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3 years ago

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First, launch the video below. You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then,

close the video window and answer the question at right. You can watch the video again at any point.

1 answer:

sesenic [268]3 years ago

3 0

Answer:

You will be asked to use your knowledge of physics to predict the outcome of an experiment. Then, close the video window and answer the question at right. You can watch the video again at any point.

Explanation:

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3 years ago

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If a galaxy has an apparent velocity of 2300 km/s, what is its distance if the Hubble constant is assumed to be 70 km/s/Mpc

prohojiy [21]

The distance of the galaxy is 32.86 Mpc.

Using the hubble law, v = H₀D where v = apparent velocity of galaxy = 2300 km/s, H = hubble constant = 70 km/s/Mpc and D = distance of galaxy.

Since we require the distance of the galaxy, we make D subject of the formula in the equation. So, we have

D = v/H₀

Substituting the values of the variables into the equation, we have

D = 2300 km/s ÷ 70 km/s/Mpc

D = 32.86 Mpc

So, the distance of the galaxy is 32.86 Mpc

Learn more about hubble law here:

brainly.com/question/18484687

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3 years ago

Calculate the total displacement if a toy car starts at 0 moves 5cm to the left then 8cm to the right then 3 cm to the left

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It’s a total of 16 cm because it starts at 0 moves then moves to left then the right then the left

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4 years ago

According to Newton's Law of Cooling, if a body with temperature T 1 is placed in surroundings with temperature T 0, different f

Mila [183]

We can substitute the given values into the equation for T, given the surrounding temperature T0 = 0, initial temperature T1 = 140, constant k = -0.0815, and time t = 15 minutes.

T = 0 + (140 - 0)e^(-0.0815*15) = 140e^(-1.2225) = 41.23°F

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3 years ago

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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°)

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3 years ago