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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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3 years ago
3.Three resistors of 25.0Ω, 30.0Ω, and 40.0Ω are in a series circuit with a 6.0-volt battery. What is the current in the circuit
AURORKA [14]

Answer:

Current = 0.063 Amperes

Explanation:

Let the three resistors be R1, R2, and R3 respectively.

Given the following data;

R1 = 25.0Ω,

R2 = 30.0Ω

R3 = 40.0Ω

Voltage = 6 Volts

First of all, we would determine the equivalent or total resistance;

Total resistance (in series) = R1 + R2 + R3

Total resistance = 25.0Ω + 30.0Ω + 40.0Ω

Total resistance = 95 Ω

Next, we find the current flowing through the circuit;

Voltage = current * resistance

Substituting into the formula, we have;

6 = current * 95

Current = 6/95

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Which stretch focuses on stretching the glutes and hamstrings while laying on your back?
WINSTONCH [101]

Answer:

knee to chest

Explanation:

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2 years ago
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Answer:

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How much charge is on each plate of a 3.00-μF capacitor when it is connected toa 15.0-V battery? b) If this same capacitor is c
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Answer:

(a) 45 micro coulomb

(b) 6 micro Coulomb

Explanation:

C = 3 micro Farad = 3 x 10^-6 Farad

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(a) q = C x V

where, q be the charge.

q = 3 x 10^-6 x 15 = 45 x 10^-6 C = 45 micro coulomb

(b)

V = 2 V, C = 3 micro Farad = 3 x 10^-6 Farad

q = C x V

where, q be the charge.

q = 3 x 10^-6 x 2 = 6 x 10^-6 C = 6 micro coulomb

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3 years ago
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