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GuDViN [60]
3 years ago
9

6.7.1 Current and Current Density A sphere of radius 10 mm that carries a charge of 8 nC =8x 10°C is whirled in a circle at the

end of an insulated string. The rotation frequency is 100x rad/s. (a) What is the basic definition of current in terms of charge? (b) What average current does this rotating charge represent? (c) What is the average current density over the area traversed by the sphere?​
Physics
1 answer:
aev [14]3 years ago
6 0

Answer:

Part a)

Rate of charge flow is known as electric current

Part b)

Average current flow is

i = 4\times 10^{-7} A

Part c)

As we know that current density is current per unit area

j = 1.27 \times 10^{-3} A/m^2

Explanation:

As we know that angular frequency of rotation is

\omega = 100\pi rad/s

now by basic definition of electric current

Part a)

Rate of charge flow is known as electric current

i = \frac{dq}{dt}

so here we have

i = \frac{Q}{T}

i = Qf

Part b)

here we know that

\omega = 2\pi f

100\pi = 2\pi f

f = 50 Hz

now we have

i = (8\times 10^{-9})(50)

i = 4\times 10^{-7} A

Part c)

As we know that current density is current per unit area

So we have

j = \frac{i}{A}

j = \frac{4\times 10^{-7}}{\pi(10\times 10^{-3})^2}

j = 1.27 \times 10^{-3} A/m^2

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What is the wavelength of a 1.28 x 10^17 Hz wave?
lisov135 [29]

Answer:

The wavelength of the wave is 2.34 nm.

Explanation:

It is required to find the wavelength of a 1.28\times 10^{17}\ Hz wave. A wave moves with a speed of light. Speed of a wave is given in terms of wavelength and frequency. So,

v=f\lambda

\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{1.28\times 10^{17}}\\\\\lambda=2.34\times 10^{-9}\ m\\\\\text{or}\\\\\lambda=2.34\ nm

So, the wavelength of the wave is 2.34 nm.

4 0
3 years ago
I have an astronomy question... Spinning up the solar nebula. The orbital speed of the material in the solar nebula at Pluto's a
attashe74 [19]
<span>The angular momentum of a particle in orbit is 

l = m v r 

Assuming that no torques act and that angular momentum is conserved then if we compare two epochs "1" and "2" 

m_1 v_1 r_1 = m_2 v_2 r_2 

Assuming that the mass did not change, conservation of angular momentum demands that 

v_1 r_1 = v_2 r_2 

or 

v1 = v_2 (r_2/r_1) 

Setting r_1 = 40,000 AU and v_2 = 5 km/s and r_2 = 39 AU (appropriate for Pluto's orbit) we have 

v_2 = 5 km/s (39 AU /40,000 AU) = 4.875E-3 km/s

Therefore, </span> the orbital speed of this material when it was 40,000 AU from the sun is <span>4.875E-3 km/s.

I hope my answer has come to your help. Thank you for posting your question here in Brainly.
</span>
3 0
3 years ago
If a ball is thrown vertically upward with a velocity of 128 ft/s, then its height after t seconds is s = 128t − 16t2.
Fed [463]

Answer:

a) 256 ft

b) 32 ft/s

c) -32 ft/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32 ft/s²

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{0^2-128^2}{2\times -32}\\\Rightarrow s=256\ ft

Maximum height = 256 ft

When s = 240 ft

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times -32\times 240+128^2}\\\Rightarrow v=32\ ft/s

Speed of the ball at 240 ft is 32 ft/s while going up

Now this will be the velocity of the ball when it reaches 240 ft, which will be considered as the initial velocity

v=u+at\\\Rightarrow 0=32-32t\\\Rightarrow t=\frac{-32}{-32}=1\ s

Now, initial velocity will be considered as zero

v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 32\times 16+0^2}\\\Rightarrow v=32\ ft/s

Speed of the ball at 240 ft is -32 ft/s while going down

3 0
3 years ago
Compared to the tropical rainforests, the temperate rainforests generally have __________.
Oxana [17]
<span>a. less biomass
b. more biomass
c. more broad-leaf trees
d. locations near the equator


i would say c
</span>
8 0
4 years ago
An object experiences an acceleration of 8.5 m/s^2 over a distance of 300 m. After that acceleration it has a velocity of 400 m/
Snezhnost [94]

Answer:

393.6m/s

Explanation:

Given parameters:

Acceleration  = 8.5m/s²

Distance  = 300m

Final velocity  = 400m/s

Unknown:

Initial velocity  = ?

Solution:

To solve this problem, we use the expression below;

             v² = u²  + 2as

v is the final velocity

u is the initial velocity

a is the acceleration

s is the distance

      So;

               v²  - 2as = u²

        u²   = v²   - 2as

        u²  = 400²   - (2 x 8.5 x 300)  

         u   = 393.6m/s

7 0
3 years ago
Read 2 more answers
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