Answer: Escaped volume = 0.0612m^3
Explanation:
According to Boyle's law
P1V1 = P2V2
P1 = initial pressure in the tire = 36.0psi + 14.696psi = 50.696psi (guage + atmospheric pressure)
P2 = atmospheric pressure= 14.696psi
V1 = volume of tire =0.025m^3
V2 = escaped volume + V1 ( since air still remain in the tire)
V2 = P1V1/P2
V2 = 50.696×0.025/14.696
V2 = 0.0862m^3
Escaped volume = 0.0862 - 0.025 = 0.0612m^3
(a) This is a freefall problem in disguise - when the ball returns to its original position, it will be going at the same speed but in the opposite direction. So the ball's final velocity is the negative of its initial velocity.
Recall that

We have
, so that

(b) The speed of the ball at the start and at the end of the roll are the same 8 m/s, so the average speed is also 8 m/s.
(c) The ball's average velocity is 0. Average velocity is given by
, and we know that
.
(d) The position of the ball
at time
is given by

Take the starting position to be the origin,
. Then after 6 seconds,

so the ball is 42 m away from where it started.
We're not asked to say in which direction it's moving at this point, but just out of curiosity we can determine that too:

Since the velocity is positive, the ball is still moving up the incline.
Answer:
t = 16.94 s
Explanation:
t is the time passes before police catch the speeder
speed of speeder Vo = V = 23.3 m/s
T = t
Police Info
Vo = 0 m/s
a = 2.75 m/s^2
t = t
Now,
displacement of the police car = displacement of the speeder.
x_{police} = Vo *t + 1/2 at^2
since Vo = 0
x police = 1/2 at^2
x police = 1/2 (2.75)(t)^2
Now the displacement of speeder is
x_{speeder} = Vt
x_{speeder} = 23.3 t
x_{speeder} = x_{police}
23.3 t = 1/2 * 2.75 t^2
23.3 t = 1.375 t^2
t = 23.3\1.375
t = 16.94
t = 16.94 s
Answer:
6 Newtons to the left.
Explanation:
We can convert this into a generic algebra equation by giving directions positive and negative values.
The 6 will be positive, and the 10 and 2 will be negative.
Add 10 and 2 to have 12.
6-12 = -6.
Therefore you have 6 newtons to the left (negative).