Answer:
6.5 m/s
Explanation:
We are given that
Distance, s=100 m
Initial speed, u=1.4 m/s
Acceleration, 
We have to find the final velocity at the end of the 100.0 m.
We know that
Using the formula
Hence, her final velocity at the end of the 100.0 m=6.5 m/s
Arterie
cwamus
capillaries
phloprant
veins
piscas
Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver
Answer:
-6461.54 N
Explanation:
From Newton's Fundamental equation,
F = m(v-u)/t.................... Equation 1
Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.
Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s
Substitute into equation 1
F = 0.75(-60-52)/0.013
F = 0.75(-112)/0.013
F = -84/0.013
F = -6461.54 N
Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.
Hence the magnitude of the average force of the wall = -6461.54 N
Answer:
A. 1254.4 J
B. 1254.4 J
C. 43904 calories
Explanation:
Parameters given:
Mass of barbell, m = 160 kg
Length of arm, d = 0.8
A. Work done by weight lifter in lifting the barbell is:
W = F * d
Where F is force
F = m * g
=> W = m * g * d
W = 160 * 9.8 *0.8
W = 1254.4 J
B. Potential energy is given as:
PE = m * g * h
Where h is height (in this case, D)
=> PE = m * g * d
PE = 160 * 9.8 * 0.8
PE = 1254.4 J
C. The number of calories the weight lifter uses after lifting the barbell 35 times is equal to 35 times the potential energy, since a calorie is the measure of the amount of potential energy a piece of food contains:
35 * 1254.4 = 43904 calories.
