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2 years ago

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A squirrel jumps into the air with a velocity of 7 m/s at an angle of 20 degrees. What is the maximum height reached by the squi

rrel?

1 answer:

Marina CMI [18]2 years ago

7 0

Answer:

.3m

Explanation:

Apex

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peach pie in a 9.00 in diameter plate is placed upon a rotating tray. Then, the tray is rotated such that the rim of the pie pla

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Explanation:

It is given that,

Diameter of the peach pie, d = 9 inches

Radius of the pie, r = 4.5 inches

The tray is rotated such that the rim of the pie plate moves through a distance of 183 inches, d = 183 inches

Let $\theta$ is the angular distance that the pie plate has moved through.

It is given by :

$\theta=\dfrac{d}{r}$

$\theta=\dfrac{183}{4.5}$

$\theta=40.66\ radian$

Since, 1 radian = 57.29 degrees

$\theta=2329.64\ degrees$

Since, 1 radian = 0.159155 revolution

$\theta=6.47124\ revolution$

Hence, this is the required solution.

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2 years ago

PLEASE HELP ME I NEED IT!!

Alla [95]

Answer:

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Explanation:

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If you break quartz to learn if it splits smoothly in a certain direction, what physical property are you testing?

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Which best explains why the shapes of a liquid can change

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An electron and a proton are held on an x axis, with the electron at x = + 1.000 m and the proton at x = - 1.000 m. Part A How m

r-ruslan [8.4K]

PART A)

Electrostatic potential at the position of origin is given by

$V = \frac{kq_1}{r_1} + \frac{kq_2}{r_2}$

here we have

$q_1 = 1.6 \times 10^{-19} C$

$q_2 = -1.6 \times 10^{-19} C$

$r_1 = r_2 = 1 m$

now we have

$V = \frac{Ke}{r} - \frac{Ke}{r}$

$V = 0$

Now work done to move another charge from infinite to origin is given by

$W = q(V_f - V_i)$

here we will have

$W = e(0 - 0) = 0$

so there is no work required to move an electron from infinite to origin

PART B)

Initial potential energy of electron

$U = \frac{Kq_1e}{r_1} + \frac{kq_2e}{r_2}$

$U = \frac{9\times 10^9(-1.6\times 10^{-19}(-1.6 \times 10^{-19})}{19} + \frac{9\times 10^9(1.6\times 10^{-19}(-1.6 \times 10^{-19})}{21}$

$U = (2.3\times 10^{-28})(\frac{1}{19} - \frac{1}{21})$

$U = 1.15\times 10^{-30}$

Now we know

$KE = \frac{1}{2}mv^2$

$KE = \frac{1}{2}(9.1\times 10^{-31}(100)^2$

$KE = 4.55 \times 10^{-27} kg$

now by energy conservation we will have

So here initial total energy is sufficient high to reach the origin

PART C)

It will reach the origin

4 0

2 years ago