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Arturiano [62]
3 years ago
8

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m3 , length 92.6 cm and diameter 2.95 cm fr

om a storage room to a machinist.
Part A) Calculate the weight of the rod, w. The acceleration due to gravity, g= 9.81 m/s^2.w= 36.9 N

Part B) Now that you know the weight of the rod, do you think that you will be able to carry the rod without a cart? Yes or No?
Physics
1 answer:
slavikrds [6]3 years ago
7 0

Answer:

48.4293354946 N

Yes

Explanation:

d = Diameter of rod = 2.95 cm

h = Length of rod = 92.6 cm

\rho = Density of rod = 7800 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

Volume of rod

V=\dfrac{1}{4}\pi d^2h\\\Rightarrow V=\dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}

Mass is given by

m=\rho V\\\Rightarrow m=7800\times \dfrac{1}{4}\times \pi\times (2.95\times 10^{-2})^2\times 92.6\times 10^{-2}\\\Rightarrow m=4.93673144695\ kg

Weight is given by

W=mg\\\Rightarrow W=4.93673144695\times 9.81\\\Rightarrow W=48.4293354946\ N

The weight of the rod is 48.4293354946 N

The mass of the rod is 4.93673144695 kg which is light. So, I will be able to carry the rod without a cart.

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Answer:

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substituting these in equation, we get:

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Also,

B/s = m³

<u>B = m³s = [L³T]</u>

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3 years ago
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A professor sits at rest on a stool that can rotate without friction. The rotational inertia of the professor-stool system is 4.
Anestetic [448]

Answer:

\omega=0.37 [rad/s]  

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Now, we just need to solve it for ω.

\omega=\frac{mvR}{I_{proffesor-stool}+mR^{2}}

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I hope it helps you!

5 0
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A rod 7.0 m long is pivoted at a point 2.0 m from the left end. A downward force of 50 N acts at the left end, and a downward fo
kicyunya [14]

If the rod is in rotational equilibrium, then the net torques acting on it is zero:

∑ τ = 0

Let's give the system a counterclockwise orientation, so that forces that would cause the rod to rotate counterclockwise act in the positive direction. Compute the magnitudes of each torque:

• at the left end,

τ = + (50 N) (2.0 m) = 100 N•m

• at the right end,

τ = - (200 N) (5.0 m) = - 1000 N•m

• at a point a distance d to the right of the pivot point,

τ = + (300 N) d

Then

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6 0
2 years ago
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charle [14.2K]

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