t = total time taken by sound to go down the well and then come back = 2 sec
d = depth of the well
D = total distance traveled by the sound
total distance traveled by sound is given as
total distance traveled = distance traveled going down + distance traveled going up
D = d + d
D = 2 d eq-1
v = speed of sound = 343 m/s
we know that distance traveled is given as
total distance traveled = speed of sound x total time
D = v t
using eq-1
2 d = v t
2 d = (343) (2)
dividing both side by 2
d = 343 x 2/2
d = 343 m
hence surface of water is 343 m deep
Hi there!
We can use the equation for the charge of a charging capacitor:
Using Capacitor equations:
Therefore, Cε equals the steady-state charge of the capacitor (the function approaches this value as t ⇒ ∞.
We can plug in the givens and solve.
Answer:
d. 6.0 m
Explanation:
Given;
initial velocity of the car, u = 7.0 m/s
distance traveled by the car, d = 1.5 m
Assuming the car to be decelerating at a constant rate when the brakes were applied;
v² = u² + 2(-a)s
v² = u² - 2as
where;
v is the final velocity of the car when it stops
0 = u² - 2as
2as = u²
a = u² / 2s
a = (7)² / (2 x 1.5)
a = 16.333 m/s
When the velocity is 14 m/s
v² = u² - 2as
0 = u² - 2as
2as = u²
s = u² / 2a
s = (14)² / (2 x 16.333)
s = 6.0 m
Therefore, If the car had been moving at 14 m/s, it would have traveled 6.0 m before stopping.
The correct option is d
