Answer:
37.5 N Hard
Explanation:
Hook's law: The force applied to an elastic material is directly proportional to the extension provided the elastic limit of the material is not exceeded.
Using the expression for hook's law,
F = ke.............. Equation 1
F = Force of the athlete, k = force constant of the spring, e = extension/compression of the spring.
Given: k = 750 N/m, e = 5.0 cm = 0.05 m
Substitute into equation 1
F = 750(0.05)
F = 37.5 N
Hence the athlete is pushing 37.5 N hard
Answer:
Explanation:
Using the equation of motion to get the acceleration due to gravoty of the rock on the planet.
S = ut+ 1/2at² where;
S is the distance of the rock above the surface of the planet = 100m
u is the initial velocity = 15m/s
a is the acceleration due to gravity
t is the time taken by the rock to reach the ground = 10s
Since the rock is thrown upward the acceleration due to gravity will be negative i.e a= -g
The equation becomes S = ut- 1/2gt²
Substituting the given value to get the time t
100 = 15(10)- 1/2g(10)²
100 = 150-50g
100-150 = -50g
-50 = -50g
g = -50/-50
g = 1m/s²
<em>Hence the acceleration due to gravity of the rock when it is on Planet XX is 1m/s²</em>
Fnety = (FT)(sin 32°) – Fg
Or the answer B, I checked it.
<span>As a sound source gets closer, both the volume and the pitch of the sound increased. Then, as the sound source passed by you, both the volume and the pitch of the sound decreased.
Hope this helps</span>
Answer:
A) 12.08 m/s
B) 19.39 m/s
Explanation:
A) Down the hill, we will apply Newton’s second law of motion in the downward direction to get:
mg(sinθ) – F_k = ma
Where; F_k is frictional force due to kinetic friction given by the formula;
F_k = (μ_k) × F_n
F_n is normal force given by mgcosθ
Thus;
F_k = μ_k(mg cosθ)
We now have;
mg(sinθ) – μ_k(mg cosθ) = ma
Dividing through by m to get;
g(sinθ) – μ_k(g cosθ) = a
a = 9.8(sin 12.03) - 0.6(9.8 × cos 12.03)
a = -3.71 m/s²
We are told that distance d = 24.0 m and v_o = 18 m/s
Using newton's 3rd equation of motion, we have;
v = √(v_o² + 2ad)
v = √(18² + (2 × -3.71 × 24))
v = 12.08 m/s
B) Now, μ_k = 0.10
Thus;
a = 9.8(sin 12.03) - 0.1(9.8 × cos 12.03)
a = 1.08 m/s²
Using newton's 3rd equation of motion, we have;
v = √(v_o + 2ad)
v = √(18² + (2 × 1.08 × 24))
v = 19.39 m/s
