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2 years ago

7

Imagine you are a water molecule, carbon atom, or nitrogen atom that is inside a cow. Write a brief explanation of how you got f

rom the atmosphere (the air) into the biosphere (the living cow). *

1 answer:

Mandarinka [93]2 years ago

6 0

Answer:

Explanation:

The processes involved in each of the molecules involved are different.

For example, water molecules that have condensed in the atmosphere (in the cloud) return back to the earth through precipitation in the form of rain water. These cows drink these water from various water bodies around them. This is one process in which water molecules move from the atmosphere into the living cow.

In the case of carbon and nitrogen atoms; the air the cow breath in consists of nitrogen, oxygen, carbon dioxide and some other gases. These gas molecules move (from the atmosphere) into the cow through the process of respiration.

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The number of significant figures on the measurement 0.050010 kg id

Alex17521 [72]

Its been some time so i might be wrong but i think the answer is 3 either or 2

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2 years ago

An experiment to measure the speed of light uses an apparatus similar to Fizeau's. The distance between the light source and the

Marina86 [1]

Answer:

$2.88*10^{8} m/s$

Explanation:

The speed of light is given by

$c=\frac {2d}{t}$ and $t=\frac {\theta}{\omega}$ hence

$c=\omega\frac {2d}{\theta}$

Speed of light is given by

$c=900\times 2\pi(\frac {2\times 10}{\frac {2\pi}{2*800}}})=2.88*10^{8} m/s$

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2 years ago

Write the equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and

yKpoI14uk [10]

Answer:

The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

Explanation:

Given that,

The equivalent formulas for velocity, acceleration, and force using the relationships covered for UCM, Newton’s Laws, and Gravitation.

We know that,

Velocity :

The velocity is equal to the rate of position of the object.

$v=\dfrac{dx}{dt}$ ....(I)

Acceleration :

The acceleration is equal to the rate of velocity of the object.

$a=\dfrac{dv}{dt}$ ....(II)

Newton’s second Laws

The force is equal to the change in momentum.

In mathematically,

$F=\dfrac{d(p)}{dt}$

Put the value of p

$F=\dfrac{d(mv)}{dt}$

$F=m\dfrac{dv}{dt}$

Put the value from equation (II)

$F=ma$

This is newton’s second laws.

Gravitational force :

The force is equal to the product of mass of objects and divided by square of distance.

In mathematically,

$F=\dfrac{Gm_{1}m_{2}}{r^2}$

Where, m₁₂ = mass of first object

m= mass of second object

r = distance between both objects

Hence, The newton’s second law is F=ma

The Gravitational force is $F=\dfrac{Gm_{1}m_{2}}{r^2}$

3 0

2 years ago

Gibbons, small Asian apes, move by brachiation, swinging below a handhold to move forward to the next handhold. A 9.4 kg gibbon

Katarina [22]

Answer:

upward force acting = 261.6 N

Explanation:

given,

mass of gibbon = 9.4 kg

arm length = 0.6 m

speed of the swing

net force must provide

$F_{branch} + F_{gravity}=F_{centripetal}$

force of gravity = - mg

$F_{branch}=F_{centripetal}-F_{gravity}$

= $\dfrac{mv^2}{r} + mg$

= $m(\dfrac{3.4^2}{0.6} +9.8)$

=9 x 29.067

= 261.6 N

upward force acting = 261.6 N

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2 years ago

A car traveling at speed v takes distance d to stop after the brakes are applied. What is the stopping distance if the car is in

Vikki [24]

49d

<h3>Further explanation</h3>

This case is about uniformly accelerated motion.

<u>Given:</u>

The initial speed was v takes distance d to stop after the brakes are applied.

<u>Question:</u>

What is the stopping distance if the car is initially traveling at speed 7.0v?

Assume that the acceleration due to the braking is the same in both cases. Express your answer using two significant figures.

<u>The Process:</u>

The list of variables to be considered is as follows.

$\boxed{u \ or \ v_i = initial \ velocity}$
$\boxed{u \ or \ v_t \ or \ v_i = terminal \ or \ final \ velocity}$
$\boxed{a = acceleration \ (constant)}$
$\boxed{d = distance \ travelled}$

The formula we follow for this problem are as follows:

$\boxed{ \ v^2 = u^2 + 2ad \ }$

a = acceleration (in m/s²)
u = initial velocity
v = final velocity
d = distance travelled

Step-1

We substitute v as the initial speed, distance of d, and zero for final speed into the formula.

$\boxed{ \ 0 = v^2 + 2ad \ }$

$\boxed{ \ v^2 = -2ad \ }$

Both sides are divided by -2d, we get $\boxed{ \ a = \Big( -\frac{v^2}{2d} \Big) \ . . . \ (Equation-1) \ }$

Step-2

We substitute 7.0v as the initial speed, zero for final speed, and Equation-1 into the formula.

$\boxed{ \ 0 = (7.0v)^2 + 2 \Big( -\frac{v^2}{2d} \Big)d' \ }$

Here d' is the stopping distance that we want to look for.

$\boxed{ \ 2 \Big( \frac{v^2}{2d} \Big)d' = (7.0v)^2 \ }$

We crossed out 2 in above and below.

$\boxed{ \ \Big( \frac{v^2}{d} \Big)d' = 49.0v^2 \ }$

We multiply both sides by d.

$\boxed{ \ v^2 d' = 49.0v^2 d \ }$

We crossed out v^2 on both sides.

$\boxed{\boxed{ \ d' = 49.0d \ }}$

Hence, by using two significant figures, the stopping distance if the car is initially traveling at speed 7.0v is 49d.

<h3>Learn more</h3>

Determine the acceleration of the stuffed bear brainly.com/question/6268248
Particle's speed and direction of motion brainly.com/question/2814900
About the projectile motion brainly.com/question/2746519

Keywords: a car traveling at speed v, takes distance d to stop after the brakes are applied, the stopping distance, if the car is initially traveling at speed 7.0v, the acceleration due to the braking is the same, two significant figures.

6 0

2 years ago

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