Answer:
a1 = 3.56 m/s²
Explanation:
We are given;
Mass of book on horizontal surface; m1 = 3 kg
Mass of hanging book; m2 = 4 kg
Diameter of pulley; D = 0.15 m
Radius of pulley; r = D/2 = 0.15/2 = 0.075 m
Change in displacement; Δx = Δy = 1 m
Time; t = 0.75
I've drawn a free body diagram to depict this question.
Since we want to find the tension of the cord on 3.00 kg book, it means we are looking for T1 as depicted in the FBD attached. T1 is calculated from taking moments about the x-axis to give;
ΣF_x = T1 = m1 × a1
a1 is acceleration and can be calculated from Newton's 2nd equation of motion.
s = ut + ½at²
our s is now Δx and a1 is a.
Thus;
Δx = ut + ½a1(t²)
u is initial velocity and equal to zero because the 3 kg book was at rest initially.
Thus, plugging in the relevant values;
1 = 0 + ½a1(0.75²)
Multiply through by 2;
2 = 0.75²a1
a1 = 2/0.75²
a1 = 3.56 m/s²
Complete question:
A diver is 10 m below the surface of water. Calculate the pressure the fluid exerted on the diver. The acceleration of gravity is 9.8 m/s2 and the density of the water is 1000 kg/m3. Answer in units of Pa. Show your work.
Answer:
Tthe pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa
Explanation:
Given;
density of water, ρ = 1000 kg/m³
diver's position below the surface of the water, h = 10 m
acceleration due to gravity, g = 9.8 m/s²
Let the atmospheric pressure, P₀ = 101325 Pa
The pressure 10 m below the surface of the water is calculated as;
P = P₀ + ρgh
P = 101325 Pa + (1000 x 9.8 x 10)Pa
P = 199325 Pa
P = 1.99 x 10⁵ Pa.
Therefore, the pressure the fluid exerted on the diver is 1.99 x 10⁵ Pa
Answer:
Newton's third law explains the generation of thrust by a rocket engine. In a rocket engine, hot exhaust gas is produced through the combustion of a fuel with an oxidizer. The hot exhaust gas flows through the rocket nozzle and is accelerated to the rear of the rocket. In re-action, a thrusting force is produced on the engine mount.
Explanation:
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