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3 years ago

What goes up and never comes down

Physics

2 answers:

Goshia [24]3 years ago

4 0

Your age hope this helped! ^^

gavmur [86]3 years ago

4 0

Age is the answer

hope this helps

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Suppose two children push horizontally, but in exactly opposite directions, on a wagon containing a third child. The first child

koban [17]

Answer:

$0.1 m/s^2$ , in the direction of F2

Explanation:

The magnitude of F2 is larger than that of F1: this means that the wagon will tend to move in the direction of F2. Therefore, the direction of friction will be opposite to the direction of motion, therefore in the direction of F1.

Let's now take the direction of the force exerted by the second child, F2, as positive direction. Then the direction of F1 and of the force of friction ( $F_f$ ) will be negative: so we have

$F_2 = +90.0 N\\F_1 = -75.0 N\\F_f = -12.0 N$

We can write the equation of the forces for the third child+wagon:

$F_1+F_2+F_f = ma$

where

m = 30.0 kg is the combined mass of the child + the wagon

Now we can solve the equation for a, to find the acceleration:

$a=\frac{F_2+F_1+F_f}{m}=\frac{90-75-12}{30}=0.1 m/s^2$

4 0

3 years ago

When is the universal theme of a story often revealed?

elena-14-01-66 [18.8K]

It is often revealed <span>at the resolution of the story, when the reader can see how the story ends.</span>

7 0

3 years ago

Read 2 more answers

An aquarium open at the top has 30-cm-deep water in it. You shine a laser pointer into the top opening so it is incident on the

Setler [38]

Answer:

You must add 8cm of water to the tank

Explanation:

In order to find how much the height is we will use the Snell Refraction law

This law relates the index of refraction of the water (n2), the index of refraction of the air (n1), the incidence angle relative to the vertical (theta1) and the refraction angle relative to the vertical (theta2) by using the next equation:

$(n1)*(sin(theta1))=(n2)*(sin(theta2))$

Then we will find the refraction angle relative to the vertical this way:

$(n1/n2)*(sin(theta1))=sin(theta2)$

$(1/1.33)*(sin(45))=sin(theta2)$

Then, theta2=32.12°

Now that we have this information we can imagine a triangle with a 30cm height and a 32.12° angle. This way we can find how much X is, this X will be the distance between the vertical line and the spot the beam hits the bottom, so we can use some trigonometry to find it, this way:

$tan(32.12)=(X/30cm)$

$X=(tan(32.12))*(30cm)$

Then, X=18.8cm, we can approximate it to 19cm

Once we have X we will add 5cm to it which is how much the beam needs to be moved, then the new X will be 24cm

Now, with the new horizontal distance we will find the new vertical distance, let´s call it Y, this way we will know how much water we must add to move the beam, then we will have a triangle with a vertical distance called Y, the same 32.12° angle will be used as we are still working with the air-water interface and a 19cm horizontal distance, then:

$tan(32.12)=(24cm/Y)$