The maximum force that the athlete exerts on the bag is equal to 1,500 N and in the opposite direction as the force that the bag exerts on the athlete.
<h3>
Newton's third law of motion</h3>
Newton's third law of motion states that action and reaction are equal and opposite.
Fa = -Fb
The force exerted by the athlete on the bag is equal to the force the bag exerted on the athlete but in opposite direction.
Thus, the maximum force that the athlete exerts on the bag is equal to 1,500 newtons and in the opposite direction as the force that the bag exerts on the athlete.
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<span>You can use the equation
V_xf = V_xi + a_x(t)
V_xf = 20.0m/s
V_xi = 0m/s
ax = 2.0
t
Thus, solve for t and get 10seconds
and then take 5 seconds to break after 20 seconds of driving
so for
a) 10 + 20 + 5 = 35 seconds
</span><span>for part b)
You can use the formula
Delta x/Delta t = average velocity
Need to find xf, knowing xi = 0
Thus, use the formula
x_f = x_i + V_xi(t) + (1/2)a_x(t)^(2)
x_f = 0 + 0(10) + (1/2)(2.0)(10)^(2)
x_f = 100m
so for the first 10 seconds the truck traveled 100ms
At a speed of 20m/s
20m/s = xm/20s
20*20 = x
x = 400
thus we have 100+400 = 500m
then it slows down from 500m to x_f
thus I use the equation
x_f = x_i + (1/2)(V_xf + V_xi)t
x_f = 500 + (1/2)(0 + 20)(5)
x_f = 500 + 50
x_f = 550
therefore the total distance traveled is 550m
</span>
<span>to calculate average velocity
550/35 = 16m/s
thus
V_xavg = 16m/s</span>
To solve this problem we will apply the concepts related to wavelength, as well as Rayleigh's Criterion or Optical resolution, the optical limit due to diffraction can be calculated empirically from the following relationship,

Here,
= Wavelength
d= Diameter of aperture
= Angular resolution or diffraction angle
Our values are given as,

The frequency of the sound is 
The speed of the sound is 
The wavelength of the sound is

Here,
v = Velocity of the wave
f = Frequency
Replacing,


The diffraction condition is then,

Replacing,

d = 0.24 m
Therefore the diameter should be 0.24m
Answer:
1.52 hour
Explanation:
M = 0.5 g, I = 3 A
Electrochemical equivalent of nickel
Z = 3.04 × 10^(-4) g/C
By use of Faraday's laws of electrolysis
M = Z I t
t = M / Z I
t = 0.5 / (3.04 × 10^-4 × 3)
t = 5482.45 second = 1.52 hour