A car accelerates from rest at 4.7m/s^2. How much time does it need to attain a speed of 6m/s?

This is a form of energy representing the motion of the molecules which make up an object. A. Thermal Energy B. Kinetic Energy C

bija089 [108]

Answer:

Kinetic energy.

Explanation:

There are many kinds of energy. Some of them are kinetic energy, potential energy, thermal energy etc.
The energy that shows the motion of the object is called its kinetic energy.
Also, the sum of kinetic energy and the gravitational potential energy is called mechanical energy.
Out of the given options, kinetic energy is the form of energy that represents the motion of the molecules which make up an object.
Hence, the correct option is (B).

4 0

3 years ago

A kangaroo can jump over an object 2.46 m high. (a) Calculate its vertical speed (in m/s) when it leaves the ground.

Elenna [48]

Answer:

a) 6.95 m/s

b) 1.42 seconds

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow -u^2=2\times -9.81\times 2.46-0^2\\\Rightarrow u=\sqrt{2\times 9.81\times 2.46}\\\Rightarrow u=6.95\ m/s$

a) The vertical speed when it leaves the ground. is 6.95 m/s

$v=u+at\\\Rightarrow t=\frac{v-u}{a}\\\Rightarrow t=\frac{0-6.95}{-9.81}\\\Rightarrow t=0.71\ s$

Time taken to reach the maximum height is 0.71 seconds

$s=ut+\frac{1}{2}at^2\\\Rightarrow 2.46=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{2.46\times 2}{9.81}}\\\Rightarrow t=0.71\ s$

Time taken to reach the ground from the maximum height is 0.71 seconds

b) Time it stayed in the air is 0.71+0.71 = 1.42 seconds

3 0

3 years ago

How do particles move differently in transverse waves and in surface water waves?

elixir [45]

The particles always move parallel and perpendicular to the waves. The waves which are in the water moves a circle. Both up and down and back and forth.

Good luck :)

3 0

3 years ago

A child is holding a wagon from rolling straight back down in a driveway that inclined at 20 degree horizontal. if the wagon wei

Alenkinab [10]

Answer:

F = 51.3°

Explanation:

The component of weight parallel to the inclined plane must be responsible for the rolling back motion of the car. Hence, the force required to be applied by the child must also be equal to that component of weight:

$F = Parallel\ Component\ of\ Weight\ of\ Wagon= WSin\theta\\$

where,

W = Weight of Wagon = 150 N

θ = Angle of Inclinition = 20°

Therefore,

$F = (150\ N)Sin\ 20^o$

8 0

2 years ago

A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (

kvv77 [185]

Answer:

v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

Em₀ = K = ½ m v²

final point. Higher

$Em_{f}$ = U = m g h

Let's use trigonometry to lock her up

cos 60 = y / L

y = L cos 60

Height is the initial length minus the length at the maximum angle

h = L - L cos 60

h = L (1- cos 60)

energy is conserved

Em₀ = Em_{f}

½ m v² = mgL (1 - cos 60)

v = 2g L (1- cos 60)

let's calculate

v² = 2 9.8 3.0 (1- cos 60)

v = 29.4 m / s

6 0

3 years ago