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Mars2501 [29]
3 years ago
9

What scientific observation did Edwin Hubble use to determine distances between galaxies?

Physics
1 answer:
Cerrena [4.2K]3 years ago
7 0

Answer: the expanding universe

Explanation:

Hope that helps!

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You push a coin across a table. The coin stops. How does this motion relate to balanced and unbalanced forces?
Snezhnost [94]
If you are pushing the coin across the table at a constant rate, the friction of the table and the horizontal force of your hand pushing are equal, and the coin itself moves at a constant rate. If you push a coin and let it go, there is no horizontal force keeping the coin going. Friction slows the coin to a stop. In both cases, the gravitational downward pull of Earth is equally but oppositely resisted by the upward push of table on the coin.
7 0
3 years ago
Use newton's law to explain the vertical acceleration of a projectile
NISA [10]

Answer:

Explained below

Explanation:

Newton's first law of motion: This law states that an object will remain at rest or continue in constant motion except it's acted upon by an external force. In projectile motion, the horizontal component of velocity will remain unchanged because we ignore air resistance since no force is acting in that horizontal direction.

Newton's second law of motion: This law states that force is the product of mass and acceleration. In projectile the force acts downwards, thus f = mg.

But g = a since internal forces will cancel out.

Thus, F = ma

5 0
3 years ago
U1=20 m/s turn it to km/h
notka56 [123]
It is 72 km/h
I hope it helps
7 0
3 years ago
Not sure if it went through last time. Please help asap!
Olin [163]
The equation for force is F=ma. Because we have the value of mass (0.42 kg) and the acceleration (14.8 m/s^2), simply plug them into the equation for force to get
0.42 \times 14.8 = 6.22
The answer is 6.22 N because newtons are the unit used to measure force.
8 0
3 years ago
Read 2 more answers
Particle A and particle B are held together with a compressed spring between them. When they are released, the spring pushes the
yaroslaw [1]

Answer:

 K_a = 8,111 J

Explanation:

This is a collision exercise, let's define the system as formed by the two particles A and B, in this way the forces during the collision are internal and the moment is conserved

initial instant. Just before dropping the particles

          p₀ = 0

final moment

          p_f = m_a v_a + m_b v_b

          p₀ = p_f

          0 = m_a v_a + m_b v_b

tells us that

          m_a = 8 m_b

         

           0 = 8 m_b v_a + m_b v_b

           v_b = - 8 v_a                    (1)

indicate that the transfer is complete, therefore the kinematic energy is conserved

starting point

           Em₀ = K₀ = 73 J

final point. After separating the body

          Em_f = K_f = ½ m_a v_a² + ½ m_b v_b²

           K₀ = K_f

           73 = ½ m_a (v_a² + v_b² / 8)

           

we substitute equation 1

           73 = ½ m_a (v_a² + 8² v_a² / 8)

           73 = ½ m_a (9 v_a²)

           73/9 = ½ m_a (v_a²) = K_a

            K_a = 8,111 J

3 0
3 years ago
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