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kogti [31]
3 years ago
13

If you started counting from 1st grade, how many years of education would the typical professional astronomer have today, at min

imum?
12
14
16
18
Physics
2 answers:
Savatey [412]3 years ago
5 0
16 is right just took the test
allochka39001 [22]3 years ago
4 0

Answer:

16 years

Explanation:

I hope this helps.

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ElenaW [278]

Answer:

Hi... Potential energy is converted to kinetic energy and kinetic energy is converted to potential energy

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Which wave interaction is responsible for echoes?
IrinaVladis [17]

Answer:Reflection

Explanation:

5 0
3 years ago
Read 2 more answers
A 6.0-kg object moving at 5.0 m/s collides with and sticks to a 2.0-kg object. After the collision the composite object is movin
gogolik [260]

Answer:

a) 23 m/s

Explanation:

  • Assuming no external forces acting during the collision, total momentum must be conserved, as follows:

       p_{o} = p_{f}  (1)

  • The initial momentum p₀, can be written as follows:

       p_{o} =  m_{1}  * v_{1o} + m_{2}* v_{2o} =   6.0 kg * 5.0 m/s + 2.0 kg * v_{2o}  (2)

  • The final momentum pf, can be written as follows:

        p_{f} = (m_{1} + m_{2} )* v_{f}  = 8.0 kg* (-2.0 m/s)  (3)

  • Since (2) and (3) are equal each other, we can solve for the only unknown that remains, v₂₀, as follows:

       v_{2o} = \frac{-6.0kg* 5m/s -8.0 kg*2.0m/s}{2.0kg}  = \frac{-46kg*m/s}{2.0kg} = -23.0 m/s  (4)

  • This means that the 2.0-kg object was moving at 23 m/s in a direction opposite to the 6.0-kg object, so its initial speed, before the collision, was 23.0 m/s.
6 0
3 years ago
A charged wire of negligible thickness has length 2L units and has a linear charge density λ. Consider the electric field E-vect
Stels [109]

Answer:

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

Explanation:

Given that

Length= 2L

Linear charge density=λ

Distance= d

K=1/(4πε)

The electric field at point P

E=2K\int_{0}^{L}\dfrac{\lambda }{r^2}dx\ sin\theta

sin\theta =\dfrac{d}{\sqrt{d^2+x^2}}

r^2=d^2+x^2

So

E=2K\lambda d\int_{0}^{L}\dfrac{dx }{(x^2+d^2)^{\frac{3}{2}}}

Now by integrating above equation

E=2K\lambda d\dfrac{L }{d^2\sqrt{L^2+d^2}}

4 0
3 years ago
Suppose a pendulum starts with a potential energy of 100 J. Assuming the pendulum has a height of 0 m at the bottom of its swing
vlabodo [156]

Answer:

100 J

Explanation:

5 0
3 years ago
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