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2 years ago

13

Is it possible to calculate the displacement based on an elapsed time from a position time graph?

1 answer:

Nataly [62]2 years ago

5 0

No I don’t think so. But it worth a try tho. Try it out.

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What is another name for the dependent variable in an experiment? A. A manipulated variable B. A responding variable C. A contro

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The answer to your question is,

A. A manipulated variable.

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A rod is made of three segments of equal length with different masses. The total mass of the rod is 6m. Will the moment of inert

bazaltina [42]

Answer: (iii) the same about both ends

Explanation:

The moment of inertia is the sum of the square of the distance from the axis of massage and rotation of each length of the rod. The moment of inertia can be expressed as I = imiri2. The connection of the rod is the product of the moment length and distance (square), so it is the same at both ends.

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3 years ago

A child is twirling a 0.0154-kg ball on a string in a horizontal circle whose radius is 0.149 m. The ball travels once around th

xeze [42]

Answer:

The answers are: a)Fcp=0,23N b)As Fcp=0,93N, it increases 4 times when the speedis doubled

Explanation:

Let´s explain what´s the centripetal force about: It´s the force applied over an object moving on a curvilinear path. This force is directed to the rotation center.

This definition is described this way:

$Fcp*r=m*V^2$ where:

Fcp is the Centripetal Force

r is the horizontal circle radius

m is the ball mass

V is the tangencial speed, same as the rotation speed

w is the angular speed

Here we need to note that the information we have talks about 1cycle/0,639s (One cycle per 0,639s). We need to express this in terms of radians/seconds. To do it we define that 1cycle is equal to 2pi, so we can find the angular speed this way:

$w=(1cycle/0,639s)*(2pi/1cycle)$

So the angular speed is $w=9,83rad/s$

Now that we have this information, we can find the tangencial speed, which will be the relation between the angular speed and the circle radius, this way:

$V=w*r$ so the tangencial speed is:

$V=(9,83rad/s)*(0,149m)$

$V=1,5m/s$

Now we have all the information to find the Centripetal Force:

$Fcp=(m*V^2)/(r)$

$Fcp=((0,0154kg)*(1,5m/s)^2)/(0,149m)$

a) So the Centripetal Force is: $Fcp=0,23N$

b) If the tangencial speed is doubled, its new value will be 3m/s. replacing this information we will get the new Centripetal Force is:

$Fcp=((0,0154kg)*(3m/s)^2)/(0,149m)$

The Centripetal Force is: $Fcp=0,93N$

Here we can see that if the speed is doubled, the Centripetal Force will increase four times.

3 0

3 years ago

A brick is dropped (zero initial speed) from the roof of a building. The brick strikes the ground in 1.90 s. You may ignore air

Gekata [30.6K]

Answer: Height of building = 17.69m, velocity of brick = 18.6m/s

Explanation: From the question, the body has a zero initial speed, thus initial velocity (u) all through the motion is zero.

By ignoring air resistance makes it a free fall motion thus making it to accelerate constantly with a value of $a = 9.8m/s^{2}$ .

Time taken to fall = 1.90s

a)

thus the height of the building is calculated using the formulae below

$H = ut + \frac{1}{2} gt^{2}$

but u = 0 , hence

$H = \frac{1}{2} gt^{2}$

$H = \frac{1}{2} *9.8* 1.9^{2} \\\\H = 17.69m$

b)

to get the value of velocity (v) as the brick hits the ground, we use the formulae below

$v^{2} = u^{2} + 2aH$

but u= 0, hence

$v^{2} = 2gH\\$

$v^{2} = 2 * 9.8 * 17.69\\v = \sqrt{2 *9.8* 17.69} \\v = 18.62m/s$

find the attachment in this answer for the accleration- time graph, velocity- time graph and distance time graph

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3 years ago