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Irina-Kira [14]
3 years ago
14

The energy of a photon was found to be

Physics
2 answers:
maria [59]3 years ago
6 0

Answer:

Green.....

Explanation:

.....

Vaselesa [24]3 years ago
4 0

now you can yourself know to which part of electromagnetic spectrum the photon belongs....

not fitting sharply in green

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The sidereal period of the moon around the Earth is 27.3 days. Suppose a satellite were placed in Earth orbit, halfway between E
icang [17]

Answer: 9.7 days

Explanation:

Applying Kepler's 3rd law, we can write the following proportion:

(Tm)² / (dem)³ = (Tsat)² / (dem/2)³

(As the satellite is placed in an orbit halfway between Erth's center and the moon's orbit).

Simplifyng common terms, and solving for Tsat, we have:

Tsat = √((27.3)²/8) = 9.7 days

7 0
3 years ago
Which instrument is used to measure depth of ocean ?
skelet666 [1.2K]
Echo sounding is a type of SONAR used to determine the depth of water by transmitting sound pulses into water. The time interval between emission and return of a pulse is recorded, which is used to determine the depth of water along with the speed of sound in water at the time.
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3 years ago
Change the following data into scientific notation.<br><br> 5 000 000 km
RoseWind [281]
To find out scientific notation, you want to make sure that number is less than 10. So do 5.000000, you don't rally need the zeros but I just want to make my point. So use 10^x meaning ten the whatever power adds zeros like 5.000000x10^6 meaning it is increasing it by six zeros moving it out of the decimals and letting become 5,000,000. 
8 0
3 years ago
An electron moves with a speed of 8.0×106m/s along the -z-axis. It enters a region where there is a uniform magnetic field B = (
Crazy boy [7]

Answer:

Acceleration, a=9.36\times 10^{18}\ m/s^2

Explanation:

It is given that,

Speed of electron, v=8\times 10^6\ m/s

Charge on an electron, q=1.6\times 10^{-19}\ C

Mass of electron, m=9.1\times 10^{-31}\ kg

Magnetic field, B=5.5i-3.7j

Magnitude, |B|=\sqrt{5.5^2+(-3.77)^2}=6.66\ T

Magnetic force is given by :

F=qvB

Also, F = ma

a=\dfrac{qvB}{m}

a=\dfrac{1.6\times 10^{-19}\times 8\times 10^6\times 6.66}{9.1\times 10^{-31}}

a=9.36\times 10^{18}\ m/s^2

So, the acceleration of the electron is 9.36\times 10^{18}\ m/s^2. Hence, this is the required solution.

5 0
3 years ago
a car moving with a speed of 10 metre per second is accelerator at the rate of 2 metre per second square find its velocity after
STatiana [176]

a = ( v(2) - v(1) ) ÷ ( t(2) - t(1) )

2 = ( v(2) - 10 ) ÷ ( 6 - 0 )

2 × 6 = v(2) - 10

v(2) = 12 + 10

v(2) = 22 m/s

7 0
3 years ago
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