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masha68 [24]
3 years ago
7

Mike recently purchased an optical telescope. Identify the part of the electromagnetic spectrum that is closest to the frequency

that Mike can observe with the help of his new tool.

Physics
2 answers:
Zepler [3.9K]3 years ago
8 0
The elctromagnetic spectrum ranges from the radiowaves to the gamma rays. The whole spectrum is shown in the attached picture. But the optical telescope can only see the visible region. So, it only covers from the 400 nm to 700 nm frequency. It follows the ROYGBIV colors, where red has the highest frequency and violet has the lowest frequency.

zalisa [80]3 years ago
7 0

Answer:

Ultraviolet and Infrared

Explanation:

Electromagnetic spectrum shows all type of EM waves basis their frequency and wavelength. It contains EM waves from radio waves having the smallest frequency to gamma rays having maximum frequency. Through an optical telescope Mike can observe only the visible part of the spectrum which is just 7% of the complete spectrum.

The closest to the frequency range of visible light lies Ultraviolet (UV) and Infrared rays. UV rays have higher frequency than visible light and IR rays have lesser frequency.

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Solution :

When the spacecraft is at halfway point, the distance from the Earth as well as Mars are same. We have to account the masses of the planets. The gravitational force that is exerted by the Earth is greater because of its combined mass with the space probe.

The mass of Earth is greater than the mass of Mars. Therefore, the force of Earth is more than Mars.

5 0
3 years ago
A force in the +x-direction with magnitude ????(x) = 18.0 N − (0.530 N/m)x is applied to a 6.00 kg box that is sitting on the ho
fiasKO [112]

Answer:

v_f=8.17\frac{m}{s}

Explanation:

First, we calculate the work done by this force after the box traveled 14 m, which is given by:

W=\int\limits^{x_f}_{x_0} {F(x)} \, dx \\W=\int\limits^{14}_{0} ({18N-0.530\frac{N}{m}x}) \, dx\\W=[(18N)x-(0.530\frac{N}{m})\frac{x^2}{2}]^{14}_{0}\\W=(18N)14m-(0.530\frac{N}{m})\frac{(14m)^2}{2}-(18N)0+(0.530\frac{N}{m})\frac{0^2}{2}\\W=252N\cdot m-52N\cdot m\\W=200N\cdot m

Since we have a frictionless surface, according to the the work–energy principle, the work done by all forces acting on a particle equals the change in the kinetic energy of the particle, that is:

W=\Delta K\\W=K_f-K_i\\W=\frac{mv_f^2}{2}-\frac{mv_i^2}{2}

The box is initially at rest, so v_i=0. Solving for v_f:

v_f=\sqrt{\frac{2W}{m}}\\v_f=\sqrt{\frac{2(200N\cdot m)}{6kg}}\\v_f=\sqrt{66.67\frac{m^2}{s^2}}\\v_f=8.17\frac{m}{s}

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3 years ago
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Answer:

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Put all the values,

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Hence, the potential difference is equal to 50 volts.

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Answer:what is the question exactly

Explanation:

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Hello Paige,

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