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2 years ago

7

Which or the following is a science ? A. Physics B.astrology C. Football statics D. Language studies

2 answers:

d1i1m1o1n [39]2 years ago

8 0

A. Physics

The study of how objects interact with each other

V125BC [204]2 years ago

5 0

Definetly Physics...........

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What is the student's kinetic energy at the bottom of the hill if he is moving

soldi70 [24.7K]

Answer:

KE = 10530 J or 10.53 KJ

Explanation:

The formula for kinetic energy is KE = 1/2 mv^2

Let's apply the formula:

KE = 1/2 mv^2

KE = 1/2 (65kg) (18m/s)^2

KE = 10530 J or 10.53 KJ

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2 years ago

after an investigation kuri determines that her hypothesis was wrong what is the best thing for kuri to do next

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: You could make slight changes to the process or consider wheather the experiment was carried out correctly

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2 years ago

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What is the kinetic energy of a 1130 kg truck that is moving with a velocity of 40 m/s?

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Answer:

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3 years ago

Two masses (5.3kg and 7.5kg) are fastened together with a small amount of explosive. They are loaded into a spring gun that is t

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8.2

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2 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°)

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2 years ago