Snow tends to sublime outside during cold, clear winter days when the atmospheric pressure is low. During this proces

Andrew is riding his bike is riding in a circle that has a radius of 10 m. He keeps

Dmitrij [34]

The acceleration that Andrew experience during his ride is 3.6m/s²

The formula for calculating centripetal acceleration is expressed as:

a = v²/r

v is the speed

r is the radius

Given the following expression

v = 6m/s

r = 10m

Substitute the given parameters into the formula

a = 6²/10

a = 36/10

a = 3.6m/s²

Hence the acceleration that Andrew experience during his ride is 3.6m/s²

Learn more here: brainly.com/question/1268866

8 0

2 years ago

It requires 3480 J to move a 675 N object how far?

goblinko [34]

5.16 meters!!!!!!!!!!!!!!!!!!!!

4 0

2 years ago

In an LC circuit at one time the charge stored by the capacitor is 10 mC and the current is 3.0 A. If the frequency of the circu

Ronch [10]

Answer:

$i_2=3.61\ A$

Explanation:

<u>LC Circuit</u>

It's a special circuit made of three basic elements: The AC source, a capacitor, and an inductor. The charge, current, and voltage are oscillating when there is an interaction between the electric and magnetic fields of the elements. The following variables will be used for the formulas:

$q, q_1, q_2$ = charge of the capacitor in any time $t, t_1, t_2$

$q_o$ = initial charge of the capacitor

$\omega$ =angular frequency of the circuit

$i, i_1, i_2$ = current through the circuit in any time $t, t_1, t_2$

The charge in an LC circuit is given by

$q(t) = q_0 \, cos (\omega t )$

The current is the derivative of the charge

$\displaystyle i(t) = \frac{dq(t)}{dt} = - \omega q_0 \, sin(\omega t).$

We are given

$q_1=10\ mc=0.01\ c, i_1=3\ A,\ q_2=6\ mc=0.006\ c\ ,\ f=\frac{1000}{4\pi}$

It means that

$q(t_1) = q_0 \, cos (\omega t_1 )=q_1\ .......[eq 1]$

$i(t_1) = - \omega q_0 \, sin(\omega t_1)=i_1.........[eq 2]$

From eq 1:

$\displaystyle cos (\omega t_1 )=\frac{q_1}{q_0}$

From eq 2:

$\displaystyle sin(\omega t_1)=-\frac{i_1}{\omega q_0}$

Squaring and adding the last two equations, and knowing that

$sin^2x+cos^2x=1$

$\displaystyle \left ( \frac{q_1}{q_0} \right )^2+\left ( \frac{i_1}{\omega q_0} \right )^2=1$

Operating

$\displaystyle \omega^2q_1^2+i_1^2=\omega^2q_o^2$

Solving for $q_o$

$\displaystyle q_o=\frac{\sqrt{\omega^2q_1^2+i_1^2}}{\omega}$

Now we know the value of $q_0$ , we repeat the procedure of eq 1 and eq 2, but now at the second time $t_2$ , and solve for $i_2$

$\displaystyle \omega^2q_2^2+i_2^2=\omega^2q_o^2$

Solving for $i_2$

$\displaystyle i_2=w\sqrt{q_o^2-q_2^2}$

Now we replace the given values. We'll assume that the placeholder is a pi for the frequency, i.e.

$\displaystyle f=\frac{1}{4\pi}\ KHz$

$w=2\pi f=500\ rad/s$

$\displaystyle q_o=\frac{\sqrt{(500)^2(0.01)^2+3^2}}{500}$

$q_0=0.01166\ c$

Finally

$\displaystyle i_2=500\sqrt{0.01166^2-.006^2}$

$i_2=5\ A$

3 0

2 years ago

Imagine you are holding an apple. a. Does this apple have energy? How do you know? b. How could you increase the potential energ

Len [333]

A). The apple has thermal energy, because its temperature is higher
than absolute zero.
It also has chemical energy, because if I eat it, I get a burst of energy
and I become ambitious for a while.
It also has gravitational potential energy, because if I drop it on my foot,
it could bruise one of my piggies.

b). I could increase its potential energy by lifting it higher, like over my head.

c). As long as I'm just holding the apple, it doesn't have any kinetic energy.
I could give it some kinetic energy by throwing it.
Or I could just drop it, and let gravity give it kinetic energy.

7 0

2 years ago

An object is moving at a constant velocity of 10 m/s for 5 seconds. What is

german

initial velocity (u)=0m/s

final velocity (v)=10m/s

time( t)=5s

acceleration (a)=v-u÷t

acceleration (a)=10-0÷5

acceleration (a)=10÷5

acceleration (a)=2

therefore acceleration (a)=2m/s

3 0

2 years ago