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Anton [14]
3 years ago
13

What is the magnitude of g at a height above Earth's surface where free-fall acceleration equals 6.5m/s^2?

Physics
1 answer:
Tom [10]3 years ago
4 0

The magnitude of 'g' at a place is expressed in terms of
the acceleration due to gravity.

You have said that the gravitational acceleration at that place
is 6.5 m/s².  So THAT's the magnitude of 'g' right there.

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PLZ EXPLAIN IM SO CONFUSED AND THIS IS DUE TONIGHT. I WILL GIVE 50 POINTS!
bezimeni [28]

When you first pull back on the pendulum, and when you pull it back really high the Potential Energy is high and the Kinetic Energy is low, But when up let go, and it gets right around the middle, that's when the Potential energy transfers to Kinetic, at that point the kinetic Energy is high and the potential Energy is low. But when it comes back up at the end. The same thing will happen, the Potential Energy is high, and the Kinetic Energy is low. Through all of that the Mechanical Energy stays the same. 

I hope this helps. :)

Brainliest?

8 0
3 years ago
Can anyone check if my answer is correct ?
ohaa [14]

I believe your answer is correct, because 8.7*10^-7 is equal to 0.00000085347.

Hope you do well!

4 0
3 years ago
A 51-kg woman runs up a vertical flight of stairs in 5.0 s. Her net upward displacement is 5.0 m. Approximately, what average po
Vika [28.1K]

Answer:

The average power the woman exerts is 0.5 kW

Explanation:

We note that power, P = The rate at which work is done = Work/Time

Work = Energy

The total work done is the potential energy gained which is the energy due to vertical displacement

Given that the vertical displacement = 5.0 m, we have

Total work done = Potential energy gained = Mass, m × Acceleration due to gravity, g × Vertical height, h

m = 51 kg

g = Constant = 9.81 m/s²

h = 5.0 m

Also, time, t = 5.0 s

Total work done = 51 kg × 9.81 m/s²× 5 m = 2501.55 kg·m²/s² = 2501.55 J

P = 2501.55 J/(5 s) = 500.31 J/s = 500.31 W ≈ 500 W = 0.5 kW.

6 0
3 years ago
A boat radioed a distress call to a Coast Guard station. At the time of the call, a vector A from the station to the boat had a
VashaNatasha [74]

Answer:

d = 39.7 km

Explanation:

initial position of the boat is 45 km away at an angle of 15 degree East of North

so we will have

r_1 = 45 sin15 \hat i + 45 cos15 \hat j

r_1 = 11.64 \hat i + 43.46\hat j

after some time the final position of the boat is found at 30 km at 15 Degree North of East

so we have

r_2 = 30 cos15\hat i + 30 sin15 \hat j

r_2 = 28.98\hat i + 7.76 \hat j

now the displacement of the boat is given as

d = r_2 - r_1

d = (28.98\hat i + 7.76 \hat j) - (11.64 \hat i + 43.46\hat j)

d = 17.34 \hat i - 35.7 \hat j

so the magnitude is given as

d = \sqrt{17.34^2 + 35.7^2}

d = 39.7 km

4 0
3 years ago
2. Can you place three forces of 5g, 6g, and 12g so they are in equilibrium. Justify your answer.
Bond [772]

Answer:

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

Explanation:

Equilibrium means their sum must be zero.

Here the forces are 5g, 6g, and 12g.

For number of forces to be in equilibrium the magnitude of largest vector should be less than sum of the magnitude of other vectors.

Here

        Magnitude of largest force = 12 g

        Sum of magnitudes of other forces = 5g + 6g = 11g

       Magnitude of largest force >   Sum of magnitudes of other forces

So this forces cannot form equilibrium.

We cannot place three forces of 5g, 6g, and 12g in equilibrium.

4 0
3 years ago
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