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3 years ago

What happens when light bounces off an object

2 answers:

7 0

When light bounces off an object, it either gets absorbed, or reflected. For example, light can get absorbed on surfaces which are black in color and they can reflect from flat surfaces such as a mirror. It all matters on the angle of incidence. The angle of incidence is also known as the angle of reflection.

Angle of reflection = Angle of incidence

The ray of light approaching the mirror is called the incident ray. The ray of light that leaves the mirror is called the reflected ray.

Hope it helps

Sever21 [200]3 years ago

3 0

Absorbtion consists of when light strikes on an object and bounces off.

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Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in

postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is $\rho = \frac{m}{V}$ , and the volume of a pyramid is (confusingly written on the proposal) $V=\frac{1}{3} Ah$ , so we can write:

$m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h$

Where s is the side of the base, being $s^2$ the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that $\frac{1m}{100cm}=1$ , and we can use this factor to convert anything written in cm to anything written in m. Example:

$500cm=500cm\frac{1m}{100cm}=5m$

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so it's not doing anything but changing the units.

We can use this tool like this:

$2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3$

Where we have used the fact that $1^3=1$ (we can elevate any conversion factor to any number and they still will be 1) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

$m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg$

And now we will convert to short tons using two conversion factors at the same time:

$m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons$

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0

3 years ago

find the gravitational force this shell exerts on a 1.60 kg point mass placed at the distance 5.01 m from the center of the shel

RideAnS [48]

The gravitational force of the shell exerts is 4.25m x 10¯¹² N.

We need to know about gravitational force to solve this problem. The gravitational force is the force caused by two masses of objects. The magnitude of gravitational force can be determined as

F = G.m1.m2 / R²

where F is the gravitational force, G is the gravitational constant (6.674 × 10¯¹¹ Nm²/kg²), m1 and m2 are the mass of the object and R is the radius.

From the question above, we know that

m1 = 1.6 kg

m2 = m

R = 5.01 m

By substituting the following parameters, we get

F = G.m1.m2 / R²

F = 6.674 × 10¯¹¹ . 1.6 . m / 5.01²

F = 4.25m x 10¯¹² N

where m is the mass of the shell

For more on gravitational force at: brainly.com/question/19050897

#SPJ4

7 0

1 year ago

In a simple electric generator, a conducting loop of wire is placed in a magnetic field. The loop of wire is then rotated. Why i

kaheart [24]

Explanation:

If a coil of wire is placed in a magnetic field and rotated, an alternating (sinusoidal) current is induced. As it rotates, sometimes it is cutting through lots of magnetic flux, and so lots of current is

induced.

At other times, it is moving parallel to

the flux, and so no flux is cut, and no current is induced. In between, some current is induced. This creates an alternating current. Either end of the coil can be connected to wires outside of the generator in order to use the current elsewhere. This would be fine for the

first few rotations, but after this, the wires would get tangled up and the generator would be useless. To avoid this, we use a commutator. In an AC generator, this is a pair of rotating conducting 'slip rings' attached to either end of the coil. Carbon brushes bring these into contact with the outside world.

5 0

3 years ago

based on how the book describes a heat pump in sec 16.3 how does a heat pump prove law of conservation of energy

sattari [20]

The law of conservation of energy states that: "energy cannot be created nor destroyed. It can only be transformed from one form to the other". A heat pump transfers heat from a colder area to a warmer area. As the conservation of energy goes, the energy that comes in, must equal the energy that comes out. Although no machine is 100% efficient, the missing energy is supplied by an external energy source, such as electricity.

4 0

4 years ago

You tie a string to the ceiling and attach a weight to the end. You hold the weight next to your face but not touching it and th

Orlov [11]

Answer:

Depends on how long the string is, how heavy the weight, and how high you let go of it.

But it will most likely hit you :)

3 0

3 years ago