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Alexxandr [17]
3 years ago
6

Match the following. Column A 1. Torque 2. Centre of gravity 3. Plumb line Column B A. Line of centre of gravity B. Maximum cons

tant velocity C. Not in motion D. Point of action of weight E. Turning effect of force F. Study of forces 4. Terminal velocity 5. Statics​
Physics
1 answer:
gregori [183]3 years ago
8 0

Answer:

1. Torque → F. Study of forces

2. C.O.G → D. Point of action of weight.

3. Plumb line → A. Line of C.O.G

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Move the objects faster to get more friction.
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What is a plasma?A.A superheated liquidB.A solid in which the particles vibrate very quicklyC.A gas in which the atoms have star
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plasma is a superheated liquid

So, a would-be the correct option.

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1 year ago
19 dm expressed in millimeters
alina1380 [7]
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1 year ago
What best explains whether bromine (Br) or neon (Ne) is more likely to form a covalent bond? On left, a purple circle labeled Br
blondinia [14]

Answer:

I believe the answer is Bromine forms covalent bonds because it has many electron shells, but neon has only two electron shells and is tightly bound to its electrons.

Explanation:

valence electrons are the outermost shell, so when you go through keeping that in mind it helps you find the right answer

6 0
3 years ago
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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