Do you remember this formula for the distance traveled while accelerated ?
<u>Distance = (initial speed) x (t) plus (1/2) x (acceleration) x (t²)</u>
I think this is exactly what we need for this problem.
initial speed = 20 m/s down
acceleration = 9.81 m/s² down
t = 3.0 seconds
Distance down = (20) x (3) plus (1/2) x (9.81) x (3)²
Distance = (60) plus (4.905) x (9)
Distance = (60) plus (44.145) = 104.145 meters
Choice <em>D)</em> is the closest one.
Answer:
10.8 s
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 0 m/s
Acceleration (a) = 5 m/s/s
Distance travelled (s) = 291 m
Time (t) taken =?
We can calculate the time taken for the car to cover the distance as follow:
s = ut + ½at²
291 = 0 × t + ½ × 5 × t²
291 = 0 + 2.5 × t²
291 = 2.5 × t²
Divide both side by 2.5
t² = 291 / 2.5
t² = 116.4
Take the square root of both side
t = √116.4
t = 10.8 s
Thus, it will take the car 10.8 s to cover the distance.
The circuit shown has a battery and two resistors, with R1>R2. Which of the two resistors dissipates the larger amount of power? Explain
Rank in order, from largest to smallest, the three currents I1 TO I3. And explain why?
The two batteries are identical and the four resistors all ahve exactly the same resistance.
a) Compare change in V(ab), V(cd), and V(ef). Are they all the same? In not, rank them in decreasing order. Explain you reasoning
b) Rank in order, from largest to smallest, the five currents I1 to I5, and ezplain why
