Work = (force) x (distance)
Work = (22 N) x (16 m)
<em>Work = 352 Joules</em>
To solve this problem, we know that:
1 Albert = 88 meters
1 A = 88 m
The first thing we have to do is to square both sides of
the equation:
(1 A)^2 = (88 m)^2
1 A^2 = 7,744 m^2
<span>Since it is given that 1 acre = 4,050 m^2, so to reach
that value, 1st let us divide both sides by 7,744:</span>
1 A^2 / 7,744 = 7,744 m^2 / 7,744
(1 / 7,744) A^2 = 1 m^2
Then we multiply both sides by 4,050.
(4050 / 7744) A^2 = 4050 m^2
0.523 A^2 = 4050 m^2
<span>Therefore 1 acre is equivalent to about 0.52 square
alberts.</span>
Answer:
692.31 N
Explanation:
Applying,
F = ma............... Equation 1
Where F = Average force required to stop the player, m = mass of the player, a = acceleration of the player
But,
a = (v-u)/t............ Equation 2
Where v = final velocity, u = initial velocity, t = time.
Substitute equation 2 into equation 1
F = m(v-u)/t............ Equation 3
From the question,
Given: m = 75 kg, u = 6.0 m/s, v = 0 m/s (to stop), t = 0.65 s
Substitute these values into equation 3
F = 75(0-6)/0.65
F = -692.31 N
Hence the average force required to stop the player is 692.31 N
Impulse = Force * time
Impulse = 500N *0.5 s =250 N*s
Answer:
Average acceleration on first part of the chunk is given as

Average acceleration on second part of the chunk is given as

Explanation:
By momentum conservation along x direction we will have

so we have


also by energy conservation






by solving above equation we will have


Average acceleration on first part of the chunk is given as


Average acceleration on second part of the chunk is given as

