Answer:
The correct option is D
Explanation:
In trying to achieve what the student wanted to see, which is to see the relationship between the weight the cord can hold and how long the cord will stretch. Since the origin of the graph is from zero, the value plotted on the vertical axis would be just the length caused by each weights. Thus, <u>the original length would have to be subtracted from the measured length to determine the actual length caused by the weight added to the cord</u>.
Answer:
the distance between the submarine and the ocean floor is 11,250 m
Explanation:
Given;
speed of the wave, v = 1500 m/s
time of motion of the wave, t = 15 s
The time taken to receive the echo is calculated as;

Therefore, the distance between the submarine and the ocean floor is 11,250 m
Answer:
4.41 m/s^2
Explanation:
(v_f)^2 - (v_i)^2 = 2a * change in distance
(21)^2 - (0)^2 = 2a * 50
a = (21^2)/(2*50)
a = 4.41 m/s^2
Answer:
103.5 meters
Explanation:
Given that a stunt person has to jump from a bridge and land on a boat in the water 22.5 m below. The boat is cruising at a constant velocity of 48.3 m/s towards the bridge. The stunt person will jump up at 6.45 m/s as they leave the bridge.
The time the person will jump to a certain spot under the bridge can be calculated by using the formula below:
h = Ut + 1/2gt^2
since the person will fall under gravity, g = 9.8 m/s^2
Also, let assume that the person jump from rest, then, U = 0
Substitute h, U and g into the formula above
22.5 = 1/2 * 9.8 * t^2
22.5 = 4.9t^2
22.5 = 4.9t^2
t^2 = 22.5/4.9
t^2 = 4.59
t = 
t = 2.143 seconds
From definition of speed,
speed = distance /time
Given that the boat is cruising at a constant velocity of 48.3 m/s towards the bridge, substitute the speed and the time to get the distance.
48.3 = distance / 2.143
distance = 48.3 * 2.143
distance = 103.5 m
Therefore, the boat should be 103.5m away from the bridge at the moment the stunt person jumps?