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3 years ago

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If 1.8 1016 electrons enter a light bulb in 3 milliseconds, what is the magnitude of the electron current at that point in the c

ircuit?

1 answer:

Mice21 [21]3 years ago

7 0

Answer:

I = 0.96 A

Explanation:

No of electrons, $n=1.8\times 10^{16}$

Time, t = 3 ms = $3\times 10^{-3}\ s$

We need to find the electric current. We know that electric charge per unit time is equal to the electric current.

$I=\dfrac{q}{t}$

q = ne (Quantization of electric charge)

$I=\dfrac{ne}{t}\\\\I=\dfrac{1.8\times 10^{16}\times 1.6\times 10^{-19}}{3\times 10^{-3}}\\\\I=0.96\ A$

So, the electric current is 0.96 A.

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A movie stunt performer is filming a scene where he swings across a river on a vine. The safety crew must use a vine with enough

Julli [10]

Answer:

1125.66956 N

Explanation:

m = Mass of stunt performer

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of the swing = 7 m/s

T = Tension

r = Radius of the swing = Length of vine = 11.5 m

From the free body diagram

$T-mg-m\frac{v^2}{r}=0\\\Rightarrow T=mg+m\frac{v^2}{r}\\\Rightarrow T=m(g+\frac{v^2}{r})\\\Rightarrow T=80(9.81+\frac{7^2}{11.5})\\\Rightarrow T=1125.66956\ N$

The minimum tension force the vine must be able to support without breaking is 1125.66956 N

7 0

3 years ago

PLZZZZZZZZZZ HELP QUICK!!!!!!!!!!!!

kifflom [539]

In an atom of hydrogen the orbit radius is given by the formula:
r = n² · α₀

where:
n = number of orbit = 15
α₀ = Bohr radius (innermost radius) = 0.529 Â

Since d = 2 · r, we can write:
d = n² · d₀
= 15² · 1.06
= 238.5 Â

Hence, the <span>diameter of the fifteenth orbit of the hydrogen atom is 238.5 </span>Â.

8 0

3 years ago

An electric ceiling fan is rotating about a fixed axis with an initial angular velocity magnitude of 0.300 rev/s . The magnitude

Salsk061 [2.6K]

1) 1.2 m/s

First of all, we need to find the angular velocity of the blade at time t = 0.200 s. This is given by

$\omega_f = \omega_i + \alpha t$

where

$\omega_i = 0.300 rev/s$ is the initial angular velocity

$\alpha = 0.895 rev/s^2$ is the angular acceleration

Substituting t = 0.200 s, we find

$\omega_f = 0.300 + (0.895)(0.200)=0.479 rev/s$

Let's now convert it into rad/s:

$\omega_f = 2\pi \cdot 0.479 rev/s=3.01 rad/s$

The distance of a point on the tip of the blade is equal to the radius of the blade, so half the diameter:

$r=\frac{0.800}{2}=0.400 m$

And so now we can find the tangential speed at t = 0.200 s:

$v=\omega_f r =(3.01)(0.400)=1.2 m/s$

2) $2.25 m/s^2$

The tangential acceleration of a point rotating at a distance r from the centre of the circle is

$a_t = \alpha r$

where $\alpha$ is the angular acceleration.

First of all, we need to convert the angular acceleration into rad/s^2:

$\alpha = 0.895 rev/s^ \cdot 2 \pi =5.62 rad/s^2$

A point on the tip of the blade has a distance of

r = 0.400 m

From the centre; so, the tangential acceleration is

$a_t = (5.62)(0.400)=2.25 m/s^2$

3) $3.6 m/s^2$

The centripetal acceleration is given by

$a=\frac{v^2}{r}$

where

v is the tangential speed

r is the distance from the centre of the circle

We already calculate the tangential speed at point a):

v = 1.2 m/s

while the distance of a point at the end of the blade from the centre is

r = 0.400 m

Therefore, the centripetal acceleration is

$a=\frac{1.2^2}{0.400}=3.6 m/s^2$

7 0

3 years ago

What is Particle-antiparticle annihilation after watching the video minutephysics Antimatter Explained

LenKa [72]

Answer:

Particle-antiparticle annihilation is the result of a direct interaction between a particle and its corresponding anti particle, like a proton and an anti-proton or an electron and a positron. The result of such interaction is the production of two powerful gamma rays, each of whom has an energy roughly equal to E = mc^2, where m is the mass of the particle.

3 0

4 years ago

A piston–cylinder assembly contains 5.0 kg of air, initially at 2.0 bar, 30 oC. The air undergoes a process to a state where th

vlada-n [284]

Answer:

Explanation:

The process is isothermic, as P V = constant .

work done = 2.303 n RT log P₁ / P₂

= 2.303 x 5 / 29 x 8.3 x 303 log 2 / 1 kJ

= 300.5k J

This energy in work done by the gas will come fro heat supplied as internal energy is constant due to constant temperature.

heat supplied = 300.5k J

specific volume is volume per unit mass

v / m

pv = n RT

pv = m / M RT

v / m = RT / p M

specific volume = RT / p M

option B is correct.

5 0

3 years ago