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2 years ago

calculate how much charge passes through the LED in 1 hour if the current flowing in the circuit below is 350mA

1 answer:

3 0

Given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

<h3>What is Current?</h3>

Current is simply the rate of flow of charged particles i.e electrons caused by EMF or voltage.

If a charge passes through the cross-section of a conductor in a given time, the current I is expressed as;

I = Q/t

Where Q is the charge and t is time elapsed.

Given the data in the question;

Time elapsed t = 1hr = 3600s

Current I = 350mA = 0.35A

Charge Q = ?

We substitute our given values into the expression above to determine the charge.

I = Q/t

Q = I × t

Q = 0.35A × 3600s

Q = 1260C

Therefore, given the the current flowing in the circuit and the elapsed time, the charge that passes through the LED is 1260 Coulombs.

Learn more about current here: brainly.com/question/3192435

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Show that rigid body rotation near the Galactic center is consistent with a spherically symmetric mass distribution of constant

irakobra [83]

To solve this problem we will use the concepts related to gravitational acceleration and centripetal acceleration. The equality between these two forces that maintains the balance will allow to determine how the rigid body is consistent with a spherically symmetric mass distribution of constant density. Let's start with the gravitational acceleration of the Star, which is

$a_g = \frac{GM}{R^2}$

Here

$M = \text{Mass inside the Orbit of the star}$

$R = \text{Orbital radius}$

$G = \text{Universal Gravitational Constant}$

Mass inside the orbit in terms of Volume and Density is

$M =V \rho$

Where,

V = Volume

$\rho =$ Density

Now considering the volume of the star as a Sphere we have

$V = \frac{4}{3} \pi R^3$

Replacing at the previous equation we have,

$M = (\frac{4}{3}\pi R^3)\rho$

Now replacing the mass at the gravitational acceleration formula we have that

$a_g = \frac{G}{R^2}(\frac{4}{3}\pi R^3)\rho$

$a_g = \frac{4}{3} G\pi R\rho$

For a rotating star, the centripetal acceleration is caused by this gravitational acceleration. So centripetal acceleration of the star is

$a_c = \frac{4}{3} G\pi R\rho$

At the same time the general expression for the centripetal acceleration is

$a_c = \frac{\Theta^2}{R}$

Where $\Theta$ is the orbital velocity

Using this expression in the left hand side of the equation we have that

$\frac{\Theta^2}{R} = \frac{4}{3}G\pi \rho R^2$

$\Theta = (\frac{4}{3}G\pi \rho R^2)^{1/2}$

$\Theta = (\frac{4}{3}G\pi \rho)^{1/2}R$

Considering the constant values we have that

$\Theta = \text{Constant} \times R$

$\Theta \propto R$

As the orbital velocity is proportional to the orbital radius, it shows the rigid body rotation of stars near the galactic center.

So the rigid-body rotation near the galactic center is consistent with a spherically symmetric mass distribution of constant density

6 0

3 years ago

A spring scale exerts a net force of 8.5 N on an object. What is the object's mass if it has an acceleration of

Amiraneli [1.4K]

Answer:

mass of the object is 2.18 kg

Explanation:

Given

Force (F) = 8.5 N = 8.5 kg.m/ $s^{2}$

acceleration (a) = 3.9 m/ $s^{2}$

Mass (m) = ?

We know that the newton's second law of motion gives the relation between mass of ab object. force acted upon and the amount the object is accelerated. It is expressed in the form of an equation:

F = ma

mass, m = F/a

= $\frac{8.5 kgms^{-2} }{3.9 ms^{-2} }$