A pool player hit a ball with a mass of 0.25kg. The ball travels at a velocity of 15 m/s for 0.75 s until it hits the side of th
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Answer:
The work required is -515,872.5 J
Explanation:
Work is defined in physics as the force that is applied to a body to move it from one point to another.
The total work W done on an object to move from one position A to another B is equal to the change in the kinetic energy of the object. That is, work is also defined as the change in the kinetic energy of an object.
Kinetic energy (Ec) depends on the mass and speed of the body. This energy is calculated by the expression:
where kinetic energy is measured in Joules (J), mass in kilograms (kg), and velocity in meters per second (m/s).
The work (W) of this force is equal to the difference between the final value and the initial value of the kinetic energy of the particle:
In this case:
- W=?
- m= 2,145 kg
- v2= 12
- v1= 25
Replacing:
W= -515,872.5 J
<u><em>The work required is -515,872.5 J</em></u>
Answer:
a) v = 0.7071 v₀, b) v= v₀, c) v = 0.577 v₀, d) v = 1.41 v₀, e) v = 0.447 v₀
Explanation:
The speed of a wave along an eta string given by the expression
v =
where T is the tension of the string and μ is linear density
a) the mass of the cable is double
m = 2m₀
let's find the new linear density
μ = m / l
iinitial density
μ₀ = m₀ / l
final density
μ = 2m₀ / lo
μ = 2 μ₀
we substitute in the equation for the velocity
initial v₀ =
with the new dough
v =
v = 1 /√2 \sqrt{ \frac{T_o}{ \mu_o} }
v = 1 /√2 v₀
v = 0.7071 v₀
b) we double the length of the cable
If the cable also increases its mass, the relationship is maintained
μ = μ₀
in this case the speed does not change
c) the cable l = l₀ and m = 3m₀
we look for the density
μ = 3m₀ / l₀
μ = 3 m₀/l₀
μ = 3 μ₀
v =
v = 1 /√3 v₀
v = 0.577 v₀
d) l = 2l₀
μ = m₀ / 2l₀
μ = μ₀/ 2
v =
v = √2 v₀
v = 1.41 v₀
e) m = 10m₀ and l = 2l₀
we look for the density
μ = 10 m₀/2l₀
μ = 5 μ₀
we look for speed
v =
v = 1 /√5 v₀
v = 0.447 v₀