Answer:
12 units
Explanation:
Original F = C q1q2/r^2 now change the parameters
New F = C 1/3 q1 2 q2 / (2r)^2
= 2/12 C q1q2/r^2 <=======the new force is 1/6 of original
1/6 * 72 = 12 units
Answer:
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Explanation:
Let L represent Moe's height during the leap.
Moe's velocity v at any point in time during the leap is;
v = dL/dt = u - gt .......1
Where;
u = it's initial speed
g = acceleration due to gravity on Mars
t = time
The determine how far the cricket was off the ground when it became Moe’s lunch.
We need to integrate equation 1 with respect to t
L = ∫dL/dt = ∫( u - gt)
L = ut - 0.5gt^2 + L₀
Where;
L₀ = Moe's initial height = 0
u = 105m/s
t = 56 s
g = 3.75 m/s^2
Substituting the values, we have;
L = (105×56) -(0.5×3.75×56^2) + 0
L = 0 m
Therefore, the cricket was 0m off the ground when it became Moe’s lunch.
Answer:
The least amount of time in which the fisherman can raise the fish to the dock without losing it is t= 2 seconds.
Explanation:
m= 5 kg
h= 2m
Fmax= 54 N
g= 9.8 m/s²
W= m * g
W= 49 N
F= Fmax - W
F= 5 N
F=m*a
a= F/m
a= 1 m/s²
h= a * t²/2
t= √(2*h/a)
t= 2 seconds
Answer:
A. increase the speed
Explanation:
For you to be able to shorten the time it takes for the care to get to its destination, you must increase the speed because if the distance is the same that means you have to increase your speed.
