Answer:
Hans Lipperhey
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Answer:
1.991 × 10^(8) N/m²
Explanation:
We are told that its volume increases by 9.05%.
Thus; (ΔV/V_o) = 9.05% = 0.0905
To find the force per unit area which is also pressure, we will use bulk modulus formula;
B = Δp(V_o/ΔV)
Making Δp the subject gives;
Δp = B(ΔV/V_o)
Now, B is bulk modulus of water with a value of 2.2 × 10^(9) N/m²
Thus;
Δp = 2.2 × 10^(9)[0.0905]
Δp = 1.991 × 10^(8) N/m²
The ratio of the distance moved by the point at which the effort is applied in a simple machine to the distance moved by the point at which the load is applied, in the same time. In the case of an ideal (frictionless and weightless) machine, velocity ratio = mechanical advantage. Velocity ratio is sometimes called distance ratio.
The answer is 3.
When a ball is thrown into the air, there is no friction (there's nothing rubbing against it), so the only force acting on the ball is gravity, which aims to pull the ball back down. Since this is the only force, the answer will equal the gravity acting on the ball. Since "weight" is defined as just that (the gravity acting on something), the answer is 3, the magnitude of the rock's weight (or the force of gravity).
Answer:
(a) 98 N
(b) 158 N
(c) 38 N
Explanation:
<h2>
Part (a)</h2>
When the acceleration is 0 m/s², the net force on the mass is 0 N. Therefore, the tension force is equal to the weight force due to Newton's Second Law:
- ∑F_y = T - w = ma_y
- ∑F_y = T - w = m(0 m/s²)
- ∑F_y = T - w = 0
- ∑F_y = T = w
Since the tension in the cable and the weight of the mass are equal to each other, we can solve for the weight force of the mass by using:
- w = mg
- w = (10 kg)(9.8 m/s²)
- w = 98 N
Since T = w, we can say that T = 98 N.
<h2>Part (b)</h2>
Let's set the upwards direction to be positive and the downwards direction to be negative. We can use Newton's Second Law to solve for the tension in the cable if the acceleration is 6 m/s² upward:
- ∑F_y = T - w = ma_y
- ∑F_y = T - mg = m(6 m/s²)
- ∑F_y = T - mg = 6m
Plug the known values into the equation and solve for T.
- T - mg = 6m
- T - (10 kg)(9.8 m/s²) = 6(10 kg)
- T - 98 = 60
- T = 158 N
The tension in the cable if the acceleration is +6 m/s² is 158 N.
<h2>
Part (c)</h2>
The process is the same, but this time acceleration is -6 m/s².
- ∑F_y = T - w = ma_y
- ∑F_y = T - mg = m(-6 m/s²)
- ∑F_y = T - mg = -6m
Plug known values into the equation and solve for T.
- T - mg = -6m
- T - (10 kg)(9.8 m/s²) = -6(10 kg)
- T - 98 = -60
- T = 38 N
The tension in the cable if the acceleration is -6 m/s² is 38 N.