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3 years ago

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Which of the following elements is commonly found in the Earth's crust, living matter, oceans, and atmosphere? A. carbon B. argo

n C. bromine D. zinc

2 answers:

jekas [21]3 years ago

8 0

The answer is carbon.

OverLord2011 [107]3 years ago

5 0

I'm pretty sure it's A. carbon

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One way to monitor global warming is to measure the average temperature of the ocean. Researchers are doing this by measuring th

Marat540 [252]

Answer:

0.07°C

Explanation:

<u>solution:</u>

the speed of a sound in water is<u>:</u>

v(T)=1480+4(T-4°C)

<u>at 4°C the travel time is:</u>

t(4◦C) = ( 7600 × 103 m ) / (1480 m/s) = 5202.7 s

<u>5°C, the travel time is:</u>

t(5◦C) = ( 7600 × 103 m ) / (1484 m/s) = 5188.7 s

<u>one degree C corresponds to a ∆t of 14 s so temperature difference is:</u>

ΔT=1 s/14 s=0.07◦C

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3 years ago

Help me please! Very much appreciated :)<br><br> Why are standards of mass necessary?

tia_tia [17]

Answer:

A measurement standard is a quantity that people agree to use as a comparison. Standards are important because they allow measurements to be compared even if different people in different parts of the world take them.

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6 0

2 years ago

Read 2 more answers

A 2000 kg roller coaster is at the top of a loop with a radius of 24 m. If its speed is 18 m/s at this point, what force does it

levacccp [35]

Answer:

$46620\ \text{N}$

Explanation:

m = Mass of roller coaster = 2000 kg

r = Radius of loop = 24 m

v = Velocity of roller coaster = 18 m/s

g = Acceleration due to gravity = $9.81\ \text{m/s}^2$

Normal force at the point will be

$N-mg=\dfrac{mv^2}{r}\\\Rightarrow N=\dfrac{mv^2}{r}+mg\\\Rightarrow N=\dfrac{2000\times 18^2}{24}+2000\times 9.81\\\Rightarrow N=46620\ \text{N}$

The force exerted on the track is $46620\ \text{N}$ .

8 0

3 years ago

Determine one way you can contribute to water in the atmosphere in your day-to-day activities pleaseeeee helppp

TEA [102]

Answer:

agricultural production of food

Explanation:

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3 years ago

Before colliding, the momentum of block A is -100 kg*m/, and block B is -150 kg*m/s. After, block A has a momentum -200 kg*m/s.

rjkz [21]

Answer:

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

Explanation:

Given

Before collision:

Momentum of block A = $p_{A1}$ = $-100\ kg\ ms^{-1}$

Momentum of block B = $p_{B1}$ = $-150\ kg\ ms^{-1}$

After collision:

Momentum of block A = $p_{A2}$ = $-200\ kg\ ms^{-1}$

Applying law of conservation of momentum to find momentum of block B after collision $p_{B2}$ .

$p_{A1}+p_{B1}=p_{A2}+p_{B2}$

Plugging in the given values and simplifying.

$-100-150=-200+p_{B2}$

$-250=-200+p_{B2}$

Adding 200 to both sides.

$200-250=-200+p_{B2}+200$

$-50=p_{B2}$

∴ $p_{B2}=-50\ kg\ ms^{-1}$

Momentum of block B after collision = $-50\ kg\ ms^{-1}$

6 0

3 years ago