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3 years ago

A 37 N object is lifted to a height of 3 meters, What is the potential energy of this object?

Physics

2 answers:

prisoha [69]3 years ago

6 0

P.e = mgh

37N × 9.8ms⁻² ˣ 3m

= 1087.8 .

Lostsunrise [7]3 years ago

5 0

Answer:

Potential energy of the object is 111 Joules.

Explanation:

It is given that,

Force of gravity acting on the object, F =W=mg= 37 N, W is its weight

It is lifted to a height of 3 meters, h = 3 m

We need to find the potential energy of this object. It is given by :

$E=m\times g\times h$

Since, mg = 37 N

So,

$E=37\ N\times 3\ m$

E = 111 Joules

So, the potential energy of this object is 111 Joules. Hence, this is the required solution.

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Explanation:

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3 years ago

Explain how heat (thermal energy) can change the motion (kinetic energy) of the particles in objects.

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Answer:

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3 years ago

Classes are canceled due to snow, so you take advantage of the extra time to conduct some physics experiments. You fasten a larg

IRINA_888 [86]

Answer:

Time : 7.96 s

Distance Traveled : 357.8 m

Explanation:

In order to solve this problem, we first consider the accelerated motion of rocket. We will be using the subscript 1 for accelerated motion.

So, for accelerated motion, we have:

Acceleration = a₁ = 14.5 m/s²

Time Period = t₁ = 3.1 s

Initial Velocity = Vi₁ = 0 m/s (Since, it starts from rest)

Final Velocity = Vf₁

Distance covered by sled during acceleration motion = s₁

Now, using 1st equation of motion:

Vf₁ = Vi₁ + (a₁)(t₁)

Vf₁ = 0 m/s + (14.5 m/s²)(3.1 s)

Vf₁ = 44.95 m/s

Now, using 2nd equation of motion:

s₁ = (Vi₁)(t) + (0.5)(a₁)(t₁)

s₁ = (0 m/s)(3.1 s) + (0.5)(14.5 m/s²)(3.1 s)

s₁ = 22.5 m

Now, we first consider the decelerated motion of rocket. We will be using the subscript 2 for decelerated motion.

So, for accelerated motion, we have:

Deceleration = a₂ = - 5.65 m/s²

Time Period = t₂ = ?

Initial Velocity = Vi₂ = Vf₁ = 44.95 m/s (Since, decelerate motion starts, where accelerated motion ends)

Final Velocity = Vf₂ = 0 m/s (Since, rocket will eventually stop)

Distance covered by sled during deceleration motion = s₂

Now, using 1st equation of motion:

Vf₂ = Vi₂ + (a₂)(t₂)

0 m/s = 44.95 m/s + (- 5.65 m/s²)(t₂)

t₂ = (44.95 m/s)/(5.65 m/s²)

t₂ = 7.96 s

Now, using 2nd equation of motion:

s₂ = (Vi₂)(t₂) + (0.5)(a₂)(t₂)

s₂ = (44.95 m/s)(7.96 s) + (0.5)(- 5.65 m/s²)(7.96 s)

s₂ = 357.8 m - 22.5 m

s₂ = 335.3 m

Thus, the total distance covered by sled will be:

Total Dustance = S = s₁ + s₂

S = 22.5 m + 335.3 m

S = 357.8 m

7 0

3 years ago

The box leaves position x=0x=0 with speed v0v0. The box is slowed by a constant frictional force until it comes to rest at posit

const2013 [10]

Answer:

fr = ½ m v₀²/x

Explanation:

This exercise the body must be on a ramp so that a component of the weight is counteracted by the friction force.

The best way to solve this exercise is to use the energy work theorem

W = ΔK

Where work is defined as the product of force by distance

W = fr x cos 180

The angle is because the friction force opposes the movement

Δk = $K_{f}$ –K₀

ΔK = 0 - ½ m v₀²

We substitute

- fr x = - ½ m v₀²

fr = ½ m v₀²/x

8 0

3 years ago