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4 years ago

Approximately what is the gravitational energy of a 100 kg person at the top of a 10 meter ladder?

Physics

1 answer:

____ [38]4 years ago

8 0

The gravitational energy is 9800 Joules.

Formula: Mass • Gravity • Height
100kg• 9.8m/s2•10m

=9800Joules

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Freely falling objects are only affected by..

Yanka [14]

they are only affected by gravity.

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3 years ago

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A 150 g egg is dropped from 3.0 meters. The egg is

Lynna [10]

Answer:

9.2 N, with significant figure rounding (2 s.f.)

Explanation:

This problem can be solved using momentum. The following equation relates momentum (mass & velocity) with force and time:

Note that where v is the final velocity and v₀ is the initial velocity. Δv just means change in velocity.

Mass of the egg is 150 g, but we need to convert to kilograms if we want to use Newtons as a unit. 150 g is equal to 0.15 kg. since 1000 g = 1kg.

m = 0.15 kg

The dropped from 3.0 meters is irrelevant as the question tells us the initial velocity of the egg: 4.4 m/s before it hits the ground.

v₀ = 4.4 m/s [down]

When it comes to a stop, the egg will have a velocity of 0.

v = 0 m/s

The time it takes for the egg to stop is 0.072 seconds.

Δt = 0.072 s

Therefore, if down is positive, then

We round to two significant figures since every quantity has two sig. figs.

We only care about the magnitude, not direction. The answer is 9.2 N.

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3 years ago

Plz, Help!!

SIZIF [17.4K]

C is the first & the second question is A

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3 years ago

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A Go Kart (m = 35 kg) has a top speed of 12 m/s . A student traveling at top speed locks the brake to avoid hitting a bus after

Sergeu [11.5K]

Answer:

350 N

Explanation:

From the question given above, the following data were obtained:

Mass (m) of Go Kart = 35 kg

Initial velocity (u) = 12 m/s

Distance (s) = 7.2 m

Force (F) =?

Next, we shall determine the acceleration of the Go Kart. This can be obtained as follow:

Initial velocity (u) = 12 m/s

Distance (s) = 7.2 m

Final velocity (v) = 0 m/s

Acceleration (a) =.?

v² = u² + 2as

0² = 12² + (2 × a × 7.2)

0 = 144 + 14.4a

Collect like terms

0 – 144 = 14.4a

– 144 = 14.4a

Divide both side by 14.4

a = – 144 / 14.4

a = – 10 m/s²

The negative sign indicate that the Go Kart is decelerating when the brake was applied.

Finally, we shall determine the force the Go Kart have when the student locked the brake. This can be obtained as follow:

Mass (m) of Go Kart = 35 kg

Acceleration (a) = 10 m/s

Force (F) =?

F = ma

F = 35 × 10

F = 350 N

Thus, the Go Kart has a force of 350 N when the student locked the brake.

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3 years ago

To pull a 50 kg crate across a horizontal frictionless floor, a worker applies a force of 210 N, directed 20° above the horizont

tangare [24]

Answer:

Explanation:

Given that,

A crate of mass M = 50kg

The crate is pulled along an horizontal floor by a string at an angle

Force pulling the crate

F = 210N

Angle θ = 20° to the horizontal

Distance moved by crate d=3m

A. Work done by force

W = FdCosθ

W = 210 × 3 × Cos20

W = 592J

B. Work done by Gravitational force?

Work is define as the dot product of force and displacement in the direction of the force.

Since the gravitational force does not cause the crate to move any distance downward

Then,

W(gravity) = mg×d

distance d=0

W(gravity) = 0 J

C. Work done by normal?

Since, the normal force did not cause the crate to move upward by any distance

Then,

W(normal) = 0J

D. The total workdone?

The only workdone on the crate is by the person

Then, the total workdone is the workdone by the person

W = 592 J

4 0

3 years ago