Help me! Btw he’s playing golf.

You are a project manager for a manufacturing company. One of the machine parts on the assembly line is a thin, uniform rod that

ra1l [238]

Answer:

a) $I=0.012\ kg.m^2$

b) $I=0.012\ kg.m^2$

Explanation:

Given that:

mass of rod, $m=0.4\ kg$

length of the rod, $l=0.6\ kg$

(a)

Moment of inertia of rod about its center and perpendicular to the rod is given as:

$I=\frac{1}{12} m.l^2$

$I=\frac{1}{12} 0.4\times 0.6^2$

$I=0.012\ kg.m^2$

(b)

Moment of inertia on bending the rod to V-shape of 60 degree angle and axis being perpendicular to the plane of V at the vertex.

We treat it as two rod with axis of rotation at the end and perpendicular to the plane of rotation. 

So, the mass and the length of the rod will become half of initial value.

$I=I_1+I_2$

$I=\frac{1}{3} \frac{m}{2} (\frac{l}{2})^2 +\frac{1}{3} \frac{m}{2} (\frac{l}{2})^2$

$I=2[ \frac{1}{3}\times 0.2\times 0.3^2]$

$I=0.012\ kg.m^2$

7 0

3 years ago

Find the frequency (in hertz) of the first overtone for a 1.75 m pipe that is closed at only one end. Use 350 m/s for the speed

serg [7]

Answer:

The first overtone frequency is 100 Hz.

Solution:

According to the question:

Length of pipe, l = 1.75 m

Speed of sound in air, v_{sa} = 350 m/s

Frequency of first overtone, $f_{1}$ is given by:

$f_{1} = \frac{v_{sa}}{2l}$

$f_{1} = \frac{350}{2\times 1.75} = 100 Hz$

Since, the frequency, as clear from the formula depends only on the speed

and the length. It is independent of the air temperature.

Thus there will be no effect of air temperature on the frequency.

7 0

3 years ago

Five ways to calculate speed without speedometer

QveST [7]

One way is speed=distance divided by time

8 0

4 years ago

Help for physics

12345 [234]

Answer:

A. 5 m/s

Explanation:

From the graph, for the first 2 seconds, the graph is a straight line meaning that the slope is a constant.

Average speed of an object is the rate of change of position. Here, the position of the object changes from 0 m to 10 m for a time interval of 2 seconds.

The change in position ( $\Delta x$ ) and time interval ( $\Delta t$ ) are given as:

$\Delta x=10-0=10\ m\\\Delta t=2-0=2\ s$

Therefore, the average speed ( $s_{avg}$ ) is given as the ratio of the total change in position and the time interval for the change.

$s_{avg}=\frac{\Delta x}{\Delta t}=\frac{10-0}{2-0}=\frac{10}{2}=5\ m/s$

Hence, the average speed is 5 m/s.

8 0

3 years ago

Read 2 more answers

A ring is attached at the center of the underside of a trampoline. A sneaky teenager crawls under the trampoline and uses the ri

IRINA_888 [86]

Answer:

E = 301.5 J

Explanation:

We have,

Mass of mother, m = 67 kg

Here, a sneaky teenager crawls under the trampoline and uses the ring to pull the trampoline slowly down. As she passes through the position at which she was before her son stretched the trampoline, her speed is 3 m/s.

It is required to find the elastic potential energy the son add to the trampoline by pulling it down. It is based on the conservation of energy.

The elastic potential energy of the mother = the elastic potential energy the son add to the trampoline.

$\dfrac{1}{2}mv^2=\dfrac{1}{2}kx^2$

So, the elastic potential energy is :

$E=\dfrac{1}{2}mv^2\\\\E=\dfrac{1}{2}\times 67\times 3^2\\\\E=301.5\ J$

So, the elastic potential energy of 301.5 J the son add to the trampoline by pulling it down.

6 0

4 years ago