Answer:
1.512 m/s
Explanation:
M₁ = 10.5
M₂ = 61.5kg
v₁ = 8.85m/s
v₂ = ?
This question involves conservation of momentum and law of conservation of momentum states that, the total momentum of a system does not change before and after impact and it's always constant.
Momentum = mass * velocity = M*V
total momentum of the system before caught the ball = total momentum after he caught the ball.
M₁V₁ = M₂V₂
10.5 * 8.85 = 61.5 * V₂
V₂ = 92.925 / 61.5
V₂ = 1.5109m/s ≈ 1.512m/s
Power is the rate at which work is being done.
Answer: The question is incomplete or some details are missing. Here is the complete question ; (a) The driver of a car slams on the brakes when he sees a tree blocking the road. The car slows uniformly with acceleration of −5.55 m/s2 for 4.05 s, making straight skid marks 63.0 m long, all the way to the tree. With what speed (in m/s) does the car then strike the tree? m/s
(b) What If? If the car has the same initial velocity, and if the driver slams on the brakes at the same distance from the tree, then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2
a ) With what speed (in m/s) does the car then strike the tree? m/s = 4.3125m/s
b) then what would the acceleration need to be (in m/s2) so that the car narrowly avoids a collision? m/s2 = -5.696m/s2
Explanation:
The detailed steps and calculation is as shown in the attached file.
Answer:
1752.14 tonnes per year.
Explanation:
To solve this exercise it is necessary to apply the concepts related to power consumption and power production.
By conservation of energy we know that:

Where,
Production of Power
Consumption of power
Where the production of power would be,

Where,
m = Total mass required
Energy per Kilogram
Efficiency
The problem gives us the aforementioned values under a production efficiency of 45%, that is,


Replacing the values we have,

Solving for m,


We have the mass in kilograms and the time in seconds, we need to transform this to tons per year, then,

tonnes per year.
As wavelength increase, frequency decrease