Ask question

Ask question

All categories

3 years ago

9

A student performs this experiment and measures the bar to have a mass of 150g and length of 36cm. What is the moment of inertia

of the bar?

1 answer:

Ugo [173]3 years ago

4 0

Answer:

The moment of inertia of the bar is $45\times10^{-4}\ kg-m^2$

Explanation:

Given that,

mass of bar = 150 g

Length l = 36 cm

We need to calculate the moment of inertia of the bar

Using formula of moment inertia

$I=\dfrac{1}{12}Ml^2$

Where,

M = mass of the bar

L = length of the bar

Put the value into the formula

$I=\dfrac{1}{12}\times150\times10^-3\times36\times10^{-2}$

$I=45\times10^{-4}\ kg-m^2$

Hence, The moment of inertia of the bar is $45\times10^{-4}\ kg-m^2$

You might be interested in

The attractive electrostatic force between the point charges +8.44 ✕ 10-6 and q has a magnitude of 0.961 n when the separation b

d1i1m1o1n [39]

The electrostatic force between two charges Q1 and q is given by
$F=k_e \frac{Q_1 q}{r^2}$
where
ke is the Coulomb's constant
Q1 is the first charge
q is the second charge
r is the distance between the two charges

Re-arranging the formula, we have
$q= \frac{F r^2}{k_e Q_1}$
and since we know the value of the force F, of the charge Q1 and the distance r between the two charges, we can calculate the value of q:
$q= \frac{(0.961 N)(0.67 m)^2}{(8.99 \cdot 10^9 N m^2 C^{-2})(+8.44\cdot 10^{-6}C)}=5.69 \cdot 10^{-6} C$

And since the force is attractive, the two charges must have opposite sign, so the charge q must have negative sign.

6 0

3 years ago

g The current through a 10 m long wire has a current density of 4 cross times 10 to the power of 6 space open parentheses bevell

olga55 [171]

Answer:

The value is $V = 2 V$

Explanation:

From the question we are told that

The length of the wire is $l = 10 \ m$

The current density is $J = 4*10^6 \ A/m^2$

The conductivity is $\sigma = 2*10^{7} \ S/m$

Generally conductivity is mathematically represented as

$\sigma = \frac{l}{RA}$

Here R is the resistance which is mathematically represented as

$R = \frac{V}{I}$

Here I is the current which is mathematically represented as

$I = J * A$

So

$R = \frac{V}{ J * A}$

And

$\sigma = \frac{l}{\frac{V}{ J * A} * A}$

=> $\sigma = \frac{l}{\frac{V}{J}}$

=> $V = \frac{l * J}{\sigma }$

=> $V = \frac{10 * 4*10^6}{2*10^{7} }$

=> $V = 2 V$

5 0

3 years ago

A rope of total mass m hnd length L is suspended vertically with an object of mass M suspended from the lower end. Find an expre

pantera1 [17]

Answer:

Part a)

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

t = 12 s

Explanation:

Part a)

Tension in the rope at a distance x from the lower end is given as

$T = \frac{m}{L}xg + Mg$

so the speed of the wave at that position is given as

$v = \sqrt{\frac{T}{\mu}}$

here we know that

$\mu = \frac{m}{L}$

now we have

$v = \sqrt{\frac{ \frac{m}{L}xg + Mg}{m/L}$

$v = \sqrt{xg + \frac{MLg}{m}}$

Part b)

time taken by the wave to reach the top is given as

$t = \int \frac{dx}{\sqrt{xg + \frac{MLg}{m}}}$

$t = \frac{1}{g}(2\sqrt{xg + \frac{MLg}{m}})$

$t = \frac{2}{9.8}(\sqrt{(39.2\times 9.8) + \frac{8(39.2)(9.8)}{1}})$

$t = 12 s$

4 0

3 years ago

How do very high density objects appear in an ultrasound?

evablogger [386]

<span>Density is entirely unrelated to an object's size. It is a property of a given</span>

7 0

3 years ago

HELP PLEASE!! GIVING BRAINLIEST!! If you answer this correctly ill answer some of your questions you have posted! (20pts)

anyanavicka [17]

Explanation:

potential energy =360800J

mass(m)=?

height (h)=25m

g=9.8m/s²

we have

potential energy =360800J

mgh=360800J

m×9.8×25=360800

m=360800/(9.8×25)=1472.653061kg

8 0

3 years ago

Read 2 more answers