Ask question

Ask question

All categories

3 years ago

15

A distance-time graph indicates that an object moves 100m in 4s and then remains at rest for 6s. What is the average speed of th

e object?

1 answer:

FromTheMoon [43]3 years ago

3 0

A distance-time graph indicates that an object moves 100m in 4s and then remains at rest for 6s. What is the average speed of the object?

What is this from?

You might be interested in

Please help don’t know how to show my work

Fudgin [204]

Answer:

64.50816

Explanation:

3 0

3 years ago

Why is gravitational forces equal for everyone?

Alik [6]

That equation is Newton's universal law of gravitation. ... Any two masses exert equal-and-opposite gravitational forces on each other. If we drop a ball, the Earth exerts a gravitational force on the ball, but the ball exerts a gravitational force of the same magnitude (and in the opposite direction) on the Earth.

7 0

4 years ago

Which statement correctly compares infrared light to ultraviolet light

allochka39001 [22]

I am fairly certain it's D. Good luck!

8 0

3 years ago

Read 2 more answers

Can someone please tell me if it’s A. Or D. Please

ale4655 [162]

D. similar types of rocks and fossils

3 0

3 years ago

Read 2 more answers

At t1 = 2.00 s, the acceleration of a particle in counterclockwise circular motion is 6.00 i + 4.00 j m/s2 . It moves at constan

Jet001 [13]

The particle moves with constant speed in a circular path, so its acceleration vector always points toward the circle's center.

At time $t_1$ , the acceleration vector has direction $\theta_1$ such that

$\tan\theta_1=\dfrac{4.00}{6.00}\implies\theta_1=33.7^\circ$

which indicates the particle is situated at a point on the lower left half of the circle, while at time $t_2$ the acceleration has direction $\theta_2$ such that

$\tan\theta_2=\dfrac{-6.00}{4.00}\implies\theta_2=-56.3^\circ$

which indicates the particle lies on the upper left half of the circle.

Notice that $\theta_1-\theta_2=90^\circ$ . That is, the measure of the major arc between the particle's positions at $t_1$ and $t_2$ is 270 degrees, which means that $t_2-t_1$ is the time it takes for the particle to traverse 3/4 of the circular path, or 3/4 its period.

Recall that

$\|\vec a_{\rm rad}\|=\dfrac{4\pi^2R}{T^2}$

where $R$ is the radius of the circle and $T$ is the period. We have

$t_2-t_1=(5.00-2.00)\,\mathrm s=3.00\,\mathrm s\implies T=\dfrac{3.00\,\rm s}{\frac34}=4.00\,\mathrm s$

and the magnitude of the particle's acceleration toward the center of the circle is

$\|\vec a_{\rm rad}\|=\sqrt{\left(6.00\dfrac{\rm m}{\mathrm s^2}\right)^2+\left(4.00\dfrac{\rm m}{\mathrm s^2}\right)^2}=7.21\dfrac{\rm m}{\mathrm s^2}$

So we find that the path has a radius $R$ of

$7.21\dfrac{\rm m}{\mathrm s^2}=\dfrac{4\pi^2R}{(4.00\,\mathrm s)^2}\implies\boxed{R=2.92\,\mathrm m}$

8 0

3 years ago