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3 years ago

9

If the swing rotates clockwise (looking from above the table), which direction is the angular velocity of the hanging mass point

ing in?
Select one:
a. clockwise
b. up
c. down
d. counterclockwise

1 answer:

AnnZ [28]3 years ago

3 0

Answer:

down

Explanation:

if you move to the right there is mass pushing there is a weight pushing down the force in which it makes it larger

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Name two objects that have a high density.

Sonbull [250]

Answer:

Iron and stone

Explanation:

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3 years ago

A student has a rectangular block. It is 2 cm wide, 2 cm tall, and 25 cm long. It has a mass of 600 g. First, calculate the volu

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Answer:

6g/cm³

Explanation:

Density is the mass per unit volume of any substance. To solve this problem:

Density = $\frac{mass}{volume}$

Since mass = 600g

Let us find the volume;

Volume = length x width x height

Volume = 25cm x 2cm x 2cm = 100cm³

Therefore;

Density = $\frac{600}{100}$ = 6g/cm³

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2 years ago

5. What is the amount of force required to accelerate a 20 kg object at a rate of 5 m/sz?

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Force required is 100 N

<u>Given that;</u>

Rate of acceleration = 5 m/s²

Mass of object = 20kg

<u>Find:</u>

Force required

<u>Computation:</u>

Force = Mass × Acceleration

Force required = Rate of acceleration × Mass of object

Force required = 20 × 5

Force required = 100 N

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2 years ago

What is dark energy?

olga2289 [7]

Explanation:

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The maximum wavelength an electromagnetic wave can have and still eject an electron from a copper surface is 264 nm .What is the

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Answer:

4.71 eV

Explanation:

For an electromagnetic wave with wavelength

$\lambda=264 nm = 2.64\cdot 10^{-7} m$

the energy of the photons in the wave is given by

$E=\frac{hc}{\lambda}=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{2.64\cdot 10^{-7}m}=7.53\cdot 10^{-19} J$

where h is the Planck constant and c the speed of light. Therefore, this is the minimum energy that a photon should have in order to extract a photoelectron from the copper surface.

The work function of a metal is the minimum energy required by the incident light in order to extract photoelectrons from the metal's surface. Therefore, the work function corresponds to the energy we found previously. By converting it into electronvolts, we find:

$E=\frac{7.53\cdot 10^{-19} J}{1.6\cdot 10^{-19} J/eV}=4.71 eV$

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2 years ago