Ask question

Ask question

All categories

3 years ago

9

On a part-time job, you are asked to bring a cylindrical iron rod of density 7800 kg/m^3 , length 92.4 cm and diameter 2.15 cm f

rom a storage room to a machinist. Calculate the weight of the rod, w. Assume the free-fall acceleration is g =9.81m/s^2

1 answer:

Zigmanuir [339]3 years ago

3 0

Answer:

25.68 N

Explanation:

From the question given above, the following data were obtained:

Density of cylindrical rod = 7800 kg/m³

Length = 92.4 cm

Diameter = 2.15 cm

Acceleration due to gravity (g) = 9.8 m/s²

Weight of rod =?

Next, we shall determine the volume of the rod. This can be obtained as follow:

Height (h) = 92.4 cm

Diameter (d) = 2.15 cm

Pi (π) = 3.14

Volume (V) =?

V = π(d/2)²h

V = 3.14 × (2.15/2)² × 92.4

V = 335.29 cm³

Next, we shall convert 335.29 cm³ to m³. This can be obtained as follow:

1 cm³ = 1×10¯⁶ m³

Therefore,

335.29 cm³ = 335.29 cm³ × 1×10¯⁶ m³ / 1 cm³

335.29 cm³ = 0.00033529 m³

Thus, 335.29 cm³ is equivalent to 0.00033529 m³.

Next, we shall determine the mass of the rod. This can be obtained as follow:

Density of rod = 7800 kg/m³

Volume of rod = 0.00033529 m³.

Mass of rod =?

Density = mass /volume

7800 = mass / 0.00033529

Cross multiply

Mass of rod = 7800 × 0.00033529

Mass of rod = 2.62 Kg

Finally, we shall determine the weight of the rod. This can be obtained as follow:

Mass (m) of rod = 2.62 Kg

Acceleration due to gravity (g) = 9.8 m/s²

Weight (W) of rod =?

W = m × g

W = 2.62 × 9.8

W = 25.68 N

Therefore, the weight of the rod is 25.68 N

You might be interested in

Air, considered an ideal gas, is contained in an insulated piston-cylinder assembly outfitted with a paddle wheel. It is initial

Maru [420]

Our data are,

State 1:

$P_1= 10psi=68.95kPa\\V_1 = 1ft^3=0.02831m^3\\T_1 = 100\°F = 310.93K$

State 2:

$P_2 =5psi=34.474kPa\\V_2 = 3ft^3=0.0899m^3$

We know as well that $3BTU=3.16kJ/K$

To find the mass we apply the ideal gas formula, which is given by

$P_1V_1=mRT_1$

Re-arrange for m,

$m= \frac{P_1V_1}{RT_1}\\m= \frac{68.95*0.02831}{(0.287)310.9}\\m=0.021893kg=0.04806lbm\\$

Because of the pressure, temperature and volume ratio of state 1 and 2, we have to

$\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Replacing,

$T_2 = \frac{P_2V_2}{P_1V_1}T_1\\T_2 =\frac{34.474*0.0844}{68.95*0.02831}*310.93\\T_2 = 464.217K=375.5\°F$

For conservative energy we have, (Cv = 0.718)

$W = m C_v = 0.718 \Delta T +dw\\dw = W - mv\Delta T\\dw = 3.16-(0.0218*0.718)(454.127-310.93)\\dw = 0.765kJ=0.72BTU$

3 0

3 years ago

You are sitting on a merry-go-round of mass 200 kg and radius 2m that is at rest (not spinning). Your mass is 50 kg. Your friend

Bogdan [553]

Answer:

a. $\tau=200J$ b. $\alpha=0.44 \frac{rad}{s^2}$ c. $\alpha=0.33\frac{rad}{s^2}$ d. The angular acceleration when sitting in the middle is larger.

Explanation:

a. The magnitude of the torque is given by $\tau=rF\sin \theta$ , being r the radius, F the force aplied and $\theta$ the angle between the vector force and the vector radius. Since $\theta=90^{\circ}, \, \sin\theta=1$ and so $\tau=rF=2m100N=200Nm=200J$ .

b. Since the relation $\tau=I\alpha$ hols, being I the moment of inertia, the angular acceleration can be calculated by $\alpha=\frac{\tau}{I}$ . Since we have already calculated the torque, all left is calculate the moment of inertia. The moment of inertia of a solid disk rotating about an axis that passes through its center is $I=\frac{1}{2}Mr^2$ , being M the mass of the disk. If we assume that a person has a punctual mass, the moment of inertia of a person would be given by $I_p=m_pr_p^{2}$ , being $m_p$ the mass of the person and $r_p^{2}$ the distance from the person to the center. Given all of this, we have

$\alpha=\frac{\tau}{I}=\frac{\tau}{I_{disk}+I_{person}}=\frac{Fr}{\frac{1}{2}Mr^2+m_pr_p^{2}}=\frac{200Nm}{\frac{1}{2}200kg*4m^2+50kg*1m^2}=\frac{200\frac{kgm^2}{s^2}}{450Nm^2}\approx 0.44\frac{rad}{s^2}$ .

c. Similar equation to b, but changing $r_p=2m$ , so

$alpha=\dfrac{200\frac{kgm^2}{s^2}}{\frac{1}{2}200*4kg\,m^2+50*4 kg\,m^2}=\dfrac{200}{600}\dfrac{1}{s^2}\approx 0.33 \frac{rad}{s^2}$ .

d. The angular acceleration when sitting in the middle is larger because the moment of inertia of the person is smaller, meaning that the person has less inertia to rotate.

5 0

3 years ago

Dahasolnce [82]

Answer:

1 million hahahahahahahahhahah

7 0

3 years ago

A runner went from 6 m/s and two seconds what was his acceleration

Radda [10]

Answer:

is it 3?

Explanation:

Im taking a guess and just dividing 6 and 2

8 0

3 years ago

A boy starts from point A and walks 3 meters toward the north, then turns around and walks 6 meters toward the south.

docker41 [41]

Answer:

Total distance traveled = 9 m

Explanation:

Given:

Distance travel towards north = 3 meter

Distance travel towards south = 6 meter

Find:

Total distance traveled

Computation:

Total distance traveled = Sum of total distance

Total distance traveled = Distance travel towards north + Distance travel towards south

Total distance traveled = 3 m + 6 m

Total distance traveled = 9 m

3 0

2 years ago