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mestny [16]
2 years ago
10

Which part of the electromagnetic spectrum has a lower frequency than visible light?

Physics
2 answers:
Mademuasel [1]2 years ago
6 0

The Correct Answer is <u>D.Infrared/</u> <em>INFARED has a lower frequency than visible light/</em>

KonstantinChe [14]2 years ago
4 0

Ultraviolet part of the electromagnetic spectrum has higher frequency than the visible light.

X-ray part of the electromagnetic spectrum has higher frequency than the visible light.

gamma rays part of the electromagnetic spectrum has higher frequency than the visible light.

infrared red part of the electromagnetic spectrum has lower frequency than the visible light.

hence the correct choice is infrared

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A boat is traveling at 80 km/hr. How many hours will it take for the boat to cover a distance of 115 km?
miskamm [114]

Answer:

Explanation:

Givens

d = 115 km

r = 80 km/hr

t = ?

Equation

d = r*T

Solution

115 = 80 * t    Divide by 80

115/80 = t

t = 1.4375 hours.

3 0
3 years ago
PLEASE TRY TO ANSWER AS MANY QUESTIONS AS YOU CAN !
suter [353]
Good luck with solving this
3 0
3 years ago
Read 2 more answers
Which term, taken from the celestial sphere, gives its name to a.m. and p.m.?
DaniilM [7]

Options are. Zenith, Great circle, Equinox, or Meridan
3 0
2 years ago
9. A 227 kg object is moved a distance of 2.4 m forward by a force. If 686 J of work is done on the object, what is the object’s
Korolek [52]

<em>1</em><em>.</em><em>259ms^2</em>

Explanation:

since, WORK DONE = FORCE*DISTANCE

AND, FORCE=MASS*ACCELERATION

SO, THE WORK DONE BECOMES=MASS*ACCELERATION*DISTANCE

ACCELERATION=WORK/(MASS*DISTANCE)

AND, WORK=686J

MASS=227kg

DISTANCE=2.4m

THEREFORE, ACCELERATION=686/(227*2.4)

=686/544.8

=1.259ms^2

4 0
3 years ago
In 1976, the SR-71A, flying at 20 km altitude (T = –56 0C), set the official jet-powered aircraft speed record of 3530 km/hr (21
Lapatulllka [165]

To solve this problem we will apply the concepts related to the calculation of the speed of sound, the calculation of the Mach number and finally the calculation of the temperature at the front stagnation point. We will calculate the speed in international units as well as the temperature. With these values we will calculate the speed of the sound and the number of Mach. Finally we will calculate the temperature at the front stagnation point.

The altitude is,

z = 20km

And the velocity can be written as,

V = 3530km/h (\frac{1000m}{1km})(\frac{1h}{3600s})

V = 980.55m/s

From the properties of standard atmosphere at altitude z = 20km temperature is

T = 216.66K

k = 1.4

R = 287 J/kg

Velocity of sound at this altitude is

a = \sqrt{kRT}

a = \sqrt{(1.4)(287)(216.66)}

a = 295.049m/s

Then the Mach number

Ma = \frac{V}{a}

Ma = \frac{980.55}{296.049}

Ma = 3.312

So front stagnation temperature

T_0 = T(1+\frac{k-1}{2}Ma^2)

T_0 = (216.66)(1+\frac{1.4-1}{2}*3.312^2)

T_0 = 689.87K

Therefore the temperature at its front stagnation point is 689.87K

6 0
3 years ago
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