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erik [133]
4 years ago
6

Why do raindrops fall with a constant speed during the later stages of their descents?

Physics
1 answer:
BartSMP [9]4 years ago
6 0
Firstly they have a acceleration downwards due the force downwards due they gravitational field acting on it's mass.
as it falls it gains speed, and as it gains speed the air Resistance which is a upward force actin on the drop increases, eventually the rain drop's upward and downward forces are balanced and hence there is no RESULTANT force therefore no acceleration, so the drops falls in constant speed (terminal verlocity is a better term)

Are you wondering that why is the raindrop still moving given that the forces are balanced? If so according to Newton's 1st law an object will keep moving or Remain at rest until a RESULTANT force acts on it.
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A net force of 15 N is applied to a cart with a mass of 2.1 kg. a. What is the acceleration of the cart? b. How long will it tak
Gennadij [26K]
The first question's answer is :
If, F=ma
Then, 15N= 2.1kg (a)
15/2.1=a
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6 0
3 years ago
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Nataly [62]

Answer:

number 4

Explanation:

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5 0
3 years ago
A block is held at rest against a wall by a force of magnitude F exerted at an angle theta from the horizontal, as shown in the
wel

Answers:

B.) F cos\theta=F_{n}

C.) F sin\theta=F_{g} \pm F_{f}

Explanation:

The image attached shows the way the force F is acting on the block. Now, if we draw a free body diagram of the situation and write the equations for the Net Force in X and Y, we will have the following:

Net Force in X:

-F_{n}+F cos\theta=0 (1)

Where:

F_{n} is the Normal force

F is the magnitude of the force exerted on the block

\theta is the angle

Net Force in Y:

F sin\theta \pm F_{f}-F_{g}=0 (2)

Where:

F_{f} is the Friction force (it is expresed with the \pm sign because this force may be up or down, we cannot know because the block is at rest)

F_{g} is the gravity force

Rewrittin (1):

F cos\theta=F_{n} (3) This is according to option B

Rewritting (2):

F sin\theta=F_{g}\pm F_{f} (3) This is according to option C

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6 0
3 years ago
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