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Arisa [49]
3 years ago
15

An object that has kinetic energy must what ?

Physics
1 answer:
11Alexandr11 [23.1K]3 years ago
4 0

An object that has kinetic energy must be <em>moving</em>.

The formula for an object's kinetic energy is

KE = (1/2) · (the object's mass) · <u><em>(the object's speed)²</em></u>

As you can see from the formula, if the object has no speed, then its kinetic energy is zero.  That's why kinetic energy is usually called the "energy of motion", and if an object HAS kinetic energy, then that tells you right away that it must be moving.

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In which condition is relative velocity known by adding the velocity od the first body and that of the second body ?​
Softa [21]

Answer:

when a two bodies A and B are moving at an angle 180° with each other then the relative velocity is the sum of bodies the velocity .i.e,

when the bodies move the opposite direction

then their relative velocity is the sum of individual velocities.

3 0
2 years ago
make a list of different types of energy sources found in our surroundings and categorize them into different form of energy.​
MrRa [10]

Answer:

hdcygcyh

Explanation:

8 0
3 years ago
Read 2 more answers
an object is moving with a speed of 35 m/s and has a kinetic energy of 1500 J, what is the mass of the object?
Elena-2011 [213]
You'd use the equation kinetic energy=mass*0.5*speed^2
So you'd rearrange this to get mass =kinetic energy /0.5 *speed^2
Which is mass= 1500J/0.5*35^2
=2.44897959183673469........kg
5 0
3 years ago
A ball is thrown straight up with an initial speed<br> of 30 m/s. What is its speed after 4.2 s?
olasank [31]

Answer:

vf = -11.16 m/s

Explanation:

vf = vi + gt

vf = 30 + (-9.8)(4.2)

vf = -11.16 m/s

3 0
2 years ago
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An electron, traveling at a speed of 5.90 × 10 6 5.90×106 m/s, strikes the target of an X-ray tube. Upon impact, the electron de
Trava [24]

Answer:

2.84\cdot 10^{-8} m

Explanation:

Due to the law of conservation of energy, the energy of the emitted X-ray photon is equal to the energy lost by the electron.

The initial kinetic energy of the electron is:

K_i = \frac{1}{2}mv_i^2 = \frac{1}{2}(9.11\cdot 10^{-31}kg)(5.90\cdot 10^6 m/s)^2=1.59\cdot 10^{-17}J

The electrons decelerates to 3/4 of its speed, so the new speed is

v_f = \frac{3}{4}v_i = \frac{3}{4}(5.90\cdot 10^6 m/s)=4.425\cdot 10^6 m/s

So the final kinetic energy is

K_f = \frac{1}{2}mv_f^2=\frac{1}{2}(9.11 \cdot 10^{-31} kg)(4.425\cdot 10^6 m/s)^2=8.9\cdot 10^{-18} J

So, the energy lost by the electron, which is equal to the energy of the emitted photon, is

E=K_i - K_f =1.59\cdot 10^{-17} J-8.9\cdot 10^{-18} J=7\cdot 10^{-18} J

The wavelength of the photon is related to its energy by

\lambda=\frac{hc}{E}

where h is the Planck constant and c the speed of light. Substituting E, we find

\lambda=\frac{(6.63\cdot 10^{-34}Js)(3\cdot 10^8 m/s)}{7\cdot 10^{-18} J}=2.84\cdot 10^{-8} m

4 0
3 years ago
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