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dlinn [17]
4 years ago
5

Four equal masses m are so small they can be treated as points, and they are equally spaced along a long, stiff wire of neglible

mass. The distance between any two adjacent masses is ℓ . m m m m ℓ ℓ ℓ What is the rotational inertia Icm of this system about its center of mass?
A.) (1/2)ml^2
B.) ml^2
C.) 5ml^2
D.) 6ml^2
E.) 2ml^2
F.) 3ml^2
G.) 7ml^2
H.) 4ml^2.
Physics
1 answer:
Mice21 [21]4 years ago
7 0

Answer:C

Explanation:

Given

Four masses are attached to the wire such that distance between two mass is L

therefore the Length of wire is 4 L

and the center of mass is at 2L

moment of inertia is distribution of mass from its rotational axis

thus moment of Inertia I is given by

I=m\times (\frac{3L}{2})^2+m\times (\frac{L}{2})^2+m\times (\frac{3L}{2})^2+m\times (\frac{L}{2})^2

I=2\times m\times (\frac{L}{2})^2+2\times m\times (\frac{3L}{2})^2

I=\frac{2mL^2}{4}+\frac{18mL^2}{4}

I=5mL^2

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If a 5.0 kg box is pulled simultaneously by a 10.0 N force and a 5.0 N force, then its acceleration must be Group of answer choi
vivado [14]

Answer:

We cannot tell from the information given

Explanation:

Given;

mass of the box, m = 5 kg

first force, F₁ = 10 N

second force, F₂ = 5 N

(I) Assuming the two forces are acting horizontally in opposite direction, the resultant force on the box is calculated as;

∑Fx = 10 N - 5 N

      = 5 N

Apply Newton's second law of motion;

∑Fx = ma

a = ∑Fx/m

a = 5 / 5

a = 1 m/s² in the direction of the 10 N force.

(II) Also, if the two forces are acting in the same direction, the resultant force is calculated as;

∑Fx = 10 N + 5 N

∑Fx = 15 N

a = 15 / 5

a = 3 m/s²

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3 years ago
A satellite travels with a constant speed |v| as it moves around a circle centered on the earth. How much work is done by the gr
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Answer:

W = 0

Explanation:

As the satellite moves in a circle the force is perpendicular to the path, therefore the work that is defined by

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W = 0

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The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and
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Complete question:

The exit nozzle in a jet engine receives air at 1200 K, 150 kPa with negligible kinetic energy. The exit pressure is 80 kPa, and the process is reversible and adiabatic. Use constant specific heat at 300 K to find the exit velocity.

Answer:

The exit velocity is 629.41 m/s

Explanation:

Given;

initial temperature, T₁ = 1200K

initial pressure, P₁ = 150 kPa

final pressure, P₂ = 80 kPa

specific heat at 300 K, Cp = 1004 J/kgK

k = 1.4

Calculate final temperature;

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}

k = 1.4

T_2 = T_1(\frac{P_2}{P_1})^{\frac{k-1 }{k}}\\\\T_2 = 1200(\frac{80}{150})^{\frac{1.4-1 }{1.4}}\\\\T_2 = 1002.714K

Work done is given as;

W = \frac{1}{2} *m*(v_i^2 - v_e^2)

inlet velocity is negligible;

v_e = \sqrt{\frac{2W}{m} } = \sqrt{2*C_p(T_1-T_2)} \\\\v_e = \sqrt{2*1004(1200-1002.714)}\\\\v_e = \sqrt{396150.288} \\\\v_e = 629.41  \ m/s

Therefore, the exit velocity is 629.41 m/s

6 0
3 years ago
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