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Angelina_Jolie [31]

2 years ago

9

a student moves a box across the floor by exerting 23.3 N of force and doing 47.2 J of work on the box. How far does the student

move the box?

1 answer:

kozerog [31]2 years ago

8 0

Work = Force x Distance
47.2J = 23.3N x d
d = 47.2/23.3
d = 2.0258 m

hope this helps :P

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Juan and Kuri are on a carousel. Juan is closer to the center of the carousel than Kuri. Which statement describes their tangent

Licemer1 [7]

Answer:

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

Explanation:

Since Juan is closer to the center and Kuri is away from the center so we can say that Juan will move smaller distance in one complete revolution

As we know that the distance moved in one revolution is given as

$d = 2\pi r$

also the time period of revolution for both will remain same as they move with the time period of carousel

Now we can say that the speed is given as

$v = \frac{2\pi r}{T}$

so Juan will have less tangential speed. so correct answer will be

Juan and Kuri complete one revolution in the same time, but Juan travels a shorter distance and has a lower speed.

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2 years ago

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Paintball guns were originally developed to mark trees for logging. A forester aims his gun directly at a knothole in a tree tha

emmasim [6.3K]

Answer:

The distance between knothole and the paint ball is 0.483 m.

Explanation:

Given that,

Height = 4.0 m

Distance = 15 m

Speed = 50 m/s

The angle at which the forester aims his gun are,

$\tan\theta=\dfrac{4}{15}$

$\tan\theta=0.266$

$\cos\theta=\dfrac{15}{\sqrt{15^2+4^2}}$

$\cos\theta=0.966$

Using the equation of motion of the trajectory

The horizontal displacement of the paint ball is

$x=(u\cos\theta)t$

$t=\dfrac{x}{u\cos\theta}$

Using the equation of motion of the trajectory

The vertical displacement of the paint ball is

$y=u\sin\theta(t)-\dfrac{1}{2}gt^2$

$y=u\sin\theta(\dfrac{x}{u\cos\theta})-\dfrac{1}{2}g(\dfrac{x}{u\cos\theta})^2$

$y=x\tan\theta-\dfrac{gx^3}{2u^2(\cos\theta)^2}$

Put the value into the formula

$y=(15\times0.266)-(\dfrac{9.8\times(15)^2}{2\times(50)^2\times(0.966)^2})$

$y=3.517\ m$

We need to calculate the distance between knothole and the paint ball

$d=h-y$

$d=4-3.517$

$d=0.483\ m$

Hence, The distance between knothole and the paint ball is 0.483 m.

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2 years ago

A car drives around a racetrack for 30 seconds. what do you need to know to calculate the average velocity of the car?

boyakko [2]

The time is given, and you want to find the average velocity. To do this, you need to know the distance covered by the driver around the racetrack in that 30 seconds. You divide this by the time, then you will obtain the average velocity in units of, say meters per second.

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zysi [14]

Answer:

Absolute error=0.006

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=0.006

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The mass of one washer is 0.0049 kg.

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