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Nataly [62]
3 years ago
7

you read a primary source and a secondary source that discuss the same experiment. There is a difference in the conclusion made

by these two sources. which should you trust more, and why?
Physics
1 answer:
Nuetrik [128]3 years ago
4 0
You should trust the primary source more.

This is because the primary source is make its conclusion from direct observation, while the secondary source is possibly making reference to another secondary source or to another primary.

The primary source should be trusted more because it is from direct observation.     
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Suppose you are standing on the edge of a dock and jump straight down. If you land on sand your stopping time is much shorter th
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In bringing you to a halt, the sand and the water exert the same impulse on you, but the sand exerts a greater average force

Explanation:

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What is the reaction force if a girl pulls on a cow?
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From newton’s third law we know that every action has a reaction force pushing back. So when the girl pulls on a cow, the cow is pulling back on her.
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A 60 N force is applied over a distance of 15 m. How much work was done?
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Answer:

900J

Explanation:

w =f×s

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_______ is a force of attraction that occurs between objects due to their mass. a. magnetism b. gravity c. weight d. deposition
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Read 2 more answers
A green block of mass m slides to the right on a frictionless floor and collides elastically with a red block of mass M which is
Charra [1.4K]

Answer:

M is equal to m

Explanation:

In case we say that the green block's mass m is less than red block's mass M, then the green block would have bounced and moved back to the left instead of coming to rest. The other case where if mass of green block's mass m would have been greater than the red block's mass M, the green block would have kept moving to the right instead of coming to rest. After collision, the red block moves to the right because of exchange of velocities. Therefore, m=M since m comes to rest and M moves to the right

In any collision, as it is asumed that no external forces can act during the collision, momentum must be conserved.

So, if we call p₁ to the momentum before collision, and p₂ to momentum after it, taking into account the information above, we can write the following:

p₁ = mv₁ + M.0 = p₂ = m.0 + Mv₂ ⇒ mv₁ = Mv₂

From the question, we also know that it was an elastic collision.

In elastic collision, added to the momentum conservation, it must be conserved the kinetic energy also.

So, if we call k₁ to the kinetic energy prior the collision, and k₂ to the one after it, we can write the following:

k₁ = 1/2 m(v₁)² + 1/2 M.0 = k₂ = 1/2m.0 + 1/2M(v₂)² ⇒ m(v₁)² = M(v₂)²

Mathematically, the only way in which both equations be true, should be with v₁ = v₂,  which is only possible if m=M too.

In this type of collision, it is said that the energy transfers from one mass to the other.

8 0
3 years ago
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