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elixir [45]
3 years ago
15

A deep space probe travels in a straight line at a constant speed of over 16,000 m/s. Assuming there is no friction in space, if

a loose part were released from the spacecraft without a push in any direction, what would happen to it? A. It would slowly coast to a stop. B. It would head off on its own trajectory. C. It would stop immediately as soon as it was no longer in contact with the probe. D. It would continue along with the spacecraft.
Physics
1 answer:
77julia77 [94]3 years ago
6 0
C because when the part gets out of the probe it would no longer stay contacted
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What formula gives the strength of an electric field, E, at a distance from a known source charge?
Deffense [45]
<h2>Hello!</h2>

The answer is: Coulomb's law equation.

<h2>Why?</h2>

The Coulomb's law states that the strength of an electric field (between two charges) can be calculated by multiplying their charges and dividing it into the square of the distance between their centers.

E=\frac{k*q*Q}{d^{2} }

Where:

E = Electric Field Strenght

k=9.0*10^{9}  \frac{N.m^{2} }{C^{2} }

q=TestCharge\\Q=SourceCharge\\

d = separation between charges (m)

Have a nice day!

5 0
4 years ago
Read 2 more answers
A girl swings on a playground swing in such a way that at her highest point she is 4.1 m from the ground, while at her lowest po
Umnica [9.8K]

Answer:

V1 =8.1 m/s

Explanation:

height at highest point (h2) = 4.1 m

height at lowest point (h1) = 0.8 m

acceleration due to gravity (g) = 9.8 m/s^{2}

from conservation of energy, the total energy at the lowest point will be the same as the total energy at the highest point. therefore

mgh1 + 0.5mV1^{2} = mgh2 + 0.5mV2^{2}

where

  • speed at highest point = V2
  • speed at lowest point = V1
  • mass of the girl and swing = m
  • at the highest point, the  speed is minimum (V1 = 0)
  • at the lowest point the speed is maximum (V2 is the maximum speed)
  • therefore the equation becomes mgh1 + 0.5mV1^{2} = mgh2

      m(gh1 + 0.5V1^{2}) = m(gh2)

      gh1 + 0.5V1^{2} = gh2

      V1 = \sqrt{\frac{gh2 - gh1}{0.5}}

now we can substitute all required values into the equation above.

V1 = \sqrt{\frac{(9.8x4.1) - (9.8x0.8)}{0.5}}

V1 = \sqrt{\frac{32.34}{0.5}}

V1 =8.1 m/s

8 0
4 years ago
A lead ball is dropped into a lake from a diving board 5.0 meters above the water. After entering the water, it sinks to the bot
xxMikexx [17]

Answer:

28.8 meters

Explanation:

We must first determine at which velocity the ball hits the water. To do so we will:

1) Assume no air resistance.

2) Use the Law of conservation of mechanical energy: E=K+P

Where

E is the mechanical energy (which is constant)

K is the kinetic energy.

P is the potential energy.

With this we have \frac{m}{2} *v^{2}  = m*g*h

Where:

m is the balls's mass <- we will see that it cancels out and as such we don't need to know it.

v is the speed when it hits the water.

g is the gravitational constant (we will assume g=9.8\frac{m}{s^{2} }.

h is the height from which the ball fell.

Because when we initially drop the ball, all its energy is potential (and P = - m*g*h) and when it hits the water, all its energy is kinetic (K=\frac{m}{2} *v^{2}. And all that potential was converted to kinetic energy.

Now, from \frac{m}{2} *v^{2}  = m*g*h we can deduce that v=\sqrt{2*g*h}

Therefore v=9.6\frac{m}{s}

Now, to answer how deep is the lake we just need to multiply that speed by the time it took the ball to reach the bottom.

So D=9.6\frac{m}{s}*3s=28.8m

Which is our answer.

7 0
3 years ago
Read 2 more answers
Helppppppppppppppppppppppppp
andrew11 [14]

Answer:

value of x=35°

y=72.5°

hope it helps you

make me brainliest plz

6 0
3 years ago
How is work affected when an object is lifted straight up instead of using a ramp? Work will increase.
kifflom [539]
Without friction, the amount of work only depends on the final height,
and is not affected by the route used to get there. 

If the ramp has no friction, then it has no effect on the total amount
of work done.  The work to lift the load straight up is the same.

If the ramp has some friction, then it takes more work to use the ramp
than to lift the load straight up.  Then the work to lift the load straight up
would be less than when the ramp is used.


3 0
3 years ago
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