Answer:
9) a = 25 [m/s^2], t = 4 [s]
10) a = 0.0875 [m/s^2], t = 34.3 [s]
11) t = 32 [s]
Explanation:
To solve this problem we must use kinematics equations. In this way we have:
9)
a)

where:
Vf = final velocity = 0
Vi = initial velocity = 100 [m/s]
a = acceleration [m/s^2]
x = distance = 200 [m]
Note: the final speed is zero, as the car stops completely when it stops. The negative sign of the equation means that the car loses speed or slows down as it stops.
0 = (100)^2 - (2*a*200)
a = 25 [m/s^2]
b)
Now using the following equation:

0 = 100 - (25*t)
t = 4 [s]
10)
a)
To solve this problem we must use kinematics equations. In this way we have:

Note: The positive sign of the equation means that the car increases his speed.
5^2 = 2^2 + 2*a*(125 - 5)
25 - 4 = 2*a* (120)
a = 0.0875 [m/s^2]
b)
Now using the following equation:

5 = 2 + 0.0875*t
3 = 0.0875*t
t = 34.3 [s]
11)
To solve this problem we must use kinematics equations. In this way we have:

10^2 = 2^2 + 2*a*(200 - 10)
100 - 4 = 2*a* (190)
a = 0.25 [m/s^2]
Now using the following equation:

10 = 2 + 0.25*t
8 = 0.25*t
t = 32 [s]
Answer:
A book on its side exerts a greater force.
Explanation:
Pressure = Force / Area
Assuming that 1kg = 10N
2kg = 20N
Area of book lying flat = 0.3m × 0.2m
= 0.6m²
Pressure of book lying flat = 20N / 0.6m²
= 30Pa (1 s.f.)
Area of book on its side = 0.2m × 0.05m
= 0.01m²
Pressure of book on its side = 20N / 0.01m²
= 2000Pa (1 s.f.)
Since 2000Pa (1 s.f.) > 30Pa (1 s.f.), a book on its side applies greater pressure than lying flat.
Thew energy stored in a capacitor of capacitance
and voltage between the plates
is
.
Substituting numerical value

Answer:
1. 
2. 
Explanation:
1. According to Newton's law of motion, the puck motion is affected by the acceleration, which is generated by the push force F.
In Newton's 2nd law: F = ma
where m is the mass of the object and a is the resulted acceleration. So in the 2nd experiment, if we double the mass, a would be reduced by half.

Since the puck start from rest, in the 1st experiment, to achieve speed of v it would take t time

Now that acceleration is halved:


You would need to push for twice amount of time 
2. The distance traveled by the puck is as the following equation:

So if the acceleration is halved while maintaining the same d:

As
, then
. Also 



So t increased by 1.14