Ask question

Ask question

All categories

2 years ago

8

A speed versus time graph is shown:

1 answer:

erma4kov [3.2K]2 years ago

6 0

In the given graph, from 4.0 s to 8.0 s, the object is at rest because the speed is zero.

In the given graph we can deduce the following;

at the time interval, 0 s to 3.5 s, the speed of the object = 1 cm/s

when the time, t= 4 s, the <em>speed</em> of the object = 0 cm/s

at the time interval, 4.0 s to 8.0 s, the<em> speed </em>of the object = 0 cm/s

When the <em>speed</em> of an object is zero (0), the object is simply at rest.

Thus, we can conclude that in the given graph, from 4.0 s to 8.0 s, the object is at rest because the speed is zero.

Learn more here:brainly.com/question/10454047

You might be interested in

Which surface has the most friction? a An ice rink b A grassy field c A paved road

vovangra [49]

Answer:

An Ice Rink

Explanation:

7 0

3 years ago

Read 2 more answers

****PLEASE HELP**** THERE ARE TWO QUESTIONS (ITS EASY)

notka56 [123]

but I think it's a and f

and

b and e

3 0

3 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

3 years ago

Boyle's Law states that when a sample of gas is compressed at a constant temperature, the pressure P and the volume V satisfy th

jarptica [38.1K]

Answer: The volume is decreasing at a rate of 80 cm3/min

Explanation: Please see the attachments below

8 0

3 years ago

Bagaimana cara untuk melatih kemampuan melempar tangkap yang baik dalam bola basket

anyanavicka [17]

Answer:

pergi ke pertandingan sepak bola dan amati

7 0

3 years ago