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amm1812
4 years ago
9

Imagine that you are in a space station in the vacuum of outer space, far from any planet. There is no air in space, so there is

no friction. There is also very little gravity. You throw a ball out the door of the space station. Use Newton's first law of motion to predict what will happen to the motion of the ball.
Physics
1 answer:
Veronika [31]4 years ago
5 0

Answer:

The ball will continue to move at a constant speed forever until another force stops it.

Explanation:

Newton's first law of motion can be seen as a law of inertia. It explains that an object at rest or in a state of uniform motion will remain in that state unless it is acted upon by an external force.

Following the above, when the ball is thrown into space, the ball will continue to move with the velocity with which it was thrown until it comes in contact with another object that stops it. If this does not happen, it will continue to drift forever at that velocity.

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I believe it’s b. :D
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HELP, what are 2 chemical properties of Tungsten.
GarryVolchara [31]

Answer:

Molybdenum and seaborgium

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What is the largest Planet known to man?
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It is the Jupiter from outer space. And the years from 170 light years away from the Earth.
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4 years ago
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A point charge with a charge q1 = 2.30 μC is held stationary at the origin. A second point charge with a charge q2 = -5.00 μC mo
Alla [95]

Answer:

W = 2.74 J

Explanation:

The work done by the charge on the origin to the moving charge is equal to the difference in the potential energy of the charges.

This is the electrostatic equivalent of the work-energy theorem.

W = \Delta U = U_2 - U_1

where the potential energy is defined as follows

U = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r^2}

Let's first calculate the distance 'r' for both positions.

r_1 = \sqrt{(x_1 - x_0)^2 + (y_1 - y_0)^2} = \sqrt{(0.170 - 0)^2 + (0 - 0)^2} = 0.170~m\\r_2 = \sqrt{(x_2 - x_0)^2 + (y_2 - y_0)^2} = \sqrt{(0.250 - 0)^2 + (0.250 - 0)^2} = 0.353~m

Now, we can calculate the potential energies for both positions.

U_1 = \frac{kq_1q_2}{r_1^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.170)^2} = -3.57~J\\U_2 = \frac{kq_1q_2}{r_2^2} = \frac{(8.99\times 10^9)(2.3\times 10^{-6})(-5\times 10^{-6})}{(0.3530)^2} = -0.829~J

Finally, the total work done on the moving particle can be calculated.

W = U_2 - U_1 = (-0.829) - (-3.57) = 2.74~J

4 0
3 years ago
Read 2 more answers
Gold has a specific heat of 0.130 J/g*C. If 195 joules of heat are added to 15 grams of gold how much does the temperature of th
Anna71 [15]

The correct answer to the question is :  100\ ^0C

EXPLANATION :

As per the question, the specific heat of gold is given as c = 0.130\ J/g^0C

The heat given to the gold dQ = 195 J

The mass of the gold is given as m = 15 gram.

We are asked to calculate the change in temperature.

Let the change in temperature is dT.

We know that dQ = mcdT

                      dT=\ \frac{dQ}{mc}

                             =\ \frac{195}{15\times 0.130}

                             =\ 100\ ^0C                   [ANS]

Hence, the change in temperature is 100 degree celsius.

5 0
4 years ago
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