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Nonamiya [84]
3 years ago
15

What do you think that astronomers mean when they use the term observable universe?

Physics
1 answer:
iren2701 [21]3 years ago
6 0
The observable universe consists of galaxies and other matter that can, principally, be seen from Earth because the light signals have had time to reach us. Not everything in the sky is the way it is when we see it, because of the distance the light travels to reach us. 

Hope this helps :)
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Explain the difference between a conductor and an insulator. Give two examples of each.
docker41 [41]
There are different kinds of conductors, most notably electrical and thermal conductors. But they are often inclusive of each other (electrical conductors are typically good thermal conductors). A conductor transmits something through its body with high efficiency while an insulator does not transmit very well. In the case of electricity, a conductor transmits electrical energy between two points while an insulator blocks the flow of electricity. Two examples of conductors are copper and silver. Two examples of insulators are wood and styrofoam.
3 0
3 years ago
Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli
serious [3.7K]

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

  • Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
  • If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

       p_{ox} = p_{oAx} + p_{oBx}  (1)

  • We can do exactly the same for the initial momentum along the y-axis:

       p_{oy} = p_{oAy} + p_{oBy}  (2)

  • The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

       p_{fx} =  (m_{A} + m_{B} ) * v_{fx}  (3)

  • We can repeat the process for the y-axis, as follows:

       p_{fy} =  (m_{A} + m_{B} ) * v_{fy}  (4)

  • Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

       v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} =  2.96 m/s (5)

  • In the same way, we can find the component of the final momentum along the y-axis, as follows:

       v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} =  3.15 m/s (6)

  • With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

      v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)

  • The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

       p_{f} = (m_{A} + m_{B})* v_{f}  = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)

  • Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
  • We can find this angle applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)

       ⇒ θ = tg⁻¹ (1.06) = 46.8º N of E

8 0
3 years ago
Calculate the acceleration due to gravity on the moon. the moon's radius is 1.74×106m and its mass is 7.35×1022 kg.
Kaylis [27]
184.44
 
7,511.7 that s the answer

7 0
3 years ago
Erik Erikson’s theory of psychosocial development emphasizes that development occurs by overcoming an emotional crisis in each o
lorasvet [3.4K]

Answer:

True

Explanation:

8 0
2 years ago
A rifle fires a 2.01 10-2-kg pellet straight upward, because the pellet rests on a compressed spring that is released when the t
Zina [86]

Answer:

The value of spring constant is 266.01 \frac{N}{m}

Explanation:

Given:

Mass of pellet m = 2.01 \times 10^{-2} kg

Height difference of pellet rise h_{f} - h_{o} = 6.03 m

Spring compression x = 9.45 \times 10^{-2} m

From energy conservation law,

Spring potential energy is stored into potential energy,

  mg(h_{f} -h_{o})  = \frac{1}{2} kx^{2}

Where k = spring constant, g = 9.8 \frac{m}{s^{2} }

  k = \frac{2mg(h_{f} -h_{o} )}{x^{2} }

  k = \frac{2 \times 9.8 \times 6.03\times 2.01 \times 10^{-2} }{(9.45\times 10^{-2} )^{2} }

  k = 266.01 \frac{N}{m}

Therefore, the value of spring constant is 266.01 \frac{N}{m}

6 0
3 years ago
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