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4 years ago

Displacement is to velocity as_____is to acceleration

Physics

2 answers:

Pavlova-9 [17]4 years ago

4 0

Answer:

Displacement is to velocity as <em>velocity </em>is to acceleration.

Explanation:

Velocity of an object is equal to the change in displacement per unit time. It is given by :

$v=\dfrac{d}{t}$

d is the displacement and t is the time taken

Acceleration of an object is given by the change in velocity per unit time. It is given by :

$a=\dfrac{v-u}{t}$

v and u are final and initial velocities

(v-u) is change in velocity

and t is time taken

So, displacement is to velocity as <em>velocity </em>is to acceleration. This is because velocity is related to displacement and similarly acceleration is related to velocity.

sergeinik [125]4 years ago

3 0

Answer:

Velocity.

Explanation:

The answer is velocity because when you make a graph, one axis is acceleration while the other axis is velocity.

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Answer:

The easiest way to do this is to realize that the time to fall is the same as the rise time (conservation of energy can easily verify this)

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2 years ago

You are comparing a reaction that produces a chemical change and one that produces a physical change. What evidence could you us

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4 years ago

You drop a single coffee filter of mass 1.5 grams from a very tall building, and it takes 49 seconds to reach the ground. In a s

Tresset [83]

(a) 0.0147 N

When the filter reaches the terminal speed, it means that its acceleration is now zero, so the net force acting on it is zero.

There are only two forces acting on the filter:

- The weight of the filter, downward: W = mg, where

$m=1.5 g = 0.0015 kg$ is the mass of the filter

$g=9.8 m/s^2$ is the acceleration due to gravity

- The air resistance, upward, $F_a$

Since the net force is zero, we have

$W-F_a =0$

and solving the equation we find the upward force of air resistance:

$F_a=W=mg=(0.0015kg)(9.8 m/s^2)=0.0147 N$

(b) 0.0441 N

The problem is exactly identical to before, but this time the mass of the stack of filters is 3 times the mass of the single filter, so

$m=3 (0.0015 kg)=0.0045 kg$

And so, the upward force of air resistance is

$F_a = mg=(0.0045 kg)(9.8 m/s^2)=0.0441 N$

(c) 28.3 s

We know that the single coffee filter takes t=49 s to reach the ground, travelling at constant terminal speed of v, so the distance covered (the height of the building) is

$d=vt$

where t = 49 s.

We also know that the air resistance is proportional to the square of the terminal speed:

$F_{air} \propto v^2$

which means

$v\propto \sqrt{F_{air}}$

For the stack of three filters, we found (b) that the air resistance is 3 times the air resistance for the single filter (a). Therefore, the terminal speed of the stack of filter will be $\sqrt{3}$ times larger than the terminal speed of the single filter.

Therefore, the time taken will be:

$t'=\frac{d}{v'}=\frac{d}{\sqrt{3} v}=\frac{t}{\sqrt{3}}$

And since t = 49 s, we have

$t'=\frac{49 s}{\sqrt{3}}=28.3 s$

8 0

3 years ago