Which process in the water cycle do plants perform

Answers

Answer 1

Answer: they preform what is known as transpiration. this is when the roots of the plant extract nutrients along with water out of the soil, this returns some of the water back into the air.

Related Questions

Determine the amount of potential energy of a 5-newton book that is moved to three different shelves on a bookcase. The height of each shelf is 1.0 meter, 1.5 meters, and 2.0 meters.
A pendulum is swinging back and forth with a period of 2.0 seconds here on Earth. This pendulum is then brought to the Moon, where the acceleration due to gravity is much smaller. What will happen to the period of the pendulum, assuming everything else about it (mass, length, initial swing height, etc) remains exactly the same? Explain your answer.
Which occurs during a disequilibrium?
Plateau period are times in life when
Calculate the average force that must be exerted on a 0.145 kg baseball in order to give it an acceleration of 130 m/s2.

Alice and Tom dive from an overhang into the lake below. Tom simply drops straight down from the edge, but Alice takes a running start and jumps with an initial horizontal velocity of 25 m/s. Neither person experiences any significant air resistance. Just as they reach the lake below.A) Alice reaches the surface the lake first. B) Tom of reaches the surface of the first. C) Alice and Tom will reach the surface of the lake at the same time. D) Information is not enough

Answers

Answer:c

Explanation:

Given

Alice launches with horizontal velocity $u=25\ m/s$

Tom simply drops straight down from the edge

Time taken by both the person is same as they have same initial vertical velocity i.e. zero so the time taken to reach the ground is zero.

Although Alice will travel more horizontal distance compared to Tom.

Thus option c is correct

Magnesium is a metal that is commonly used in products that needed to be light weight. Suppose a 2.00kg piece of magnesium has 8160 j of entertainment added to it it's temperature increases 4K what is the specific heat of magnesium

Answers

Given that,

Mass of magnesium, m = 2 kg

Heat added to it, Q = 8160 J

Increase in temperature, $\Delta T=4\ K$

To find,

The specific heat of magnesium.

Solution,

Th formula that is used to find the heat required to raise the temperature in terms of specific heat is given by :

$Q=mc\Delta T\n\nc=(Q)/(m\Delta T)\n\nc=(8160\ J)/(2\ kg* 2\ K)\n\nc=2040\ J/kg-K$

So, the specific heat of magnesium is $2040\ J/kg-K$ .

A bus slows down uniformly from 21 m/s to a complete stop in 21 seconds. How far did the bus travel before stopping?

Answers

$velocity_(final)-velocity_(initial)=(distance)/(time)\n\n v_(initial)=21(m)/(s)\n v_(final)=0(m)/(s)\n time=21s\n\n \Delta v *time=distance\n\n distance=|(0-21)*21|=441m \n\nBus\ traveled\ 441meters.$

A cellist tunes the C string of her instrument to a fundamental frequency of 65.4 Hz . The vibrating portion of the string is 0.600 m long and has a mass of 14.4 g .What is μ, the mass per unit length of the string?
To determine the wave speed from purely kinematic quantities, you need to know the wavelength of the wave. What is the wavelength λ of the fundamental mode in the C string of the cello?

Answers

Explanation:

Given that,

Fundamental frequency of the string, f = 65.4 Hz

Length of the string, l = 0.6 m

Mass, m = 14.4 g = 0.0144 kg

(a) Let $\mu$ is the mass per unit length of the string. It can be calculated as :

$\mu=(m)/(l)$

$\mu=(0.0144\ kg)/(0.6\ m)$

$\mu=0.024\ kg/m$

(b) If f is the fundamental frequency of the string, the wavelength of the fundamental mode is given by :

$l=(n\lambda)/(2)$

$\lambda=(2l)/(n)$

n = 1

$\lambda=2l=2* 0.6\ m$

$\lambda=1.2\ m$

Hence, this is the required solution.

The mass per unit length of the string is 0.024 kg/m, and the wavelength is 1.2 meters.

What is the frequency?

It is defined as the number of waves that crosses a fixed point in one second known as frequency. The unit of frequency is per second.

We have:

Fundamental frequency = 65.4 Hz

Length of the vibrating string portion = 0.6 meter

Mass of the vibrating string portion = 144 grams

We know the formula for mass per unit length:

$\rm \mu = (m)/(l)$

$=\rm (0.0144 \ kg)/(0.6 \ meter)$ ( m = 144 grams ⇒ 0.0144 kg)

$\rm \mu = 0.024 \ kg/m$

The wavelength of the fundamental mode is given by:

$\rm l =(n\lambda)/(2)$

$\rm \lambda = (2l)/(n)$

$\rm \lambda = 2* 0.6 \Rightarrow 1.2 meter$ (n = 1)

Thus, the mass per unit length of the string is 0.024 kg/m, and the wavelength is 1.2 meters.

Learn more about the frequency here:

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A disk-shaped space station 175 m in diameter spins uniformly about an axis perpendicular to the plane of the disk through its center. How many rpm (rev/min) must this disk make so that the acceleration of all points on its rim is g/2?

Answers

The angular velocity of the disk must be 2.25 rpm

Explanation:

The centripetal acceleration of an object in circular motion is given by

$a=\omega^2 r$

where

$\omega$ is the angular velocity

r is the distance of the object from the axis of rotation

For the space station in this problem, we have

$a=(g)/(2)=(9.8)/(2)=4.9 m/s^2$ is the centripetal acceleration

The diameter of the disk is

d = 175 m

So the radius is

$r=(175)/(2)=87.5 m$

So, a point on the rim has a distance of 87.5 m from the axis of rotation. Therefore, we can re-arrange the previous equation to find the angular velocity:

$\omega = \sqrt{(a)/(r)}=\sqrt{(4.9)/(87.5)}=0.237 rad/s$

And this is the angular velocity of any point along the disk. Converting into rpm,

$\omega=0.236 (rad)/(s)\cdot (60 s/min)/(2\pi rad/rev)=2.25 rpm$

Learn more about circular motion:

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Anita is running to the right at 5 m/s , as shown in (Figure 1) . Balls 1 and 2 are thrown toward her by friends standing on the ground. According to Anita, both balls are approaching her at 16 m/s .According to her friends, with what speed was ball 1 thrown?

According to her friends, with what speed was ball 2 thrown?

Answers

Final answer:

In the context of relative velocity, Ball 1 was thrown with a speed of 21 m/s and Ball 2 was thrown with a speed of 11 m/s according to her friends.

Explanation:

The problem presented involves the concept of relative velocity. The speed of the balls relative to Anita is 16 m/s and she is running at a speed of 5 m/s.

If ball 1 is thrown in the same direction as Anita is running, then the friends on ground would see the speed of the ball as the sum of its velocity relative to Anita and the speed of Anita. So, the speed of ball 1 would be 16 m/s + 5 m/s = 21 m/s. Ball 1 was thrown with a speed of 21 m/s according to her friends.

For Ball 2, if it's thrown in the opposite direction to which Anita is running, then according to her friends the speed of Ball 2 would be 16 m/s (speed of the ball relative to Anita) - 5 m/s (Anita's speed) = 11 m/s. Therefore, Ball 2 was thrown with a speed of 11 m/s according to her friends.

Learn more about relative velocity here:

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A 1200 kg car starts from rest and travels 100m in a time of 10 seconds . A) what is the acceleration of the car? B) what force must have acted on the car during this time? C) what is the velocity of the car after 10 seconds? D) what work was done on the car? E) what minimum power did the engine produce?

The Hubble telescope’s orbit is 5.6 × 105 meters above Earth’s surface. The telescope has a mass of 1.1 × 104 kilograms. Earth exerts a gravitational force of 9.1 × 104 Newtons on the telescope. The magnitude of Earth’s gravitational field strength at this location is

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