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cricket20 [7]
3 years ago
12

I have two questions

Physics
2 answers:
Anettt [7]3 years ago
8 0

Answer:

1)The energy transported by a wave is directly proportional to the square of the amplitude. So whatever change occurs in the amplitude, the square of that effect impacts the energy. This means that a doubling of the amplitude results in a quadrupling of the energy

2)Just as wavelength and frequency are related to light, they are also related to energy. The shorter the wavelengths and higher the frequency corresponds with greater energy. So the longer the wavelengths and lower the frequency results in lower energy.

Explanation:

Vesna [10]3 years ago
4 0
The correct answer to the question is : C) A pendulum swings back and forth.
EXPLANATION:
Before answering this question, first we have to understand the law of conservation of energy.
As per law of conservation of energy, energy can neither be created not be destroyed. It can only change from one form to another form, and the total energy of the universe is always constant.
As per the question, we have a pendulum which is moving back and forth.
Let us consider a pendulum which is taken to some height from its equilibrium or mean point. The energy possessed by the pendulum at this height is gravitational potential energy. When the pendulum is released, the potential energy is converted into kinetic energy. At mean point, whole of its potential energy is converted into kinetic energy.
Due to inertia, the pendulum reaches at the other extreme point. During its movement from mean point to extreme point, the kinetic energy is converted into potential energy. At extreme point, whole of its kinetic energy must have converted into potential energy. The same process will be repeated.
Hence, it obeys law of conservation of energy.
Hence, swinging of pendulum back and forth is the best example to demonstrate the law of conservation of energy.
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The charges that are free to move in a metallic conducting wire and that are responsible for the flow of electric current are
Anna007 [38]

Answer:Negatively charged particle called Free Electrons

Explanation:

Current is the flow of charged particles called Free electrons. Electrons are free to move from one atom to another and we call them a sea of de-localized electrons. In absence of any externally applied emf, these electrons are randomly moving but with the onset of emf, these electrons flow in a particular direction.    

6 0
3 years ago
Michael's house is 5.0 km away from his school. How long would it take him to go ti school, riding a bus, if its velocity is 25
Anvisha [2.4K]

Answer:

12 mins

Explanation:

The distance covered is 5 km, divide this by 25 to get the fraction of an hour it takes. Doing this you get .2, times this by 60 min (1 hour) to get how many mins it takes

8 0
3 years ago
Difference between kilogram and kilometre in points​
skelet666 [1.2K]

Answer:

Kilogram(kg) is the SI unit for mass while kilometre(km) is a unit for length. They are both similar in that they are 10^3 of a unit, thus kilo. As kilogram represents mass, it is a measure of how much matter is present in an object. While kilometre is a measure of distance/how long or short an object is.

7 0
3 years ago
Candice is examining a cell under a microscope. She has identified a cell wall, a nucleus, and a chloroplast. What type of organ
RoseWind [281]

Answer:

A plant

Explanation:

because animals don't have cell walls, and fungus and bacteria dont have chloroplasts

7 0
3 years ago
The left end of a long glass rod 8.00 cm in diameter, with an index of refraction 1.60, is ground to a concave hemispherical sur
sp2606 [1]

Answer:

a) q = -9.23 cm, b)  h’= 0.577 mm , c) image is right and virtual

Explanation:

This is an optical exercise, where the constructor equation should be used

        1 / f = 1 / p + 1 / q

Where f is the focal length, p the distance to the object and q the distance to the image

A) The cocal distance is framed with the relationship

       1 / f = (n₂-1) (1 /R₁ -1 /R₂)

In this case we have a rod whereby the first surface is flat R1 =∞ and the second surface R2 = -4 cm, the sign is for being concave

       1 / f = (1.60 -1) (1 /∞ - 1 / (-4))

       1 / f = 0.6 / 4 = 0.15

        f = 6.67 cm

We have the distance to the object p = 24.0 cm, let's calculate

       1 / q = 1 / f - 1 / p

       1 / q = 1 / 6.67 - 1/24

       1 / q = 0.15 - 0.04167 = 0.10833

       q = -9.23 cm

distance to the negative image is before the lens

B) the magnification of the lenses is given by

       M = h ’/ h = - q / p

        h’= - q / p h

        h’= - (-9.23) / 24.0 0.150

        h’= 0.05759 cm

        h’= 0.577 mm

C) the object is after the focal length, therefore, the image is right and virtual

6 0
3 years ago
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