As the volume of the container increases the pressure inside will decrease because the atoms have more room to move around in.
Answer:
Electric field at a distance of 1.45 cm will be 
Explanation:
We have given the distance d = 1.45 cm = 0.0145 m
And the potential difference 
There is a relation between potential difference and electric field
Electric field at a distance d due to a potential difference is given by
, here E is electric field, V is potential difference and d is distance
So 
Answer:
The charge on the ball bearing 4.507 × 10^-8 C
Explanation:
From Coulomb's law
F = kq1q2/r²
make q2 the subject
q2 = Fr²/kq1
q2 = (1.8×10^-2 × 0.026²) ÷ (9×10^9 × 30×10^-9)
q2 = 4.507 × 10^-8 C
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:
The force exerted on the
is 
Explanation:
From the question we are told that
The area is 
The magnitude of charge placed on them is 
The charge placed between the plate is 
The electric field generated around the plate is mathematically represented as

Substituting values


The force exerted the charge
is mathematically represented as

Substituting values

