Answer:
35.3 N
Explanation:
U = 0, V = 0.61 m/s, s = 0.39 m
Let a be the acceleration.
Use third equation of motion
V^2 = u^2 + 2 as
0.61 × 0.61 = 0 + 2 × a × 0.39
a = 0.477 m/s^2
Force = mass × acceleration
F = 74 × 0.477 = 35.3 N
<h2>
Answer:</h2>
0.126m
<h2>
Explanation:</h2>
According to Hooke's law, the force (F) acting on a spring to cause an extension or compression (e) is given by;
F = k x e -------------------(i)
Where;
k = the spring's constant.
From the question, the force acting on the spring is the weight(W) of the mass. i.e
F = W -----------------------(ii)
<em>But;</em>
W = m x g;
where;
m = mass of the object
g = acceleration due to gravity [usually taken as 10m/s²]
<em>From equation (ii), it implies that;</em>
F = W = m x g
<em>Now substitute F = m x g into equation(i) as follows;</em>
F = k x e
m x g = k x e ------------------(iii)
<em>From the question;</em>
m = m1 = 3.5kg
k = 278N/m
<em>Substitute these values into equation (iii) as follows;</em>
3.5 x 10 = 278 x e
35 = 278e
<em>Now solve for e;</em>
e = 35/278
e = 0.126m
Therefore, the distance the spring is stretched from its unstretched length (which is the same as the extension of the spring) is 0.126m
B- the acceleration is greater for the more massive rock
Answer:
4.7 s
Explanation:
The complete question is presented in the attached image to this solution.
v(t) = 61 - 61e⁻⁰•²⁶ᵗ
At what time will v(t) = 43 m/s?
We just substitute 43 m/s into the equation for the velocity of the diver and solve for t.
43 = 61 - 61e⁻⁰•²⁶ᵗ
- 61e⁻⁰•²⁶ᵗ = 43 - 61 = -18
e⁻⁰•²⁶ᵗ = (18/61) = 0.2951
In e⁻⁰•²⁶ᵗ = In 0.2951 = -1.2205
-0.26t = -1.2205
t = (1.2205/0.26) = 4.694 s = 4.7 s to the nearest tenth.
Hope this Helps!!!
