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Greeley [361]
3 years ago
12

A current of 2.0A flows through a light bulb. What is the amount of charge flowing through the light bulb in 40 seconds?

Physics
1 answer:
lidiya [134]3 years ago
5 0

Answer:

Quantity of charge = 80 Coulombs

Explanation:

Given the following data;

Current = 2 A

Time = 40 seconds

To find the amount of charge flowing through the light bulb;

Mathematically, the quantity of charge passing through a conductor is given by the formula;

Quantity of charge = current * time

Substituting into the formula, we have;

Quantity of charge = 2 * 40

Quantity of charge = 80 Coulombs

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A mass attached to a 50.0 cm long string starts from rest and is rotated 40 times in one minute before reaching a final angular
ch4aika [34]

To solve this problem it is only necessary to apply the kinematic equations of angular motion description, for this purpose we know by definition that,

\theta = \frac{1}{2}\alpha t^2 +\omega_0 t + \theta_0

Where,

\theta = Angular Displacement

\alpha =Angular Acceleration

\omega_0 = Angular velocity

\theta_0 =Initial angular displacement

For this case we have neither angular velocity nor initial angular displacement, then

\theta = \frac{1}{2}\alpha t^2

Re-arrange for \alpha,

\alpha = \frac{2\theta}{t^2}

Replacing our values,

\alpha = \frac{2(40rev*\frac{2\pi rad}{1rev})}{60^2}

\alpha = 0.139rad/s

Therefore the ANgular acceleration of the mass is 0.139rad/s^2

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3 years ago
Help me, please with this queastion
sergejj [24]

Answer:

Kinetic energy is the energy due to motion. Potential energy is energy stored in matter. The joule (J) is the SI unit of energy and equals (kg×m2s2) ( kg × m 2 s 2 ) .

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2 years ago
A jet accelerates from rest down a runway at 1.75m/s² for a distance of 1500 m before takeoff.
babunello [35]
A. Using the third equation of motion:
v2 = u2 + 2as
from the question;
the jet was initially at rest
hence u = 0
a = 1.75m/s2
s = 1500m
v2 = 02 + 2(1.75)(1500)
v2 = 5250
v = √5250
v = 72.46m/s
hence it moves with a velocity of 72.46m/s.
b. s = ut + 1/2at2
1500 = 0(t) + 1/2(1.75)t2
1500 × 2 = 2× 1/2(1.75)t2
3000 = 1.75t2
1714.29 = t2
41.4 = t
hence the time taken for the plane to down the runway is 41.4s.


Read more on Brainly.com - brainly.com/question/18743384#readmore
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How are hot spots used to track plate motion
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