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user100 [1]
3 years ago
11

Please help!! giving a lot of points

Physics
1 answer:
Readme [11.4K]3 years ago
7 0

Question 1.

  • mass = 4500 kg
  • potential energy (p.e) = 67500 J

now, we know :

=》

p.e =  mgh

=》

67500 = 4500 \times 10 \times h

=》

67500 = 45000 \times h

=》

h =  \dfrac{67500}{45000}

=》

h = 1.5 \: m

note : if we take acceleration due to gravity as 9.8, then height = 1.53 m

Question 2.

  • mass = 4500 kg
  • kinetic energy = 63000 j

we know,

=》

k.e =  \dfrac{1}{2} mv {}^{2}

=》

63000 =  \dfrac{1}{2}  \times 4500 \times  {v}^{2}

=》

{v}^{2}  =  \dfrac{63000 \times 2}{4500}

=》

{v}^{2}  = 28

=》

v =  \sqrt{28}

=》

v = 2 \sqrt{7} \:  \:  ms {}^{ - 1}

or

=》

5.29 \:  \: ms {}^{ - 1}

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A 28.0 kg child plays on a swing having support ropes that are 2.30 m long. A friend pulls her back until the ropes are 45.0 ∘ f
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Answer

A)184.9J

B)=3.63m/s

C) Zero

Explanation:

A)potential energy of the child at the initial position, measured relative the her potential energy at the bottom of the motion, is

U=Mgh

Where m=28kg

g= 9.8m/s

h= difference in height between the initial position and the bottom position

We are told that the rope is L = 2.30 m long and inclined at 45.0° from the vertical

h=L-Lcos(x)= L(1-cosx)=2.30(1-cos45)

=0.674m

Her Potential Energy will now

= 28× 9.8×0.674

=184.9J

B)we can see that at the bottom of the motion, all the initial potential energy of the child has been converted into kinetic energy:

E= 0.5mv^2

where

m = 28.0 kg is the mass of the child

v is the speed of the child at the bottom position

Solving the equation for v, we find

V=√2k/m

V=√(2×184.9/28

=3.63m/s

C)we can find work done by the tension in the rope is given using expresion below

W= Tdcosx

where W= work done

T is the tension

d = displacement of the child

x= angle between the directions of T and d

In this situation, we have that the tension in the rope, T, is always perpendicular to the displacement of the child, d. x= 90∘ and cos90∘=0 hence, the work done is zero.

6 0
3 years ago
What happens when the elevator is in free-fall, that is, what is the value for FN and what is the sensation experienced by the p
Komok [63]

Answer:

Weightlessness

Explanation:

When the elevator is in free fall, this can only occur when the cord of the elevator breaks.

The acceleration of the elevator will be equal to the acceleration due to gravity. That is:

a = g

Then, the body will experience what we called WEIGHTLESSNESS

Where the normal reaction N of the person will tend to zero. That is

N = 0

The value of FN < 0 because the person inside the lift will experience weightlessness.

7 0
3 years ago
A 0.0780 kg lemming runs off a
attashe74 [19]

Answer:

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Explanation:

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its potential energy (PE) when it lands

The potential energy PE, relative to the ground, will be zero, because the lemming is at the ground level.

HOWEVER, a much better question would be:

A 0.0780 kg lemming runs off a 5.36 m high cliff at 4.84 m/s. What is its kinetic energy (KE) when it lands?

Let’s review the 4 basic kinematic equations of motion for constant acceleration (this is a lesson – suggest you commit these to memory):

s = ut + ½at^2 …. (1)

v^2 = u^2 + 2as …. (2)

v = u + at …. (3)

s = (u + v)t/2 …. (4)

where s is distance, u is initial velocity, v is final velocity, a is acceleration and t is time.

In this case, u = 0, a = 9.81m/s^2, s = 5.36m

So we find v using equation (2)

v^2 = u^2 + 2as

v^2 = 0 + 2(9.81)(5.36) = 105.1632

So the kinetic energy resulting from the vertical drop is Ekv = ½mv^2

Ekv = ½(0.078)(105.16) = 4.101J

BUT we need to add in the kinetic energy resulting from the horizontal velocity, which did not change during the vertical drop.

Ekh = ½(0.078)(4.84^2) = 0.914J

So the total kinetic energy is Ek = Ekv + Ekh = 4.101 + 0.914 = 5.015J

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An ice skater applies a horizontal force to a 150 N block on frictionless, level ice, causing the block to accelerate uniformly
Molodets [167]

Answer:

<em>The friction force has a magnitude of 54 N and points to the left</em>

Explanation:

<u>Friction Force</u>

When an object is moving and encounters friction in the air or rough surfaces, it loses acceleration because the friction force opposes motion.

The friction force when an object is moving on a horizontal surface is calculated by:

Fr=\mu N

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The friction force is:

Fr=0.36*150

Fr = 54 N

The friction force has a magnitude of 54 N and points to the left

5 0
3 years ago
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