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Svetllana [295]
3 years ago
5

Distinguish

Physics
1 answer:
yaroslaw [1]3 years ago
4 0

Answer:

Velocity and speed both are continuously increasing.

Acceleration is constant.

Explanation:

Speed is defined as length of path covered by a body per unit time. Speed is a scalar quantity that consist of magnitude only and not direction.

Velocity is defined as the displacement per unit times. Displacement is the shortest distance between the two points. It is a vector quantity and  hence has a direction in the direction of displacement along with its own magnitude.

  • Both velocity and speed have same unit of measure which is meter per second in S.I. During <em>free fall</em> in the absence of any air resistance the velocity and speed both will be having a vertical downward direction with continuously increasing magnitude. Tough we are not concerned about the direction when discussing about speed but here both are equal since the motion is linear.

Acceleration is the rate of change in velocity of a body which is a vector quantity. For speed we are concerned about instantaneous acceleration since for a short period of time it may have a specific direction.

  • During free fall the acceleration is of a body is equal to the acceleration due to gravity and constant when the height of fall is much lesser than the radius of the earth.
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Answer:

1. at least two charged interacting parts

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4 0
3 years ago
A boy standing throws a penny horizontally at 7.25 m/s out of the window of his apparent buliding. If the window is 10.0 m above
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Answer:

Explanation:

This is a 2D problem (parabolic) so we have to think that way. We have to split up the problem into its 2 dimensions to solve it. Think "y-stuff" and "x-stuff".

In the y-stuff category:

v₀ = 0 (initial upwards velocity is 0 since we are told the penny is thrown horizontally)

Δx = -10.0 m (this displacement is negative because the penny lands 10.0 m below the point from which it was thrown)

a = -9.8 m/s/s

t = ? (we need to find the time in this dimension so we can use it in the x dimension to find the displacement, our unknown)

In the "x-stuff" category:

v₀ = 7.25 m/s (this is given)

Δx = ???

a = 0 (acceleration in this dimension is ALWAYS 0)

t = (we will solve for this in the y-dimension and plug it in here).

In the y dimension:

Δx = v₀t + \frac{1}{2}at^2 and plugging in from the y-dimension info:

-10.0=0t+\frac{1}{2}(-9.8)t^2 which simplifies to

-10.0=-4.9t^2 so

t=\sqrt{\frac{-10.0}{-4.9} } which, to 2 significant digits is

t = 1.4 seconds

Now we will do the same in the x-dimension, using t = 1.4:

Δx = v₀t + \frac{1}{2}at^2 and filling in the x-stuff:

Δx = 7.25(1.4)+\frac{1}{2}(0)(1.4)^2 Notice that the stuff after the + sign goes to 0 cuz of the multiplication of 0, so what we are left with is another form of the d = rt equation:

Δx = 7.25(1.4) + 0 so

Δx = 1.0 × 10¹ m (That's rounded correctly to 2 sig dig's: 10 m from the base of the building).

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3 years ago
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