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3 years ago

13

What is potential energy

1 answer:

aivan3 [116]3 years ago

6 0

Answer:

mechanical energy, stored energy, or energy caused by its position.

Explanation:

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Because of the curvature of the earth, the maximum distance D that you can see from the top of a tall building of height h is es

mojhsa [17]

Answer:

D = 9.9 10⁶ mi

Explanation:

In the exercise they give the expression for maximum viewing distance

D = 2 r h + h²

Ask for this distance for a height of 1100 feet

Let's calculate

D = 2 3960 1100 + 1100²

D = 8.712 10⁶ + 1.21 10⁶

D = 9.92 10⁶ mi

D = 9.9 10⁶ mi

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3 years ago

Two astronauts are 2.00 m apart in their spaceship. One speaks to the other. The conversation is transmitted to earth via electr

Eduardwww [97]

Answer:

$D=1693742.7m$

Explanation:

For sound waves we have v=d/t where v is the speed of sound and d the distance between the astronauts, while for electromagnetic waves we have c=D/t where c is the speed of light and D the distance between the spaceship and Earth. <em>We have written both times as the same</em> because is what is imposed by the problem, so we have t=d/v=D/c, which means:

$D=\frac{dc}{v}$

And for our values:

$D=\frac{(2m)(299792458m/s)}{354m/s}=1693742.7m$

5 0

3 years ago

Points A (-5,6), B (2,-2), and C (-6,-3) are placed in three different quadrants of a Cartesian coordinate system. Convert each

AURORKA [14]

Answer: A ( $\sqrt{61}$ ,309.8°)

B (2 $\sqrt{2}$ , 315°)

C ( $3\sqrt{5}$ , 26.56°)

Explanation: To transform rectangular coordinates into polar coordinates use:

$r=\sqrt{x^{2}+y^{2}}$ and $\theta=tan^{-1}(\frac{y}{x})$

For point A:

$r=\sqrt{(-5)^{2}+6^{2}}$

$r=\sqrt{61}$

$\theta=tan^{-1}(\frac{6}{-5})$

$\theta=tan^{-1}(-1.2)$

$\theta=-50.2$ °

Point A is in the II quadrant, so we substract the angle for 360° since it is in degrees:

$\theta=360-50.2$

$\theta=$ 309.8°

Polar coordinates for point A is ( $\sqrt{61}$ , 309.8°)

For point B:

$r=\sqrt{2^{2}+(-2)^{2}}$

$r=\sqrt{8}$

$r=2\sqrt{2}$

$\theta=tan^{-1}(\frac{-2}{2} )$

$\theta=tan^{-1}(1)$

$\theta=-45$ °

Point B is in IV quadrant, so:

$\theta=360-45$

$\theta=$ 315°

Polar coordinates for point B is ( $2\sqrt{2}$ , 315°)

For point C:

$r=\sqrt{(-6)^{2}+(-3)^{2}}$

$r=\sqrt{45}$

$r=3\sqrt{5}$

$\theta=tan^{-1}(\frac{-3}{-6} )$

$\theta=tan^{-1}(0.5)$

$\theta=$ 26.56°

Polar coordinates for point C is ( $3\sqrt{5}$ , 26.56°)

3 0

3 years ago

A car initially at rest, accelerates at a constant rate of 4.0 m/s for 6s. How fast will the car be traveling at 6s

Katen [24]

It will be traveling exactly 24 miles per hour <span />

7 0

3 years ago

1. A student demonstrates electromagnetic induction using a

Katen [24]

answer is :D it would be a great answer

4 0

3 years ago