Answer:
51.85m/s
Explanation:
Given parameters:
Mass of ball = 0.0459kg
Force = 2380N
Time taken = 0.001s
Unknown:
Speed of the ball afterwards = ?
Solution:
To solve this problem, we use Newton's second law of motion:
F = m x 
F is the force
m is the mass
v is the final velocity
u is the initial velocity
t is the time taken
2380 = 0.0459 x 
0.0459v = 2.38
v = 51.85m/s
Answer:
The direction of the B-field is in the +y-direction.
Explanation:
The corresponding formula is
This means, we should use right-hand rule.
Our index finger is pointed towards +x-direction (direction of velocity),
our middle finger should point towards the direction of the B-field,
and our thumb should point towards the +z-direction (direction of the force).
Since our middle finger in this situation points towards +y-direction, the B-field should be in +y-direction.
Answer: 361° C
Explanation:
Given
Initial pressure of the gas, P1 = 294 kPa
Final pressure of the gas, P2 = 500 kPa
Initial temperature of the gas, T1 = 100° C = 100 + 273 K = 373 K
Final temperature of the gas, T2 = ?
Let us assume that the gas is an ideal gas, then we use the equation below to solve
T2/T1 = P2/P1
T2 = T1 * (P2/P1)
T2 = (100 + 273) * (500 / 294)
T2 = 373 * (500 / 294)
T2 = 373 * 1.7
T2 = 634 K
T2 = 634 K - 273 K = 361° C
Answer:
The charge stored in the capacitor will stay the same. However, the electric potential across the two plates will increase. (Assuming that the permittivity of the space between the two plates stays the same.)
Explanation:
The two plates of this capacitor are no longer connected to each other. As a result, there's no way for the charge on one plate to move to the other. 
, the amount of charge stored in this capacitor, will stay the same.
The formula 
relates the electric potential across a capacitor to:
- , the charge stored in the capacitor, and
, the capacitance of this capacitor.
While stays the same, moving the two plates apart could affect the potential 
by changing the capacitance of this capacitor. The formula for the capacitance of a parallel-plate capacitor is:
,
where
is the permittivity of the material between the two plates.
is the area of each of the two plates.
is the distance between the two plates.
Assume that the two plates are separated with vacuum. Moving the two plates apart will not affect the value of . Neither will that change the area of the two plates.
However, as (the distance between the two plates) increases, the value of will become smaller. In other words, moving the two plates of a parallel-plate capacitor apart would reduce its capacitance.
On the other hand, the formula can be rewritten as:
.
The value of (charge stored in this capacitor) stays the same. As the value of becomes smaller, the value of the fraction will become larger. Hence, the electric potential across this capacitor will become larger as the two plates are moved away from one another.
Answer: D
Rs = 10.0 m/s
The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s
Explanation:
Since the boat is moving perpendicular to the current of the river, the speed of the boat has two components.
i. 8.0m/s in the direction perpendicular to the current
ii. 6.0m/s in the direction of the current.
So, the resultant speed can be derived by using the equation;
Rs = √(Rx^2 + Ry^2)
Taking
Ry = 8.0m/s
Rx = 6.0m/s
Substituting into the equation, we have;
Rs = √(6.0^2 + 8.0^2)
Rs = √(36+64) = √100
Rs = 10.0 m/s
The speed of the boat relative to an observer standing on the shore as it crosses the river is 10.0m/s
