Answer:
The skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Explanation:
To solve the problem it is necessary to go back to the theory of conservation of momentum, specifically in relation to the collision of bodies. In this case both have different addresses, consideration that will be understood later.
By definition it is known that the conservation of the moment is given by:
Our values are given by,
As the skater 1 run in x direction, there is not component in Y direction. Then,
Skate 1:
Skate 2:
Then, if we applying the formula in X direction:
m_1v_{x1}+m_2v_{x2}=(m_1+m_2)v_{fx}
75*5.45-75*1.41=(75+75)v_{fx}
Re-arrange and solving for v_{fx}
v_{fx}=\frac{4.04}{2}
v_{fx}=2.02m/s
Now applying the formula in Y direction:
Therefore the skater 1 and skater 2 have a final speed of 2.02m/s and 2.63m/s respectively.
Answer:
Yes.
Explanation:
That is also a law violation.
Answer:
kwkwskdkmfmfmdkakksjddndkkakamsjjfjdmakisnddnianbfdnnskake
now
Answer:
Statement 2 is wrong
Explanation:
To check the statements in this exercise, let's describe the main properties of electromagnetic waves. Let's describe the characteristics
* they are transverse waves
* formed by the oscillations of the electric and magnetic fields
* the speed of the wave is the speed of light
with these concepts let's review the final statements
1) True. Formed by the oscillation of the two fields
2) False. They are transverse waves
3) True. Can travel by vacuum as they are supported by oscillations of the electric and magnetic fields
4) True. They all have the same speed of light
Statement 2 is wrong
Answer : Relatively hot objects
Explanation : We know that, the temperature of the objects is inversely proportional to their wavelengths. The objects emitting radiation in the visible region have short wavelength and hence are relatively hotter.
We know the range of wavelength of the visible spectrum is from 400 nm to 780 nm.
