Answer:
160J
Explanation:
Given force = 8N and total distance = 20 meters
Workdone = force x distance
= 8 x 20
= 160J
Therefore, workdone by Riley in pulling the hoover is 160J
Answer:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.
Explanation:
Answer: Given m = 10 kg and . F = 20 N. Thus, the force required to accelerate the object upward direction is 20 N.
Answer:
The speed of the ball was, v = 3 m/s
Explanation:
Given data,
The time period of the ball, t = 8 s
The distance the ball rolled, d = 24 m
The velocity of an object is defined as the object's displacement to the time taken. The formula for the velocity is,
v = d / t m/s
Substituting the given values in the above equation,
v = 24 / 8
= 3 m/s
Hence, the speed of the ball was, v = 3 m/s
Nuclear power plants, wind farms, water farms, and geothermal heating
Answer:
Vi = 8.28 m/s
Explanation:
This problem is related to the projectile motion.
As we know there are two components of motion associated with this, the horizontal component and vertical component.
The horizontal distance covered by the ball is
Vx*t = x
Vx*t = 5.3
Vx = 5.3/t eq. 1
Also we know that
Vx = Vicos(60)
Vx = Vi*0.5 eq. 2
equate eq. 1 and eq. 2
5.3/t = Vi*0.5
5.3/0.5 = Vi*t
Vi*t = 10.6 eq. 3
The vertical distance is
Vy = y1 + Vyi*t - 0.5gt²
also we know that
Vyi = Visin(60)
Vyi = Vi*0.866
It is given that V1 = 1.9 m and and Vy = 3 m is the vertical distance
3 = 1.9 + Vi*0.866*t - 0.5gt²
3 = 1.9 + Vi*0.866*t - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
3 = 1.9 + 0.866(Vi*t) - 0.5(9.8)t²
1.1 = 0.866(Vi*t) - 4.9t²
0.866(Vi*t) = 4.9t² + 1.1
substitute Vi*t = 10.6 in above equation
0.866(10.6) = 4.9t² + 1.1
9.18 = 4.9t² + 1.1
4.9t² = 8.08
t² = 8.08/4.9
t² = 1.648
t = 1.28 sec
Finally, initial speed can be found by substituting the value of t into eq. 3
Vi*t = 10.6
Vi = 10.6/t
Vi = 10.6/1.28
Vi = 8.28 m/s
