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Novosadov [1.4K]
3 years ago
15

Which constellation is marked by a W or M? a. Cassiopeia b. Draco c. Ursa Minor d. Ursa Major

Physics
2 answers:
yuradex [85]3 years ago
8 0

Answer:

Ursa Minor Constellation

Explanation:

Ursa Minor constellation lies in the northern sky. The constellation's name means “the smaller bear,” or “the lesser bear,” in Latin. The Great Bear constellation is represented by its larger neighbor Ursa Major.

Anestetic [448]3 years ago
6 0

I believe it is the Cassiopeia. It is visible during winter in the early part of the night. It look like a distorted letter W or M.

I am not very sure of it though, I apologize if this is not what you meant.

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Answer:

a=(v-u)÷t

a= (2-10)÷2

a= -8÷2

a= -4 m/s²

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On a velocity vs. time graph, what does it mean when the line crosses over the x-axis?
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On a velocity - time graph, if the line crosses the x - axis it depicts that the object has started moving in the opposite direction.

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T or F-Climate is directly related to the amount of energy from the sun or solar energy that an area receives
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Answer:

true

Explanation:

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8 0
3 years ago
BRAINLIEST AND MAY DOUBLE POINTS!!!
raketka [301]

Given that:

 Energy of bulb  (Work ) = 30 J,

 Time (t) = 3 sec

The power consumption =  ?

 We know that, Power can be defined as rate of doing work

              Power (P) = Work(Energy supplied) ÷ time

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<em> The power consumption is 10 W.</em>


3 0
3 years ago
In a nuclear physics experiment, a proton (mass 1.67×10^(−27)kg, charge +e=+1.60×10^(−19)C) is fired directly at a target nucleu
Arte-miy333 [17]

The given question is incomplete. The complete question is as follows.

In a nuclear physics experiment, a proton (mass 1.67 \times 10^(-27)kg, charge +e = +1.60 \times 10^(-19) C) is fired directly at a target nucleus of unknown charge. (You can treat both objects as point charges, and assume that the nucleus remains at rest.) When it is far from its target, the proton has speed 2.50 \times 10^6 m/s. The proton comes momentarily to rest at a distance 5.31 \times 10^(-13) m from the center of the target nucleus, then flies back in the direction from which it came. What is the electric potential energy of the proton and nucleus when they are 5.31 \times 10^{-13} m apart?

Explanation:

The given data is as follows.

Mass of proton = 1.67 \times 10^{-27} kg

Charge of proton = 1.6 \times 10^{-19} C

Speed of proton = 2.50 \times 10^{6} m/s

Distance traveled = 5.31 \times 10^{-13} m

We will calculate the electric potential energy of the proton and the nucleus by conservation of energy as follows.

  (K.E + P.E)_{initial} = (K.E + P.E)_{final}

 (\frac{1}{2} m_{p}v^{2}_{p}) = (\frac{kq_{p}q_{t}}{r} + 0)

where,    \frac{kq_{p}q_{t}}{r} = U = Electric potential energy

     U = (\frac{1}{2}m_{p}v^{2}_{p})

Putting the given values into the above formula as follows.

        U = (\frac{1}{2}m_{p}v^{2}_{p})

            = (\frac{1}{2} \times 1.67 \times 10^{-27} \times (2.5 \times 10^{6})^{2})

            = 5.218 \times 10^{-15} J

Therefore, we can conclude that the electric potential energy of the proton and nucleus is 5.218 \times 10^{-15} J.

4 0
3 years ago
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