Answer:
n the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
Explanation:
Velocity is a vector therefore it has magnitude and direction, a change in either of the two is the consequence of an acceleration on the system.
In the case of linear motion, the change occurs in the magnitude of the velocity, the direction remaining constant.
= (v₂-v₁)/Δt
In the case of circular motion, the magnitude of the velocity remains constant, the change in its direction occurring.
= v2/R
In the general case, both the module and the address change
a = Ra ( a_{t}^2 + a_{c}^2)
Answer:
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Answer:
k = 1 700.7 N/m
v0 = 9.8 m/s^2
Explanation:
Hello!
We can answer this question using conservation of energy.
The potential energy of the spring (PS) will transform to kinetic energy (KE) of the ball, and eventually, when the velocity of the ball is zero, all that energy will be potential gravitational (PG) energy.
When the kinetic energy of the ball is zero, that is, when it has reached its maximum heigh, all the potential energy of the spring will be equal to the potential energy of the gravitational field.
PS = (1/2) k x^2 <em>where x is the compresion or elongation of the spring</em>
PG = mgh
a)
Since energy must be conserved and we are neglecting any energy loss:
PS = PG
Solving for k
k = (2mgh)/(x^2) = ( 2 * 1.7 * 9.81 * 4.9 Nm)/(0.31^2 m^2)
k = 1 700.7 N/m
b)
Since the potential energy of the spring transfors to kinetic energy of the ball we have that:
PS = KE
that is:
(1/2) k x^2 = (1/2) m v0^2
Solving for v0
v0 = x √(k/m) = (0.31 m ) √( 1 700.7 N/m / 1.7kg)
v0 = 9.8 m/s^2
The answers are 3.3kWh, 1.2kWh, and 120kWh.
Multiply the amount of kW used per hour by the amount of time it was powered and you receive your answer.
