Answer:
a) a = 3.72 m / s², b) a = -18.75 m / s²
Explanation:
a) Let's use kinematics to find the acceleration before the collision
v = v₀ + at
as part of rest the v₀ = 0
a = v / t
Let's reduce the magnitudes to the SI system
v = 115 km / h (1000 m / 1km) (1h / 3600s)
v = 31.94 m / s
v₂ = 60 km / h = 16.66 m / s
l
et's calculate
a = 31.94 / 8.58
a = 3.72 m / s²
b) For the operational average during the collision let's use the relationship between momentum and momentum
I = Δp
F Δt = m v_f - m v₀
F = 
F = m [16.66 - 31.94] / 0.815
F = m (-18.75)
Having the force let's use Newton's second law
F = m a
-18.75 m = m a
a = -18.75 m / s²
Answer:
The total electrical power we are using is: 1316 W.
Explanation:
Using the ohm´s law 
and the formula for calculate the electrical power, we can find the total electrical power that we are using. First we need to find each electrical power that is using every single component, so the radio power is:
, so the radio power is: 
, then we find the pop-corn machine power as: 
and finally there are three light bulbs of 110(W) so: P=3*110(W)=330(W) and the total electrical power is the adding up every single power so that: P=330(W)+770(W)+216(W)=1316(W).
I believe this would be an example of Mary's velocity. We have her speed and direction which is all you need to find velocity.
Answer: Released
Explanation: Energy is released in this reaction possibly in the form of heat thus it is an exergonic and or exothermic reaction.
Answer:
Explanation:
Let's analyze the situation presented in order to know which answer is correct.
When the stick collides with the puck, it exerts a force for a certain time and discants. / After this time the horizontal force decreases to zero and the disk continues to move by the action of the initial velocity on the x axis and the acceleration of gravity on the y axis.
Therefore, after the collision, the only force that acts on the disk is the gravitational attractive force (WEIGHT), directed on the axis and in a negative direction.
The correct answer is:
C) Since there is no frictional force exerted on the puck, a normal force is not exerted on the puck, but the gravitational force is exerted on the puck
