Kinetic energy = (1/2) (mass) x (speed)²
At 7.5 m/s, the object's KE is (1/2) (7.5) (7.5)² = 210.9375 joules
At 11.5 m/s, the object's KE is (1/2) (7.5) (11.5)² = 495.9375 joules
The additional energy needed to speed the object up from 7.5 m/s
to 11.5 m/s is (495.9375 - 210.9375) = <em>285 joules</em>.
That energy has to come from somewhere. Without friction, that's exactly
the amount of work that must be done to the object in order to raise its
speed by that much.
Answer:
a)
a = 2 [m/s^2]
b)
a = 1.6 [m/s^2]
c)
xt = 2100 [m]
Explanation:
In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.
a)
When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:

where:
Vf = final velocity = 40 [m/s]
Vi = initial velocity = 0 (starting from rest)
a = acceleration [m/s^2]
t = time = 20 [s]
40 = 0 + (a*20)
a = 2 [m/s^2]
The distance can be calculates as follows:

where:
x1 = distance [m]
40^2 = 0 + (2*2*x1)
x1 = 400 [m]
Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.
v = x2/t2
where:
x2 = distance [m]
t2 = 30 [s]
x2 = 40*30
x2 = 1200 [m]
b)
Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.

Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.
0 = 40 - (a *25)
a = 40/25
a = 1.6 [m/s^2]
The distance can be calculates as follows:

0 = (40^2) - (2*1.6*x3)
x3 = 500 [m]
c)
Now we sum all the distances calculated:
xt = x1 + x2 + x3
xt = 400 + 1200 + 500
xt = 2100 [m]
A. The particles are packed more tightly in materials with more density which causes the vibrations to bounce of the partials more rapidly which makes them go faster
To minimize neutron leakage from a reactor, the ratio of the surface area to the volume should be a minimum. For a given volume V the ratio of the sphere will be
.
We know that the surface area and volume of the sphere is given by:

Therefore, the ratio between the surface area and the volume for the sphere will be:

Equating the volume to the constant c, we will find the value of
.

Substituting the value of r in the ration between surface area and volume, we get:

Calculating the constants, we get:

Hence, the ration between surface area and volume is 
To learn more about surface area and volume of sphere, refer to:
brainly.com/question/4387241
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