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1 year ago

15

What is the mass in rams of 30.00 ml of a liquid that has a density of 1.21 g/ml?

1 answer:

Murljashka [212]1 year ago

3 0

Answer:

36.3 g

Explanation:

1.21 g/ml * 30 ml = 36.3 g ( see how the 'ml' units cancel ?)

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A football player at practice pushes a 60 kg blocking sled across the field at a constant speed. The coefficient of kinetic fric

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Answer:

The force must he apply to the sled is of F= 764.4 N.

Explanation:

m= 60 kg

g= 9.8 m/s²

μ=0.3

W= m*g

W= 588 N

Fr= μ*W

Fr= 176.4 N

F= W + Fr

F= 764.4 N

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The impact that an earthquake can have on an area?

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2 years ago

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Having aced your Physics 2111 class, you get a sweet summer-job working in the International Space Station. Your room-mate, Cosm

Sphinxa [80]

Answer:

a

The speed is $s = 5.857 m/s$

b

The distance is $D = 22.4 \ m$

Explanation:

From the question we are told that

The speed of the banana is $v = 16 \ m/s$

The distance from my location is $d = 8.2 \ m$

The time taken is $t = 1.4 \ s$

The speed of the ice cream is

$s = \frac{d}{t}$

substituting values

$s = \frac{8.4}{1.4}$

$s = 5.857 m/s$

The distance of separation between i and Valdimir is the same as the distance covered by the banana

So

$D = v * t$

substituting values

$D = 16 * 1.4$

$D = 22.4 \ m$

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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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3 years ago