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klemol [59]
3 years ago
15

On the surface of Earth, a spacecraft has a mass of 2.00 x 104 kilograms. What is the mass of the spacecraft at a distance of on

e Earth radius above Earth's surface?
Physics
1 answer:
natta225 [31]3 years ago
4 0

Answer:

Mass will be same as 2\times 10^4kg

Explanation:

We have given mass of the spacecraft m=2\times 10^4kg kg on te earth

We have to find the the mass of the spacecraft at a distance one earth radius from the surface of earth

As we know that mass is always conserved it does nit depend on the gravity when spacecraft goes above the earth only acceleration due to gravity is change and it has no effect on mass

So mass will remain same

So mass will be 2\times 10^4kg

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At the same moment, one rock is dropped and one is thrown downward with an initial velocity of 29m/s from the top of a building
Inessa [10]

Answer:

The thrown rock strike 2.42 seconds earlier.

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This is an uniformly accelerated motion problem, so in order to find the arrival time we will use the following formula:

x=vo*t+\frac{1}{2} a*t^2\\where\\x=distance\\vo=initial velocity\\a=acceleration

So now we have an equation and unkown value.

for the thrown rock

\frac{1}{2}(9.8)*t^2+29*t-300=0

for the dropped rock

\frac{1}{2}(9.8)*t^2+0*t-300=0

solving both equation with the quadratic formula:

\frac{-b\±\sqrt{b^2-4*a*c} }{2*a}

we have:

the thrown rock arrives on t=5.4 sec

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6 0
3 years ago
Which of the following options is correct and why?
Dimas [21]

Answer:

Option (e) = The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere.

Explanation:

So, we are given the following set of infomation in the question given above;

=> "spherical Gaussian surface of radius R centered at the origin."

=> " A charge Q is placed inside the sphere."

So, the question is that if we are to maximize the magnitude of the flux of the electric field through the Gaussian surface, the charge should be located where?

The CORRECT option (e) that is " The charge can be located anywhere since flux does not depend on the position of the charge as long as it is inside the sphere." Is correct because of the reason given below;

REASON: because the charge is "covered" and the position is unknown, the flux will continue to be constant.

Also, the Equation that defines Gauss' law does not specify the position that the charge needs to be located, therefore it can be anywhere.

6 0
3 years ago
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