All categories

2 years ago

The acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

1 answer:

6 0

We have that for the Question "the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?"

it can be said that the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

From the question we are told

the acceleration of the object at time t = 0.7 s is most nearly equal to which of the following?

Generally the equation for the Force is mathematically given as

F=\frac{F}{dx}

Therefore

F=-kdx

k=600Nm^{-1}

now

K.E=0.5x ds^2

K.E=600*(-0.1^2)

K.E=3J

Therefore

the acceleration of the object at time t = 0.7 s is most nearly equal to the slope of the line connecting the origin and the point where the graph where the graph crosses the 0.7s grid line

For more information on this visit

brainly.com/question/23379286

You might be interested in

What are two benefits of scientists using a diagram to model the water cycle?

Sidana [21]

Explanation:

it can be used to show how the parts of the cycle relate to one another

8 0

2 years ago

Is (2, 2) a solution of y < 4x-6<br> Help

vesna_86 [32]

Answer:

B) No.

Explanation:

Okay,so,

this is equation is y=mx +b

mx represents the slope

and b represents the y-intercept

in order to figure this out you need to plot the y-intercept first

that makes its (0,-6) because the 6 is negative in the equation

4x is also equal to 4/1 since we dont know what x is

we have to do rise over run for this

you go up 4 spots on the y intercept from -6 because 4 is positive

then you go to the right 1 time because 1 is positive.

this leaves you at (1,-2)

so, (2,2) is NOT a solution

7 0

3 years ago

What is the electric potential, i.e. the voltage, 0.30 m from a point charge of 6.4 x 10-C?

Gwar [14]

Answer:

V = 192 kV

Explanation:

Given that,

Charge, $q=6.4\times 10^{-6}\ C$

Distance, r = 0.3 m

We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

$V=\dfrac{kq}{r}\\\\V=\dfrac{9\times 10^9\times 6.4\times 10^{-6}}{0.3}\\\\V=192000\ V\\\\V=192\ kV$

So, the required electric potential is 192 kV.

3 0

2 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. in 6.00 s, it rotates 44.5 rad. du

Klio2033 [76]

a. The disk starts at rest, so its angular displacement at time $t$ is

$\theta=\dfrac\alpha2t^2$

It rotates 44.5 rad in this time, so we have

$44.5\,\mathrm{rad}=\dfrac\alpha2(6.00\,\mathrm s)^2\implies\alpha=2.47\dfrac{\rm rad}{\mathrm s^2}$

b. Since acceleration is constant, the average angular velocity is

$\omega_{\rm avg}=\dfrac{\omega_f+\omega_i}2=\dfrac{\omega_f}2$

where $\omega_f$ is the angular velocity achieved after 6.00 s. The velocity of the disk at time $t$ is

$\omega=\alpha t$

so we have

$\omega_f=\left(2.47\dfrac{\rm rad}{\mathrm s^2}\right)(6.00\,\mathrm s)=14.8\dfrac{\rm rad}{\rm s}$

making the average velocity

$\omega_{\rm avg}=\dfrac{14.8\frac{\rm rad}{\rm s}}2=7.42\dfrac{\rm rad}{\rm s}$

Another way to find the average velocity is to compute it directly via

$\omega_{\rm avg}=\dfrac{\Delta\theta}{\Delta t}=\dfrac{44.5\,\rm rad}{6.00\,\rm s}=7.42\dfrac{\rm rad}{\rm s}$

c. We already found this using the first method in part (b),

$\omega=14.8\dfrac{\rm rad}{\rm s}$

d. We already know

$\theta=\dfrac\alpha2t^2$

so this is just a matter of plugging in $t=12.0\,\mathrm s$ . We get

$\theta=179\,\mathrm{rad}$

Or to make things slightly more interesting, we could have taken the end of the first 6.00 s interval to be the start of the next 6.00 s interval, so that

$\theta=44.5\,\mathrm{rad}+\left(14.8\dfrac{\rm rad}{\rm s}\right)t+\dfrac\alpha2t^2$

Then for $t=6.00\,\rm s$ we would get the same $\theta=179\,\rm rad$ .

7 0

3 years ago

A 3.00 μF capacitor is charged to 480 V and a 4.00 μF capacitor is charged to 500 V . Part A These capacitors are then disconnec

pogonyaev

Q before connected = Q after connected C1V1+C2V2 = (C1+C2) V

C1= 3×10^-6 F

V1= 480v

C2= 4×10^-6 F

V2= 500v

(3×10^-6)×(480) + (4×10^-6)×(500) = (3×10^-6 + 4×10^-6) × V

Simplifying the above, we get:

( 1440× 10^-6) + (2000 ×10^-6) = (7 × 10^-6) × V.

Further simplified as:

3440 × 10^-6 = 7 × 10^-6 × V

Making V the subject

V = 491.43volts

Therefore the potential difference across each capacitor is 491.43v

4 0

3 years ago