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3 years ago

Can anyone help me with this

Physics

1 answer:

Murrr4er [49]3 years ago

8 0

Answer:

event s

Explanation:

the cow gets oxygen from the tree and proceeds to respirate.

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A car travels 290 m over a time of 5s. What was the speed of the car?

Elanso [62]

Answer: 58

Explanation:

I did 290/5

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3 years ago

How did the old view of the solar system look like

zhuklara [117]

The old view of the solar system was geocentric, which means that people believed that the Earth is the center of the universe. However, with the discoveries of Galileo, Newton, and Copernicus, evidence was brought forth that the solar system is actually heliocentric, which means that the sun is the center of the universe.

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3 years ago

Alice drops a rock to a well. She hears the splash of the rock 4.1s later. The speed of sound is 340m/s. How deep is the well

Maksim231197 [3]

Answer:

697 m

Explanation:

From the question above,

Note: Alice hears the splash of the rock in the well through the help of echo.

Applying,

v = 2x/t........................ Equation 1

Where v = speed of sound, x = depth of the well, t = time.

make x the subject of the formula in equation 1

x = vt/2................... Equation 2.

Given: v = 340 m/s, t = 4.1 s

Substitute these values into equation 2

x = 340(4.1)/2

x = 697 m.

3 0

3 years ago

A hot-air balloon of diameter 12 mm rises vertically at a constant speed of 14 m/s. A passenger accidentally drops his camera fr

fgiga [73]

Answer:

<em>The balloon is 66.62 m high</em>

Explanation:

<u>Combined Motion </u>

The problem has a combination of constant-speed motion and vertical launch. The hot-air balloon is rising at a constant speed of 14 m/s. When the camera is dropped, it initially has the same speed as the balloon (vo=14 m/s). The camera has an upward movement for some time until it runs out of speed. Then, it falls to the ground. The height of an object that was launched from an initial height yo and speed vo is

$\displaystyle y=y_o+v_o\ t-\frac{g\ t^2}{2}$

The values are

$\displaystyle y_o=15\ m$

$\displaystyle v_o=14\ m/s$

We must find the values of t such that the height of the camera is 0 (when it hits the ground)

$\displaystyle y=0$

$\displaystyle y_o+v_o\ t-\frac{g\ t^2}{2}=0$

Multiplying by 2

$\displaystyle 2y_o+2v_ot-gt^2=0$

Clearing the coefficient of $t^2$

$\displaystyle t^2-\frac{2\ V_o}{g}\ t-\frac{2\ y_o}{g}=0$

Plugging in the given values, we reach to a second-degree equation

$\displaystyle t^2-2.857t-3.061=0$

The equation has two roots, but we only keep the positive root

$\displaystyle \boxed {t=3.69\ s}$

Once we know the time of flight of the camera, we use it to know the height of the balloon. The balloon has a constant speed vr and it already was 15 m high, thus the new height is

$\displaystyle Y_r=15+V_r.t$