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igomit [66]
3 years ago
5

Mars has a surface gravity that is only 0.38 (a little over one-third) that of Earth's. Thus, the weight of an object will also

be 0.38 times that of Earth. What is the G.P.E. of a climber who weighs 675 newtons on Earth at the top of 26,000-meter Olympus Mons?
Physics
1 answer:
Gemiola [76]3 years ago
3 0

Answer:

Potential energy on surface of Mars is 6669000 J

Explanation:

As we know that the weight of the climber on the surface of Earth is given as 675 N

So weight of the climber on the Mars is given as

W = 0.38 (675)

W = mg = 256.5 N

Now we know that gravitational Potential energy is given as

U = mgH

U = (256.5)(26000)

U = 6669000 J

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4 years ago
To test the hypothesis that the population mean mu=3. 6, a sample size n=14 yields a sample mean 4. 007 and sample standard devi
PolarNik [594]

The P value for the given data set is 25127. For finding P value, we have to must find the Z value.

<h3>How to get the z scores?</h3>

If we've got a normal distribution, then we can convert it to standard normal distribution and its values will give us the z score.

The Z value is calculated as;

Z = \dfrac{X - \mu}{\sigma})

Z = (X - μ) / σ

Z = (4.007 - 3.6) / 0.607

Z = 0.67051

The P value for the given data set is 25127.

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5 0
3 years ago
A car moving in a straight line starts at X=0 at t=0. It passesthe point x=25.0 m with a speed of 11.0 m/s at t=3.0 s. It passes
Agata [3.3K]

Answer:

Average velocity v = 21.18 m/s

Average acceleration a = 2 m/s^2

Explanation:

Average speed equals the total distance travelled divided by the total time taken.

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

Average acceleration equals the change in velocity divided by change in time.

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

Where;

v1 and v2 are velocities at time t1 and t2 respectively.

And x1 and x2 are positions at time t1 and t2 respectively.

Given;

t1 = 3.0s

t2 = 20.0s

v1 = 11 m/s

v2 = 45 m/s

x1 = 25 m

x2 = 385 m

Substituting the values;

Average speed v = ∆x/∆t = (x2-x1)/(t2-t1)

v = (385-25)/(20-3)

v = 21.18 m/s

Average acceleration a = ∆v/∆t = (v2-v1)/(t2-t1)

a = (45-11)/(20-3)

a = 2 m/s^2

8 0
4 years ago
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