So momentum is just velocity times mass, this means Momentum = Velocity x Mass.
We can rearrange this to be Velocity = Momentum/Mass.
Since we know momentum and mass we can now solve.
Velocity = 264/(45+2.5)
= 5.56 m/s
Answer:
a = 10.07m/s^2
Their acceleration in meters per second squared is 10.07m/s^2
Explanation:
Acceleration is the change in velocity per unit time
a = ∆v/t
Given;
∆v = 50.0miles/hour - 0
∆v = 50.0miles/hours × 1609.344 metres/mile × 1/3600 seconds/hour
∆v = 22.352m/s
t = 2.22 s
So,
Acceleration a = ∆v/t = 22.352m/s ÷ 2.22s
a = 10.07m/s^2
Their acceleration in meters per second squared is 10.07m/s^2
The same amount of force that the cart is using nothing more nothing less
To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

The kinetic energy can be written as,

Where,
Gravitational Universal Constant
Mass of Earth
Height
Radius of Earth
From the conservation of energy:

Rearranging to find the velocity,
Escape velocity at a certain height from the earth
If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,


Replacing the values we have that


Therefore the escape velocity is 3.6km/s
This question is incomplete, the complete question is;
A block of mass m begins at rest at the top of a ramp at elevation h with whatever PE is associated with that height. The block slides down the ramp over a distance d until it reaches the bottom of the ramp.
How much of its original total energy (in J) survives as KE when it reaches the ground? m = 9.9 kg h = 4.9 m d = 5 m μ = 0.3 θ = 36.87°
Answer:
the amount of its original total energy (in J) that survives as KE when it reaches the ground will is 358.975 J
Explanation:
Given that;
m = 9.9 kg
h = 4.9 m
d = 5 m
μ = 0.3
θ = 36.87°
Now from conservation of energy, the energy is;
Et = mgh
we substitute
Et = 9.9 × 9.8 × 4.9
= 475.398 J
Also the loss of energy i
E_loss = (umg cosθ) d
we substitute
E_loss = 0.3 × 9.9 × 9.8 × cos36.87° × 5
= 116.423 J
so the amount of its original total energy (in J) that survives as KE when it reaches the ground will be
E = Et - E_loss
E = 475.398 J - 116.423 J
E = 358.975 J