Answer:
0.66 s
Explanation:
∆x = v∆t → 2 × 115 = 350 ∆t → ∆t = 230/350 = 0.66 s
Answer:
94.13 ft/s
Explanation:
<u>Given:</u>
= time interval in which the rock hits the opponent = 10 s - 5 s = 5 s
= distance to be moved by the rock long the horizontal = 98 yards
= displacement to be moved by the rock during the time of flight along the vertical = 0 yard
<u>Assume:</u>
= magnitude of initial velocity of the rock
= angle of the initial velocity with the horizontal.
For the motion of the rock along the vertical during the time of flight, the rock has a constant acceleration in the vertically downward direction.
Now the rock has zero acceleration along the horizontal. This means it has a constant velocity along the horizontal during the time of flight.
On dividing equation (1) by (2), we have
Now, putting this value in equation (2), we have
Hence, the initial velocity of the rock must a magnitude of 94.13 ft/s to hit the opponent exactly at 98 yards.
Answer:
<h2>A. 180 miles</h2><h2>B. 60 miles</h2><h2 />
Explanation:
In this problem, we are required to solve for the total distance that the car travelled. and the displacement
A) the distance travelled by car
this can be gotten by summing all the distances the car has travelled.
i,e total distance= 60 miles+120 miles
total distance= 180 miles
B) the displacement of the car
the displacement can be gotten by subtracting the final distance from the initial distance
final distance = 120 miles
initial distance= 60 miles
displacement= 120-60= 60 miles
20N•m or 20J. Work is equal to force•distance, and 5N•4m is 20N•m, or J
