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3 years ago

What happens to the kinetic energy of an object, if its velocity is doubled?

Physics

2 answers:

denis-greek [22]3 years ago

8 0

As Kinetic Energy=half of Mass into Velocity square so as velocity is doubled the K.E of the object is doubled..

crimeas [40]3 years ago

4 0

If the velocity is doubled in a moving object that would cause that moving object to develop more kinetic energy.<span />

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How long have advances in astronomy been occurring? Only for the past few years

solniwko [45]

I'd have to say that the list of choices doesn't go far enough.

Advances in Astronomy have been occurring for at least the past two millennia (2000 years). Maybe longer.

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3 years ago

What causes the weight of an object to change depending on if it is on Earth or the Moon?

Gnoma [55]

The acceleration due to gravity

Brainliest please

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2 years ago

A 0.6 kg block attached to a spring of force constant 13.6 N/m oscillates with an amplitude of 9 cm. Find the maximum speed of t

mash [69]

Answer:

1) 0.43 meters per second

2) 0.21 meters per second

3) 1.02 $\frac{m}{s^{2}}$

4) 0.66 seconds

Explanation:

part 1

By conservation of energy, the maximum kinetic energy (K) of the block is at equilibrium point where the potential energy is zero. So, at the equilibrium kinetic energy is equal to maximum potential energy (U):

$K=U$

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}$

With m the mass, v the speed, k the spring constant and xmax the maximum position respect equilibrium position. Solving for v

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.09m)^2}{0.6}}=0.43\frac{m}{s}$

part 2

Again by conservation of energy we have kinetic energy equal potential energy:

$\frac{mv^2}{2}=\frac{kx_{max}^2}{2}=$

$v=\sqrt{\frac{kx_{max}^2}{m}}=\sqrt{\frac{(13.6)(0.045m)^2}{0.6}}=0.21\frac{m}{s}$

part 3

Acceleration can be find using Newton's second law:

$F=ma$

with F the force, m the mass and a the acceleration, but elastic force is -kx, so:

$-kx=ma$

$a= -\frac{kx}{m}=-\frac{(13.6)(0.045)}{0.6}=-1.02\frac{m}{s^{2}}$

part 4

The period of an oscillator is the time it takes going from one extreme to the other one, that is going form 4.5 cm to -4.5 cm respect the equilibrium position. That period is:

$T=2\pi\sqrt{\frac{m}{k}}=T=2\pi\sqrt{\frac{0.6}{13.6}}=1.32s$

So between 0 and 4.5 cm we have half a period:

$t=\frac{T}{2}=0.66s$

7 0

3 years ago

Place the following types of electromagnetic radiation in order of increasing wavelength. ultraviolet light gamma rays radio wav

kvv77 [185]

Answer: gamma rays < ultraviolet light < radio waves

Explanation: We have to explain the ordering we take into account a relationship for electromagnetic wave, which is given by:

c=λ*ν c is the speed of light. λ and ν are the wavelegth and the frequency of the waves.

Then we also know that the high frequency of these radiation correspond to the gamma ray (10^18 1/s) while the lowest correspond to radio wave (10^8 1/s) so:

The ultraviolet ligth is ranged at frequencies of 10^16 1/s located bewteen gamma and radio radiation.

as large frequency lower wavelength we can order these radiations from short to long wavelengths as follows:

gamma rays < ultraviolet light < radio waves

3 0

3 years ago

. A new programming language requires variable names to start with a letter (upper or lowercase), which can be followed by lette

cupoosta [38]

Answer:

Explanation:

Total letters available on the keyboard = 26

Doubling this amount to account for both upper and lower case letters.

Therefore, Total characters (Upper + Lower Case) = 52 characters

Total numbers on keyboard (0, 1, 2, 3, 4, 5, 6, 7, 8, 9) = 10 characters

Underscore = 1 character

Thus, Total allowed characters = 52 + 10 + 1 = 63

This is a problem of the permutation as order of the characters results in a different name.

Our file name should be of atleast 3 and smaller than 7 (i.e. 6) characters

Therefore, total possible permutations can be obtained by subtraction permutation of 2 characters from the permutations of 6 characters.

$Total\; Permutations\; =\; (^{63}P_{6}) \;-\;(^{63}P_{2}) \\\\Total\; Permutations\; =\;48920775120\;-\;3906\\\\Total\; Permutations\; =\;48920771214$

7 0

3 years ago

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