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Usimov [2.4K]
3 years ago
10

The star nearest to our sun is Proxima Centauri, at a distance of 4.3 light-years from the sun. How far away, in km, is Proxima

Centauri from the sun?
Physics
2 answers:
Cloud [144]3 years ago
5 0

Answer: 4.068(10)^{13} km

Explanation:

A light year is a unit of length and is defined as <em>"the distance a photon would travel in vacuum during a Julian year at the speed of light at an infinite distance from any gravitational field or magnetic field. </em>"

In other words: It is the distance that the light travels in a year.

This unit is equivalent to 9.461(10)^{12}km, which mathematically is expressed as:

1Ly=9.461(10)^{12}km

Doing the conversion:

4.3Ly.\frac{9.461(10)^{12}km}{1Ly}=4.068(10)^{13}km

DaniilM [7]3 years ago
5 0

The Answer: The star closest to our sun is 4.2 light years away from the sun.

Explanation:

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A 5000 kg African elephant has a resting metabolic rate of 2500 W. On a hot day, the elephant's environment is likely to be near
alina1380 [7]

Answer:

36 kg

Explanation:

To answer this question, a few assumptions have to be made:

  • That the temperature on the day is 35 °C
  • That all the heat from the elephant is goes to warming/evaporating the water on the surface of the elephant

Energy released per hour = 2500 J/s * 3600 s = 9 000 000 J

Q = mcΔT

9 000 000 J= m *4.186 J/g-K * (373K - 308K) + m*2260 J/g

m =  36 000 g = 36 kg

3 0
3 years ago
a car initially at rest move with the constant accerates along straght line read after it's spread increase and finally related
nasty-shy [4]

Answer:

32km per hour

Explanation:

Explanation:

In first case v = a t

==> a t = 40 km p h

Now distance covered S1 + S2 + S3

S1 = 1/2 a t^2 and S3 = 1/2 a t^2

But S2 = 3t * 40 = 120 t km

Hence total distance = at^2 + 120 t

Time taken (total) = t + 3t + t = 5 t

Hence average speed = at^2 + 120 t / 5 t

Cancelling t we have at + 120 / 5 = 40 + 120 / 5 = 160/5 = 32 km per hour

8 0
3 years ago
When a moving object collides with an object that isn't moving, what happens to the kinetic energy of each object?
a_sh-v [17]

Answer:

Since the objects are all motionless after the collision, the final kinetic energy is also zero; the loss of kinetic energy is a maximum. Such a collision is said to be perfectly inelastic.

Explanation:

6 0
3 years ago
Read 2 more answers
Two teams are playing tug-of-war. The team on the right is pulling with 4320 N. The team on the left is pulling with 4380 N. Whi
zmey [24]
60 N to the left because 4380-4320is 60
6 0
3 years ago
A moving particle encounters an external electric field that decreases its kinetic energy from 9970 eV to 6340 eV as the particl
marusya05 [52]

Answer:

q = -1.61x10⁻¹⁷ C

Explanation:

The charge of the particle can be found using the definition of the work done by electric force:  

W = -q\Delta V         (1)

<u>Where</u>:

q: is the charge

ΔV: is the difference in electric potential

The work is also equal to:

W = E_{p_{A}} - E_{p_{B}}    (2)

<u>Where</u>:

E_{p_{A}} and E_{p_{B}} are the electric potential energy of the points A and B, respectively.

Now, by conservation of energy we have:

K_{A} + E_{p_{A}} = K_{B} + E_{p_{B}}     (3)

<u>Where</u>:  

K_{A} and K_{B} are the kinetic energy of the points A and B, respectively.

Rearranging equation (3):  

K_{B} - K_{A} = E_{p_{A}} - E_{p_{B}}      

K_{B} - K_{A} = W

K_{B} - K_{A} = -q\Delta V

Solving the above equation for q:

q = -\frac{K_{B} - K_{A}}{V_{B} - V_{A}} = -\frac{6340 eV - 9970 eV}{19.0 V - 55.0 V} = -100.83 e \cdot \frac{1.6 \cdot 10^{-19} C}{1 e} = -1.61 \cdot 10^{-17} C                                              

Therefore, the charge of the particle is -1.61x10⁻¹⁷ C.

I hope it helps you!

7 0
3 years ago
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