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liubo4ka [24]
3 years ago
14

Obviously, we can make rockets to go very fast, but what is a reasonable top speed? assume that a rocket is fired from rest at a

space station in deep space, where gravity is negligible. assume that c = 3.00 108 m/s.
Physics
1 answer:
steposvetlana [31]3 years ago
3 0

 

I’ve answered this problem before and there were 2 parts in this problem.

The solution would be like this for this specific problem:

<span>A.    </span><span>Vf = Vi + Vex*ln(Mi / Mf) </span><span>
<span>0.002 * 3e8m/s = 0 + 2000m/s * ln(Mi / Mf) </span>
<span>300 = ln(Mi / Mf) </span>
<span>1.9e130 = Mi / Mf </span></span>

 

<span>B.    </span><span>4000m/s = 2000m/s * ln(Mi / Mf) </span><span>
<span>2 = ln(Mi / Mf) </span>
<span>7.389 = Mi / Mf </span>
<span>Mf = Mi / 7.389 = 0.135*Mi<span> </span></span></span>

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Answer:

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4 0
3 years ago
3. Which planet, other than Earth, is confirmed to have water on it?
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Mars.

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2 years ago
A force of 42N is needed to start a box sliding across the floor. The weight of the box is 55N.
Bezzdna [24]
I think the right answer is c
4 0
3 years ago
Read 2 more answers
if gas in a sealed container has a pressure of 50kpa at 300k what will the pressure be if the tempature rises to 360k
Zarrin [17]
The pressure law states that pressure is directly proportional to temperature.
p=kt where p is pressure, k is a constant, and t is temperature.

p=kt -- substitute
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3 0
3 years ago
A spring with a spring constant of 50 N/m is stretched 15cm. What is the force and energy associated with this stretching?
Olenka [21]
Data:
F (force) = ? (Newton)
k (<span>Constant spring force) = 50 N/m
x (</span>Spring deformation) = 15 cm → 0.15 m

Formula:
F = k*x

Solving: 
F = k*x
F = 50*0.15
\boxed{\boxed{F = 7.5\:N}}\end{array}}\qquad\quad\checkmark

Data:
E (energy) = ? (joule)
k (Constant spring force) = 50 N/m
x (Spring deformation) = 15 cm → 0.15 m

Formula:
E = \frac{k*x^2}{2}

Solving:(Energy associated with this stretching)
E = \frac{k*x^2}{2}
E =  \frac{50*0.15^2}{2}
E =  \frac{50*0.0225}{2}
E =  \frac{1.125}{2}
\boxed{\boxed{E = 0.5625\:J}}\end{array}}\qquad\quad\checkmark

7 0
3 years ago
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