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Kipish [7]
3 years ago
12

What are some indicators of energy transformations?

Physics
1 answer:
Annette [7]3 years ago
5 0

Explanation:

Contact, vision, sound, flavor, and smell are all markers of energy transformations. The most basic example would be when we notice something has begun to pass through vision. Whenever an entity accelerates or slows down, energy is constantly transformed.

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A 385-g tile hangs from one end of a string that goes over a pulley with a moment of inertia of and a radius of 15.0 cm. A mass
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If a motorcycle accelerates uniformly from rest at 5 m / s2, how long does it take for it to reach a speed of 20 m / s.
irina1246 [14]

Answer:

4 seconds

Explanation:

Given:

v₀ = 0 m/s

v = 20 m/s

a = 5 m/s²

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v = at + v₀

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t = 4 s

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Which best describes cosmic microwave background radiation? radiation thought to exist but not yet measured radiation predicted
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Answer:  leftover radiation from the big bang

Microwave background radiation is a form of electromagnetic radiation that is present in the whole universe and completely fills it.

Its frequency belongs to the microwaves range and is one of the main demonstrations of the of the Big Bang theory model.

It is important to note the Big Bang occurred 400,000 years before the events related to the microwave cosmic radiation, this means the Big Bang occurred first.

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Xplain the difference between aerobic and anaerobic exercise and give 1 example of each?
Lilit [14]

Answer:

The term "aerobic" refers to the body's ability to provide energy through the use of oxygen.  The term anaerobic refers to the body's ability to provide energy without the use of oxygen.

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May you help me answer this​
Firdavs [7]

1) See three Kepler laws below

2a) Acceleration is 2.2 m/s^2

2b) Tension in the string: 27.4 N

3a) Kinetic energy is the energy of motion, potential energy is the energy due to the position

3b) The kinetic energy of the object is 2.25 J

Explanation:

1)

There are three Kepler's law of planetary motion:

  1. 1st law: the planets orbit the sun in elliptical orbits, with the Sun located at one of the 2 focii
  2. 2nd law: a segment connecting the Sun with each planet sweeps out equal areas in equal time intervals. A direct consequence of this is that, when a planet is further from the sun, it travels slower, and when it is closer to the sun, it travels faster
  3. 3rd law: the square of the period of revolution of a planet around the sun is directly proportional to the cube of the semi-major axis of its orbit. Mathematically, T^2 \propto r^3, where T is the period of revolution and r is the semi-major axis of the orbit

2a)

To solve the problem, we have to write the equation of motions for each block along the direction parallel to the incline.

For the block on the right, we have:

M g sin \theta - T = Ma (1)

where

Mg sin \theta is the component of the weight of the block parallel to the incline, with

M = 8.0 kg (mass of the block)

g=9.8 m/s^2 (acceleration of gravity)

\theta=35^{\circ}

T = tension in the string

a = acceleration of the block

For the block on the left, we have similarly

T-mg sin \theta = ma (2)

where

m = 3.5 kg (mass of the block)

\theta=35^{\circ}

From (2) we get

T=mg sin \theta + ma

Substituting into (1),

M g sin \theta - mg sin \theta - ma = Ma

Solving for a,

a=\frac{M-m}{M+m}g sin \theta=\frac{8.0-3.5}{8.0+3.5}(9.8)(sin 35^{\circ})=2.2 m/s^2

2b)

The tension in the string can be calculated using the equation

T=mg sin \theta + ma

where

m = 3.5 kg (mass of lighter block)

g=9.8 m/s^2

\theta=35^{\circ}

a=2.2 m/s^2 (acceleration found in part 2)

Substituting,

T=(3.5)(9.8)(sin 35^{\circ}) +(3.5)(2.2)=27.4 N

3a)

The kinetic energy of an object is the energy due to its motion. It is calculated as

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

The potential energy is the energy possessed by an object due to its position in a gravitational field. For an object near the Earth's surface, it is given by

U=mgh

where

m is the mass of the object

g is the strength of the gravitational field

h is the heigth of the object relative to the ground

3b)

The kinetic energy of an object is given by

K=\frac{1}{2}mv^2

where

m is the mass of the object

v is its speed

For the object in this problem,

m = 500 g = 0.5 kg

v = 3 m/s

Substituting, we find its kinetic energy:

K=\frac{1}{2}(0.5)(3)^2=2.25 J

Learn more about acceleration and forces:

brainly.com/question/11411375

brainly.com/question/1971321

brainly.com/question/2286502

brainly.com/question/2562700

And about kinetic energy:

brainly.com/question/6536722

#LearnwithBrainly

7 0
3 years ago
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