Answer:
Yes, yes it would since we need light
Explanation:
Answer:
I = Δq / t
Explanation:
The quantity of electricity i.e charge is related to current and time according to the equation equation:
Q = It
Δq = It
Where:
Q => is the quantity of electricity i.e charge
I => is the current.
t => is the time.
Thus, we can rearrange the above expression to make 'I' the subject. This is illustrated below:
Δq = It
Divide both side by t
I = Δq / t
Assuming Earth's gravity, the formula for the flight of the particle is:
<span>s(t) = -16t^2 + vt + s = -16t^2 + 144t + 160. </span>
<span>This has a maximum when t = -b/(2a) = -144/[2(-16)] = -144/(-32) = 9/2. </span>
<span>Therefore, the maximum height is s(9/2) = -16(9/2)^2 + 144(9/2) + 160 = 484 feet. </span>
Answer:
the final velocity of the car is 59.33 m/s [N]
Explanation:
Given;
acceleration of the car, a = 13 m/s²
initial velocity of the car, u = 120 km/h = 33.33 m/s
duration of the car motion, t = 2 s
The final velocity of the car in the same direction is calculated as follows;
v = u + at
where;
v is the final velocity of the car
v = 33.33 + (13 x 2)
v = 59.33 m/s [N]
Therefore, the final velocity of the car is 59.33 m/s [N]
Answer:
t = 8 s
Explanation:
In order to find the time taken by the dragster we will use equations of motion. Here, we will use second equation of motion:
s = Vi t + (1/2)at²
where,
s = distance covered = 320 m
Vi = Initial Velocity = 0 m/s (Since, dragster starts from rest)
t = time taken = ?
a = acceleration of dragster = 10 m/s²
Therefore,
320 m = (0 m/s)t + (1/2)(10 m/s²)t²
t² = (320 m)(2)/(10 m/s²)
t = √(64 s²)
<u>t = 8 s</u>
