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2 years ago

I am the only non-metal in Group 14. Who am I?

Physics

1 answer:

Nana76 [90]2 years ago

3 0

Answer:

it's either carbon or silicone

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How much power is used by a car engine if it does 48496 J of work in 4 min?

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3 years ago

Read 2 more answers

Which is not a source of electrical energy in the United States

xenn [34]

Every practical source of energy that you can imagine, as well as a few impractical ones, are used somewhere in the USA.

From whale oil in Alaska, to nuclear energy, to coal, petroleum, natural gas, solar energy, wind energy, and biomass.

Oh ! Geothermal energy and tidal energy aren't too popular, but I'll bet if you looked, you'd find these used too, SOMEwhere in the 50 states.

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2 years ago

Jellyfish, earthworms, grasshoppers, and humans are all classified in the same......................

Pepsi [2]

They are all classified in the same kingdom

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2 years ago

Suppose you hear a clap of thunder 14.4 s after seeing the associated lightning strike. The speed of light in air is 3.00 ✕ 108

Marina86 [1]

Answer

Given,

Time to hear the clap = 14.4 s

speed of the light = 3 x 10⁸ m/s

Speed of sound = 343 m/s

a) distance where lightning strike

D = s x t

D = 14.4 x 343

D = 4939.2 m

b) No, we do not need to know the value of speed of light. Because we need to calculate the distance where we hear the sound. To calculate that we need to know the speed of sound.

5 0

3 years ago

A ball is dropped from rest from the top of a cliff that is 24 m high. From ground level, a second ball is thrown straight upwar

sesenic [268]

Answer:

6.0 m below the top of the cliff

Explanation:

We can find the velocity at which the ball dropped from the cliff reaches the ground by using the SUVAT equation

$v^2-u^2 = 2gd$

where

u = 0 (it starts from rest)

g = 9.8 m/s^2 (acceleration of gravity, we assume downward as positive direction)

h = 24 m is the distance covered

Solving for h,

$v=\sqrt{2gh}=\sqrt{2(9.8)(24)}=21.7 m/s$

So the ball thrown upward is launched with this initial velocity:

u = 21.7 m/s

From now on, we take instead upward as positive direction.

The vertical position of the ball dropped from the cliff at time t is

$y_1 = h - \frac{1}{2}gt^2$

While the vertical position of the ball thrown upward is

$y_2 = ut - \frac{1}{2}gt^2$

The two balls meet when

$y_1 = y_2\\h-\frac{1}{2}gt^2 = ut - \frac{1}{2}gt^2 \\h = ut \rightarrow t = \frac{h}{u}=\frac{24}{21.7}=1.11 s$

So the two balls meet after 1.11 s, when the position of the ball dropped from the cliff is

$y_1 = h -\frac{1}{2}gt^2 = 24-\frac{1}{2}(9.8)(1.11)^2=18.0 m$

So the distance below the top of the cliff is

$d=24.0 - 18.0 = 6.0 m$

4 0

3 years ago