7 kg is the mass of the block

g=9.8 m/s^2 is the acceleration of gravity

h_i = 0.54 is the initial height

u = 0 is the initial speed

h_f = 0.25 m is the final height of the block

v is the final speed when the block is at a height of 0.25 m

Solving for v,

v=\sqrt{u^2+2g(h_i-h_f)}=\sqrt{0+2(9.8)(0.54-0.25)}=2.38 m/s

b)

The total mechanical energy of the block can be calculated from the initial conditions, and it is

E=K_i + U_i = 0 + mgh_i = (2.7)(9.8)(0.54)=14.3 J

At the bottom of the ramp, the gravitational potential energy has become zero (because the final heigth is zero), and all the energy has been converted into kinetic energy. However, then the block compresses the spring, and the maximum compression of the spring occurs when the block stops: at that moment, all the energy of the block has been converted into elastic potential energy of the spring. So we can write

E=E_e = \frac{1}{2}kx^2

where

k = 453 N/m is the spring constant

x is the compression of the spring

And solving for x, we find

x=\sqrt{\frac{2E}{k}}=\sqrt{\frac{2(14.3)}{453}}=0.25 m

c)

If there is no friction acting on the block, we can apply again the law of conservation of energy. This time, the initial energy is the elastic potential energy stored in the spring:

E=E_e = 14.3 J

while the final energy is the energy at the point of maximum height, where all the energy has been converted into gravitational potetial energy:

E=U_f = mg h_f

where h_f is the maximum height reached. Solving for this quantity, we find

h_f = \frac{E}{mg}=\frac{14.3}{(2.7)(9.8)}=0.54 m

which is the initial height: this is correct, because the total mechanical energy is conserved, so the block must return to its initial position.

Learn more about kinetic and potential energy:

brainly.com/question/1198647

brainly.com/question/10770261

brainly.com/question/6536722

#LearnwithBrainly

5 0
3 years ago
Two planets have the same surface gravity, but planet b has twice the radius of planet
nevsk [136]
The formula for the acceleration due to gravity is:

a = Gm/r²
where
G is the universal gravitational constant = 6.6726 x 10⁻¹¹ N-m²/kg²
m is the mass of planet
r is the radius of planet

So, if they have the same a:

m₁/r₁² = m₂/r₂²
So, if m₁ = m and r₂ = 2r₁,
m/r₁² = m₂/(2r₁)²
m₂ = 4m

<em>Thus, the answer is D.</em>
8 0
3 years ago
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