Answer: Just a few examples are the tension in the rope on a tether ball, the force of Earth's gravity on the Moon, friction between roller skates and a rink floor, a banked roadway's force on a car, and forces on the tube of a spinning centrifuge. Any net force causing uniform circular motion is called a centripetal force.

6 0

3 years ago

Read 2 more answers

A fisherman notices that his boat is moving up and down periodically without any horizontal motion, owing to waves on the surfac

leva [86]

Answer:

A. 5.600 m

B. 5.800 s

C. 0.966 m/s

D. 0.315 m

Explanation:

A. The wavelength is the distance between 2 crests, which is 5.600 m

B. Period of the wave is the time for the wave to complete 1 circle (highest point to next highest point). Since it takes 2.9s to travel from highest point to lowest point, it would take another 2.9 to travel from lowest point to the next highest point. So the total time is 2.9 + 2.9 = 5.8 s,

C. The wavespeed is wavelength over unit of time:

5.6 / 5.8 = 0.966 m/s

D. The amplitude would be half the length of highest point to lowest point, which is 0.63 / 2 = 0.315 m

5 0

3 years ago

The Voltmeter reads 16 V. The ammeter reads 0.25 A. The resistance must be

harina [27]

Answer:

64 ohms =R

Explanation:

R is directly proportional to V

R is inversely proportional to I(Current

4 0

3 years ago

Chapter 38, Problem 001

Norma-Jean [14]

Answer:

a) $\lambda=2.95x10^{-6}m$

b) infrared region

Explanation:

Photon energy is the "energy carried by a single photon. This amount of energy is directly proportional to the photon's electromagnetic frequency and is inversely proportional to the wavelength. If we have higher the photon's frequency then we have higher its energy. Equivalently, with longer the photon's wavelength, we have lower energy".

Part a

Is provide that the smallest amount of energy that is needed to dissociate a molecule of a material on this case 0.42eV. We know that the energy of the photon is equal to:

$E=hf$

Where h is the Planck's Constant. By the other hand the know that $c=f\lambda$ and if we solve for f we have:

$f=\frac{c}{\lambda}$

If we replace the last equation into the E formula we got:

$E=h\frac{c}{\lambda}$

And if we solve for $\lambda$ we got:

$\lambda =\frac{hc}{E}$

Using the value of the constant $h=4.136x10^{-15} eVs$ we have this:

$\lambda=\frac{4.136x10^{15}eVs (3x10^8 \frac{m}{s})}{0.42eV}=2.95x10^{-6}m$

$\lambda=2.95x10^{-6}m$

Part b

If we see the figure attached, with the red arrow, the value for the wavelenght obtained from part a) is on the infrared region, since is in the order of $10^{-6}m$

4 0

3 years ago