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finlep [7]
3 years ago
13

Define are sources of energy​

Physics
1 answer:
OleMash [197]3 years ago
6 0

\boxed{\sf{QUESTION:}}

Define sources of energy

\boxed{\sf{ANSWER:}}

Energy is the capacity of a body to do work.

We Classify the sources of energy on the basis of:

<u>Occurence:</u>

  1. Natural Sources: Natural energy sources are those which are made available to us by nature. Solar energy, wind energy, energy from water (hydro energy) are some of such natural sources of Energy.
  2. Synthetic Sources: Synthetic energy are those that use man-made materials as sources of energy. Fir example, chemical energy, stored in the batteries, (used in calculators, watches, etc.) is a synthetic source of energy.

<u>Physical</u><u> </u><u>State:</u>

  1. Solid: Firewood, Charcoal, coal are examples of solid fuels.
  2. Liquid: Kerosene, Petrol & diesel are all liquid fuels.
  3. Gas: Petroleum gas, commonly used as LPG (Liquified Petroleum Gas), & natural gas, also used CNG (Compressed Natural Gas), are examples of gaseous fuels

<u>Availability</u><u>:</u>

  1. Renewable: A renewable source of energy is a natural resource that can replenish itself naturally over a short period of time. Wind, sun, biomass (from plants) & hydropower (from water) are all renewable sources of energy. These arr inexhaustible natural resources.
  2. Non-renewable: Energy sources which get us & cannot be replaced or replenished in a short period of time are called non-renewable sources of energy. These are also called exhaustible natural resources. Fossil fuels, (like petroleum, natural gas & coal), are non-renewable sources of energy.

<h2>Hope It's Helped! :D</h2>
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You place a 55.0 kg box on a track that makes an angle of 28.0 degrees with the horizontal. The coefficient of static friction b
Radda [10]

Answer:

\theta=34 \textdegree

Explanation:

From the question we are told that:

Mass m=55kg

Angle \theta =28.0

Coefficient of static friction \alpha =0.680

Generally, the equation for Newtons second Law is mathematically given by

For

\sum_y=0

N=mgcos \theta

for

\sum_x=0

F_{s}=mgsin\theta

Where

F_{s}=\alpha*N\\\\F_{s}=\alpha*m*gcos \theta

F_{s}=0.68*55*9.8*cos 28

F_{s}=323.62N

Therefore

\alpha mgcos \theta=mg sin \theta

\theta=tan^{-1}(0.68)

\theta=34 \textdegree

6 0
3 years ago
the proper execution of suggested physical fitness tests is important inachieving the accuracyof the assessment results of one's
ankoles [38]

Answer:

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5 0
3 years ago
I got part c right but idk why the other parts are wrong HELP!
dedylja [7]

a) The impulse is 76.5 Ns

b) The average force is 546.4 N

c) The final speed is 31.5 m/s

Explanation:

a)

The impulse exerted on an object is defined as

J=\int F\Delta t

where

F is the magnitude of the force exerted on the object

\Delta t is the time interval during which the force is applied

If we consider a graph of the force applied vs time, it follows that the impulse exerted is equal to the area under the graph.

Therefore, in this problem, we can calculate the impulse by computing the area under the graph. We have a trapezium, whose bases are

B=0.14-0 = 0.14s\\b=8-5=3s

and whose height is

h=900 N

Therefore, the area (and the impulse) is

J=\frac{(B+b)h}{2}=\frac{(0.14+0.03)(900)}{2}=76.5 Ns

b)

In this problem, the force applied is not constant. However, we can rewrite the impulse also as

J=F_{avg} \Delta t

where

F_{avg} is the average force exerted during the whole time \Delta t

In this problem we have

J = 76.5 Ns is the impulse (calculated in part a)

\Delta t = 0.14 s is the time interval

Solving for the average force, we find

\Delta t = \frac{J}{F_{avg}}=\frac{76.5}{0.14}=546.4 N

c)

According to the impulse theorem, the impulse exerted on an object is equal to the change in momentum of the object:

J=\Delta p = m(v-u)

where

m is the mass of the object

v is the final velocity

u is the initial velocity

In this problem, we have

J = 76.5 Ns

m = 3.0 kg is the mass

u = 6.0 m/s is the initial velocity

Solving for v, we find the final velocity (and speed):

v=u+\frac{J}{m}=6.0+\frac{76.5}{3}=31.5 m/s

Learn more about impulse and momentum:

brainly.com/question/9484203

#LearnwithBrainly

6 0
3 years ago
What does activation energy has to do with a chemical reaction.
asambeis [7]

Explanation:

Activation energy and reaction rate

The activation energy of a chemical reaction is closely related to its rate. Specifically, the higher the activation energy, the slower the chemical reaction will be. ... The released energy helps other fuel molecules get over the energy barrier as well, leading to a chain reaction.

6 0
3 years ago
Why are electromagnets used in metal scrap yards?
Andrej [43]
Because the electromagnets can pick up magnetic material and move it around, hope this helps
8 0
3 years ago
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