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Olenka [21]
4 years ago
10

What is the volume of a 72 g of copper if copper has a density of 9 g/cm3

Physics
1 answer:
azamat4 years ago
6 0

Answer:

8 centimetre cube

Explanation:

because volume is equal to mass per densiry

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<h3><em>If two objects with the same charge are brought towards each other the force produced will be repulsive, it will push them apart. If two objects with opposite charges are brought towards each other the force will be attractive, it will pull them towards each other.</em></h3><h3><em>hope it helps.... thank you....</em></h3>
7 0
3 years ago
Please complete it if you know the answer. "The active region of a transistor is for.........
zubka84 [21]

Answer:

the active region is bound by cutoff region and saturation or power dissipation region.

Explanation:

5 0
3 years ago
The solid rod shown has a radius of 0.75 in. if it is subjected to the force the 500lb determine the max normal stress developed
Kipish [7]
The normal stress follows the formula written below:

σ = F/A

There are two types of stress, axial and tangential. Since we are only given with the dimension of the radius (and not the length), the possible stress is axial. So, the area is,

A = πr² = π(0.75 in)² = 1.767 in²

So,
σ = F/A = 500 lb/1.767 in² = <em>282.94 psi</em>
7 0
3 years ago
An isolated conducting sphere has a 10 cm radius. One wire carries a current of 1.000 002 0 A into it. Another wire carries a cu
romanna [79]

Answer:

t = 5.56 ms

Explanation:

Given:-

- The current carried in, Iin = 1.000002 C

- The current carried out, Iout = 1.00000 C

- The radius of sphere, r = 10 cm

Find:-

How long would it take for the sphere to increase in potential by 1000 V?

Solution:-

- The net charge held by the isolated conducting sphere after (t) seconds would be:

                                   qnet = (Iin - Iout)*t

                                   qnet = t*(1.000002 - 1.00000) = 0.000002*t

- The Volt potential on the surface of the conducting sphere according to Coulomb's Law derived result is given by:

                                   V = k*qnet / r

Where,                        k = 8.99*10^9   ..... Coulomb's constant

                                   qnet = V*r / k

                                   t = 1000*0.1 / (8.99*10^9 * 0.000002)

                                   t = 5.56 ms

                                   

7 0
3 years ago
What are the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz? It is open at both ends. (The overt
vekshin1

Answer:

f_{2}=180Hz,f_{3}=270Hz,f_{4}=360Hz\\

Explanation:

Given data

Frequency f=90 Hz

To find

First three overtones of bassoon

Solution

The fundamental frequency of bassoon is found by substituting n=1 in below equation

f=v/λ=nv/2L

f_{1}=v/2L

The first overtone of bassoon is found by substituting n=2

So

f_{2}=2v/2L\\f_{2}=2(v/2L)\\as \\f_{1}=v/2L\\So\\f_{2}=2f_{1}\\f_{2}=2(90Hz)\\f_{2}=180Hz

The second overtone of bassoon is found by substituting n=3

So

f_{3}=3v/2L\\f_{3}=3(v/2L)\\as \\f_{1}=v/2L\\So\\f_{3}=3f_{1}\\f_{3}=3(90Hz)\\f_{3}=270Hz

The third overtone of bassoon is found by substituting n=4

So

f_{4}=4v/2L\\f_{4}=4(v/2L)\\as \\f_{1}=v/2L\\So\\f_{4}=4f_{1}\\f_{4}=4(90Hz)\\f_{4}=360Hz

5 0
3 years ago
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