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masha68 [24]
2 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]2 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]2 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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A kangaroo jumps to a vertical height of 2.8 m. How long was it in the air before returning to earth
BaLLatris [955]
The answer would be 2.8m height on earth takes 
2.8=1/2*9.8*t^2 => <span>s = ut +1/2at^2 </span>
8 0
2 years ago
The reaction energy of a reaction is the amount of energy released by the reaction. It is found by determining the difference in
solmaris [256]

Answer: the correct answer is 7.8026035971 x 10^(-13) joule

Explanation:

Use Energy Conservation. By ``alpha decay converts'', we mean that the parent particle turns into an alpha particle and daughter particles. Adding the mass of the alpha and daughter radon, we get

m = 4.00260 u + 222.01757 u = 226.02017 u .

The parent had a mass of 226.02540 u, so clearly some mass has gone somewhere. The amount of the missing mass is

Delta m = 226.02540 u - 226.02017 u = 0.00523 u ,

which is equivalent to an energy change of

Delta E = (0.00523 u)*(931.5MeV/1u)

Delta E = 4.87 MeV

Converting  4.87 MeV to Joules

1 joule [J] = 6241506363094 mega-electrón voltio [MeV]

4 mega-electrón voltio = 6.40870932 x 10^(-13) joule

4.87 mega-electrón voltio = 7.8026035971 x 10^(-13) joule

5 0
2 years ago
Kimonoski takes a 9-minute shower every day. The shower uses about 1.8 gal per minute of water. He also uses 23 gallons of hot w
ioda

Answer:

Q_{week} = 458884.6\, BTU

Explanation:

The weekly water consumption of Kimonoski is:

m_{bath,week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (1.8\,\frac{gal}{min} )\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (\frac{1\,min}{60\,s} )\cdot (9\,min)\cdot (\frac{60\,s}{1\,min} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{bath.week} = 948.205\,lbm

m_{others, week} = (62.4\,\frac{lbm}{ft^{3}})\cdot (23\,gal)\cdot (\frac{0.134\,ft^{3}}{1\,gal} )\cdot (7\,\frac{days}{week} )\cdot (1\,week)

m_{others, week} = 1346.218\,lbm

m_{week} = m_{bath,week} + m_{others, week}

m_{week} = 2294.423\,lbm

The total energy required per week for hot water is:

Q_{week} = m_{week}\cdot c_{p,water}\cdot \Delta T

Q_{week} =(2294.423\,lbm)\cdot (1\,\frac{BTU}{lbm\cdot ^{\textdegree}F} )\cdot (50^{\textdegree}F)

Q_{week} = 458884.6\, BTU

3 0
1 year ago
A shot putter releases the shot some distance above the level ground with a velocity of 12.0 m/s, 51.0 ∘above the horizontal. Th
alina1380 [7]

A) Zero

The motion of the shot is a projectile's motion: this means that there is only one force acting on the projectile, which is gravity. However, gravity only acts in the vertical direction: so, there are no forces acting in the horizontal direction. Therefore, the x-component of the acceleration is zero.

B) -9.8 m/s^2

The vertical acceleration is given by the only force acting in the vertical direction, which is gravity:

F=mg

where m is the projectile's mass and g is the gravitational acceleration. Therefore, the y-component of the shot's acceleration is equal to the acceleration due to gravity:

a_y = g = -9.8 m/s^2

where the negative sign means it points downward.

C) 7.6 m/s

The x-component of the shot's velocity is given by:

v_x = v_0 cos \theta

where

v_0 = 12.0 m/s is the initial velocity

\theta=51.0^{\circ} is the angle of the shot

Substituting into the equation, we find

v_x = (12.0 m/s)(cos 51^{\circ})=7.6 m/s

D) 9.3 m/s

The y-component of the shot's velocity is given by:

v_y = v_0 sin \theta

where

v_0 = 12.0 m/s is the initial velocity

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Substituting into the equation, we find

v_y = (12.0 m/s)(sin 51^{\circ})=9.3 m/s

E) 7.6 m/s

We said at point A) that the acceleration along the x-direction is zero: therefore, the velocity along the x-direction does not change, so the x-component of the velocity at the end of the trajectory is equal to the x-velocity at the beginning:

v_x = 7.6 m/s

F) -11.1 m/s

The y-component of the velocity at time t is given by:

v_y(t) = v_y + at

where

v_y = 9.3 m/s is the initial y-velocity

a = g = -9.8 m/s^2 is the vertical acceleration

t is the time

Since the total time of the motion is t=2.08 s, we can substitute this value into the equation, and we find:

v_y(2.08 s)=9.3 m/s + (-9.8 m/s^2)(2.08 s)=-11.1 m/s

where the negative sign means the vertical velocity is now downward.

3 0
2 years ago
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