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3 years ago

In the chemical equation above, the small number after the O in 1202 represent —

Physics

1 answer:

Murrr4er [49]3 years ago

7 0

Answer:

G.) The number atoms of that element in the molecule

Explanation:

F is incorrect because the coefficient represents the amount of one type of molecule, not the subscript

G is correct because subscripts represent how many atoms of that element are present in that single molecule

H is incorrect because energy is not represented in this simple type of equation

J is incorrect because it doesn't even make sense

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What is the momentum of a 50 at a speed of 5m/s

aleksandr82 [10.1K]

50*5=250
Momentum will be 250kgm/s^2

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3 years ago

Amber is a (a) metal (b) rubber (c) resin (d) sugar

zlopas [31]

Answer:

c. Resin.

Explanation:

Amber is a resin.

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2 years ago

Ethan and Rebecca are riding on a merry-go-round. Ethan rides on a horse at the outer rim of the circlar platform, twice as far

xz_007 [3.2K]

Answer:

Ethan's tangential speed is twice as Rebecca's.

Explanation:

Let $r$ be the distance from the center of the platform to Rebecca's horse, $v_R$ Rebecca's tangential speed, $v_E$ the Ethan's tangential speed and $\omega$ the merry-go-round angular speed. The distance from the center of the platform to Ethan's horse is $2r$ . The angular speed is related to the tangential speed by the equation:

$\omega =\frac{v}{R}$

Since the angular speed is the same for Ethan and Rebecca, we have that:

$\frac{v_E}{2r}=\frac{v_R}{r}$

Now, solving for $v_E$ we get:

$v_E=2v_R$

It means that Ethan's tangential speed is twice as Rebecca's.

3 0

3 years ago

A biker pedals hard for 3 seconds. What is his initial velocity if he accelerated by 4m/s2 until he's going 20m/s. (Which equati

____ [38]

Answer:

answer is option 4

Explanation:

you have to use option 4 because u need to find out initial velocity (Vi)

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3 years ago

A certain frictionless simple pendulum having a length L and mass M swings with period T. If both L and M are doubled, what is t

Mama L [17]

Answer:

c. √2T

Explanation:

The period of a simple pendulum is given by;

$T = 2\pi \sqrt{\frac{L}{g} } \\\\\frac{T}{2\pi} = \sqrt{\frac{L}{g}} \\\\\frac{T^2}{4\pi^2} = \frac{L}{g}\\\\\frac{g}{4\pi^2} = \frac{L}{T^2}\\\\ \frac{L_1}{T_1^2}= \frac{L_2}{T_2^2}\\\\T_2^2 = \frac{L_2T_1^2}{L_1}\\\\L_2 = 2L_1\\\\ T_2^2 = \frac{2L_1T_1^2}{L_1}\\\\ T_2^2 =2T_1^2\\\\T_2 = \sqrt{2T_1^2}\\\\T_2 = T_1\sqrt{2}$

Thus, the the new period will be √2T

4 0

3 years ago