Answer:
This came to mind
Explanation:
when a cannon fires (in real life or in the movies) have noticed that the cannon recoils, sliding backwards after the explosion. Again, a non-zero net force on the cannon changes its momentum.
Answer:
![\lambda=6.83\times 10^{-5}\ m](https://tex.z-dn.net/?f=%5Clambda%3D6.83%5Ctimes%2010%5E%7B-5%7D%5C%20m)
Explanation:
Given that,
An infrared telescope is tuned to detect infrared radiation with a frequency of 4.39 THz.
We know that,
1 THz = 10¹² Hz
So,
f = 4.39 × 10¹² Hz
We need to find the wavelength of the infrared radiation.
We know that,
![\lambda=\dfrac{c}{f}\\\\\lambda=\dfrac{3\times 10^8}{4.39\times 10^{12}}\\\\=6.83\times 10^{-5}\ m](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bc%7D%7Bf%7D%5C%5C%5C%5C%5Clambda%3D%5Cdfrac%7B3%5Ctimes%2010%5E8%7D%7B4.39%5Ctimes%2010%5E%7B12%7D%7D%5C%5C%5C%5C%3D6.83%5Ctimes%2010%5E%7B-5%7D%5C%20m)
So, the wavelength of the infrared radiation is
.
Answer:
Orbital period, T = 1.00074 years
Explanation:
It is given that,
Orbital radius of a solar system planet, ![r=4\ AU=1.496\times 10^{11}\ m](https://tex.z-dn.net/?f=r%3D4%5C%20AU%3D1.496%5Ctimes%2010%5E%7B11%7D%5C%20m)
The orbital period of the planet can be calculated using third law of Kepler's. It is as follows :
![T^2=\dfrac{4\pi^2}{GM}r^3](https://tex.z-dn.net/?f=T%5E2%3D%5Cdfrac%7B4%5Cpi%5E2%7D%7BGM%7Dr%5E3)
M is the mass of the sun
![T^2=\sqrt{9.96\times 10^{14}}\ s](https://tex.z-dn.net/?f=T%5E2%3D%5Csqrt%7B9.96%5Ctimes%2010%5E%7B14%7D%7D%5C%20s)
T = 31559467.6761 s
T = 1.00074 years
So, a solar-system planet that has an orbital radius of 4 AU would have an orbital period of about 1.00074 years.
The emf is induced in the wire will be 1.56 ×10 ⁻³ V. The induced emf is the product of the magnetic field,velocity and length of the wire.
<h3>What is induced emf?</h3>
Emf is the production of a potential difference in a coil as a result of changes in the magnetic flux passing through it.
When the flux coupling with a conductor or coil changes, electromotive Force, or EMF, is said to be induced.
The given data in the problem is;
B is the magnitude of the magnetic field,= 5.0 ×10⁻⁵ T
V(velocity)=125 M/SEC
L(length)=25 cm=0.25 m
The maximum emf is found as;
E=VBLsin90°
E=125 × 5.0 × 10⁻⁵ ×0.25
E=1.56 ×10 ⁻³ V
Hence, the emf is induced in the wire will be 1.56 ×10 ⁻³ V
To learn more about the induced emf, refer to the link;
brainly.com/question/16764848
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