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Dahasolnce [82]
3 years ago
12

An 8 kg cat is running 4 m/s. How much kinetic energy does it have?

Physics
1 answer:
Zepler [3.9K]3 years ago
7 0

Answer: Kinetic energy = (1/2) (mass) (speed)²

Kinetic energy = (1/2) (8 kg) (4 m/s)² =

 

                                 (4 kg) (16 m²/s²) =

                                  64  (kg-m/s²) (m) = 64 newton-m = 64 joules

  (about 17.6 pounds)

Explanation:

You might be interested in
Which is an example of transforming potential energy to kinetic energy? Select two options.
Rasek [7]

<u>Complete Question:</u>

Which is an example of transforming potential energy to kinetic energy? Select two options.

changing thermal energy to electrical energy

changing chemical energy to thermal energy

changing nuclear energy to radiant energy

changing radiant energy to electrical energy

changing mechanical energy to chemical energy

<u>Correct Answer:</u>

The examples of transforming potential energy to kinetic energy are changing chemical energy to thermal energy and changing nuclear energy to radiant energy.

<u>Explanation:</u>

As stated by the conservation of energy law, any form of energy is usually transferred to another form. The basic kinds of energy is potential and kinetic energy. Potential energy is the energy stored for the objects at rest and kinetic energy is the energy utilized by the objects for motion.

So in the given options, chemical energy is the energy stored by the chemical bonds to make a stable compound and that energy is converted to thermal energy when the bonds get broken. So the stored energy or the energy required to keep the bonds intact is chemical energy and it is thus a form of potential energy.

And when these bonds get broken, the electrons use the thermal energy released by this breakage as their kinetic energy. So one form of transforming potential energy to kinetic energy is by changing chemical energy to thermal energy.

Similarly, the nuclear energy is exhibited by the elementary particles in an atom. So it is similar to potential energy and the radiant energy is released whenever there is an excitation. So the radiant energy will be similar to kinetic energy.

Thus, the changing of chemical energy to thermal energy and the changing of nuclear energy to radiant energy are the examples of transforming potential energy to kinetic energy.

5 0
3 years ago
Two point charges of equal magnitude q are held a distance d apart. Consider only points on the line passing through both charge
NemiM [27]

let us consider that the two charges are of opposite nature .hence they will constitute a dipole .the separation distance is given as d and magnitude of each charges is q.

the mathematical formula for potential is V=\frac{1}{4\pi\epsilon} \frac{q}{d}

for positive charges the potential is positive and is negative for negative charges.

the formula for electric field is given as-E=\frac{1}{4\pi\epsilon} \frac{q}{r^2}

for positive charges,the line filed is away from it and for negative charges the filed is towards it.

we know that on equitorial line the potential is zero.hence all the points situated on the line passing through centre of the dipole and perpendicular to the dipole length is zero.

here the net electric field due to the dipole can not be zero  between the two charges,but we can find the points situated on the axial  line but  outside of charges where the electric field is zero.

now let the two charges of same nature.let these are positively charged.

here we can not find a point between two charges and on the line joining  two charges  where the potential is zero.

but at the mid point of the line joining two charges the filed is zero.

5 0
3 years ago
If two planets orbit a star, but planet B is twice as far from the star as planet A, planet A will receive ____ times the flux t
Tresset [83]

Answer:

The nearest plant (A) receives 4 times more radiation from the farthest plant

Explanation:

The energy emitted by the star is distributed on the surface of a sphere, whereby intensity received is the power emitted between the area of ​​the sphere

                I = P / A

               P = I A

The area of ​​the sphere is

               A = 4π r²

Since the amount of radiation emitted by the star is constant, we can write this expression for the position of the two planets

               P = I₁ A₁ = I₂ A₂

               I₁ / I₂ = A₂ / A₁

 Suppose index 1 corresponds to the nearest planet,

            r2 = 2 r₁

            I₁ / I₂ = r₁² / r₂²

            I₁ / I₂ = r₁² / (2r₁)²

            I₁ / I₂ = ¼

           4 I₁ = I₂

The nearest plant (A) receives 4 times more radiation from the farthest plant

7 0
3 years ago
A simple experiment to measure the speed of sound doesn't involve a stopwatch. You can fill up along tube with water and put a t
Serhud [2]

Answer:

Explanation:

In order to answer this problem you have to know the depth of the column, we say R, this information is important because allows you to compute some harmonic of the tube. With this information you can compute the depth of the colum of air, by taking tino account that the new depth is R-L.

To find the fundamental mode you use:

f_n=\frac{nv_s}{4L}

n: mode of the sound

vs: sound speed

L: length of the column of air in the tube.

A) The fundamental mode id obtained for n=1:

f_1=\frac{v_s}{4L}

B) For the 3rd harmonic you have:

f_3=\frac{3v_s}{4L}

C) For the 2nd harmonic:

f_2=\frac{2v_s}{4L}

7 0
3 years ago
What are the first three overtones of a bassoon that has a fundamental frequency of 90.0 Hz? It is open at both ends. (The overt
vekshin1

Answer:

f_{2}=180Hz,f_{3}=270Hz,f_{4}=360Hz\\

Explanation:

Given data

Frequency f=90 Hz

To find

First three overtones of bassoon

Solution

The fundamental frequency of bassoon is found by substituting n=1 in below equation

f=v/λ=nv/2L

f_{1}=v/2L

The first overtone of bassoon is found by substituting n=2

So

f_{2}=2v/2L\\f_{2}=2(v/2L)\\as \\f_{1}=v/2L\\So\\f_{2}=2f_{1}\\f_{2}=2(90Hz)\\f_{2}=180Hz

The second overtone of bassoon is found by substituting n=3

So

f_{3}=3v/2L\\f_{3}=3(v/2L)\\as \\f_{1}=v/2L\\So\\f_{3}=3f_{1}\\f_{3}=3(90Hz)\\f_{3}=270Hz

The third overtone of bassoon is found by substituting n=4

So

f_{4}=4v/2L\\f_{4}=4(v/2L)\\as \\f_{1}=v/2L\\So\\f_{4}=4f_{1}\\f_{4}=4(90Hz)\\f_{4}=360Hz

5 0
3 years ago
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