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3 years ago

How fast is the car traveling if it has driven a total of 200 miles in 5.5 hours?

1 answer:

6 0

Average speed = (distance covered) / (time to cover the distance)

   = (200 miles) / (5.5 hours)

   = (200/5.5) mile/hour

   = 36.4 miles per hour (rounded)

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Which of the following properties is the same for all electromagnetic radiation in a vacuum?

Crazy boy [7]

It is wavelength i think so!!!!

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3 years ago

If an unknown element has a mass of 17 and contains 6 neutrons, how many protons does it have ?

lbvjy [14]

The mass number is the total number of protons and neutrons within an atom and since we know that the unknown element has 6 neutrons, we can simply subtract the number of neutrons from the mass number to get the number of protons.

17 - 6 = 11

There are 11 protons in this unknown element.

Extra:

The number of protons (+) and electrons (-) are equal in a neutral atom so since you know that there are 11 protons you also know that there are 11 electrons. On the periodic table, the element with 11 electrons is Na or Sodium.

Hope this helps! :)

6 0

3 years ago

Starting from rest, a disk rotates about its central axis with constant angular acceleration. In 1.00 s, it rotates 21.0 rad. Du

ELEN [110]

With constant angular acceleration $\alpha$ , the disk achieves an angular velocity $\omega$ at time $t$ according to

$\omega=\alpha t$

and angular displacement $\theta$ according to

$\theta=\dfrac12\alpha t^2$

a. So after 1.00 s, having rotated 21.0 rad, it must have undergone an acceleration of

$21.0\,\mathrm{rad}=\dfrac12\alpha(1.00\,\mathrm s)^2\implies\alpha=42.0\dfrac{\rm rad}{\mathrm s^2}$

b. Under constant acceleration, the average angular velocity is equivalent to

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

where $\omega_f$ and $\omega_i$ are the final and initial angular velocities, respectively. Then

$\omega_{\rm avg}=\dfrac{\left(42.0\frac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)}2=42.0\dfrac{\rm rad}{\rm s}$

c. After 1.00 s, the disk has instantaneous angular velocity

$\omega=\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)=42.0\dfrac{\rm rad}{\rm s}$

d. During the next 1.00 s, the disk will start moving with the angular velocity $\omega_0$ equal to the one found in part (c). Ignoring the 21.0 rad it had rotated in the first 1.00 s interval, the disk will rotate by angle $\theta$ according to

$\theta=\omega_0t+\dfrac12\alpha t^2$

which would be equal to

$\theta=\left(42.0\dfrac{\rm rad}{\rm s}\right)(1.00\,\mathrm s)+\dfrac12\left(42.0\dfrac{\rm rad}{\mathrm s^2}\right)(1.00\,\mathrm s)^2=63.0\,\mathrm{rad}$

5 0

4 years ago

Describe the Compton Effect

Nana76 [90]

<h3>Compton scattering, discovered by Arthur Holly Compton, is the scattering of a photon after an interaction with a charged particle, usually an electron. If it results in a decrease in energy of the photon, it is called the Compton effect. Part of the energy of the photon is transferred to the recoiling electron.</h3>

7 0

3 years ago

Read 2 more answers

How do I know if i’m doing number 2 right?

Ksju [112]

We are given an object that is speeding up on a level ground.

Let's remember that the gravitational energy depends on the change in height, therefore, if the object is not changing its height it means that the gravitational energy remains constant.

The kinetic energy depends on the velocity. If the velocity is increasing this means that the kinetic energy is also increasing.

Now, every change in velocity requires acceleration and acceleration requires a force. The force and the distance that the object moves are equivalent to the work that is transferred to the object and therefore, the change in kinetic energy. This means that the total energy of the system increases as work is transferred to the mass.

We have that the total energy of the system increases in the form of kinetic energy and that the gravitational potential energy remains constant. Therefore, the diagrams should look like pie charts that grow but the area of the segment of the potential energy stays the same. It should look similar to the following.

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1 year ago