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asambeis [7]
3 years ago
7

If a 50 kg student is standing on the edge of a cliff. Find the student’s gravitational potential energy if the cliff is 80 m hi

gh.
Physics
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

Student’s(50 kg) gravitational potential energy if the cliff is 80 m high = 39.2 kJ

Explanation:

   Potential energy is given by the expression, PE =mgh, m is the mass, g is the acceleration due to gravity value and h is the height.

 Here mass of student, m = 50 kilogram

           Height of cliff , h = 80 meter

           Acceleration due to gravity = 9.8m/s^2

 Potential energy, PE = 50 * 9.8 * 80 = 39200 J

                                    = 39.2 kJ

Student’s(50 kg) gravitational potential energy if the cliff is 80 m high = 39.2 kJ

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If we add 50 Joules of thermal energy to a heat engine, and that heat engine does 30 Joules of work, how much thermal energy is
Natalka [10]

Answer:

The correct answer should be

A. 20 Joules

Explanation:

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7 0
2 years ago
You stand on a merry-go-round which is spinning at f = 0.25 revolutions per second. You are R = 200 cm from the center. (a) Find
wariber [46]

Answer:

a) ω = 9.86 rad/s

b) ac = 194. 4 m/s²

c) minimum coefficient of static friction, µs = 19.8

Explanation:

a) angular speed, ω = 2πf, where f is frequency of revolution

1 rps = 6.283 rad/s, π = 3.142

ω = 2 * 3.14 * 0.25 * 6.28

ω = 9.86 rad/s

b) centripetal acceleration, a = rω²

where r is radius in meters; r = 200 cm or 2 m

a = 2 * 9.86²

a = 194. 4 m/s²

c) µs = frictional force/ normal force

frictional force = centripetal force = ma; where a is centripetal acceleration

normal force = mg; where g = 9.8 m/s²

µs = ma/mg = a/g

µs = 194.4 ms⁻²/9.8 ms⁻²

c) minimum coefficient of static friction, µs = 19.8

5 0
2 years ago
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Tems11 [23]
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3 years ago
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Do the math: How many seconds would it take an echo sounder’s ping to make the trip from a ship to the Challenger Deep (10,994 m
Lera25 [3.4K]

Answer:

<h2>14.66secs</h2>

Explanation:

Given the formula for calculating the depth in metres expressed as

depth in meters = ½ (1500 m/sec × Echo travel time in seconds)

Given depth of the challenger = 10, 994 meters, we will substitute this given value into the formula given to calculate the time take for the echo to travel.

10, 994 = depth in meters = ½ * 1500 m/sec × Echo travel time in seconds

10,994 = 750 * Echo travel time in seconds

Dividing both sides by 750;

Echo travel time in seconds = 10,994 /750

Echo travel time in seconds ≈ 14.66secs (to two decimal places)

Therefore, it would take an echo sounder’s ping 14.66secs to make the trip from a ship to the Challenger Deep and back

6 0
2 years ago
in the derivation of the time period of a pendulum in electric field when considering the fbd of bob to find the g effective why
Neko [114]

Answer:

we learned that an object that is vibrating is acted upon by a restoring force. The restoring force causes the vibrating object to slow down as it moves away from the equilibrium position and to speed up as it approaches the equilibrium position. It is this restoring force that is responsible for the vibration. So what forces act upon a pendulum bob? And what is the restoring force for a pendulum? There are two dominant forces acting upon a pendulum bob at all times during the course of its motion. There is the force of gravity that acts downward upon the bob. It results from the Earth's mass attracting the mass of the bob. And there is a tension force acting upward and towards the pivot point of the pendulum. The tension force results from the string pulling upon the bob of the pendulum. In our discussion, we will ignore the influence of air resistance - a third force that always opposes the motion of the bob as it swings to and fro. The air resistance force is relatively weak compared to the two dominant forces.

The gravity force is highly predictable; it is always in the same direction (down) and always of the same magnitude - mass*9.8 N/kg. The tension force is considerably less predictable. Both its direction and its magnitude change as the bob swings to and fro. The direction of the tension force is always towards the pivot point. So as the bob swings to the left of its equilibrium position, the tension force is at an angle - directed upwards and to the right. And as the bob swings to the right of its equilibrium position, the tension is directed upwards and to the left. The diagram below depicts the direction of these two forces at five different positions over the course of the pendulum's path.

that's what I know so far

8 0
3 years ago
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