How do water and minerals get to each cell in a plant?

A 0.20 mass on a horizontal spring is pulled back a certain distance and released. The maximum speed of the mass is measured to

olga55 [171]

Answer:

d

Explanation:

Ya gon find the Kenitic Energy first

K=½mv²===> K=½×0.2×(0.2)²===> 0.1(0.04)===> 0.004

and now the replacement:

0.004=½×0.4V²====> v²=0.02===> V=0.14m/s

4 0

3 years ago

In the first stage of a two-stage Carnot engine, energy is absorbed as heat Q1 at temperature T1 = 500 K, work W1 is done, and e

Nataly [62]

Answer:

Efficiency = 52%

Explanation:

Given:

First stage

heat absorbed, Q₁ at temperature T₁ = 500 K

Heat released, Q₂ at temperature T₂ = 430 K

and the work done is W₁

Second stage

Heat released, Q₂ at temperature T₂ = 430 K

Heat released, Q₃ at temperature T₃ = 240 K

and the work done is W₂

Total work done, W = W₁ + W₂

Now,

The efficiency is given as:

$\eta=\frac{\textup{Total\ work\ done}}{\textup{Energy\ provided}}$

or

Work done = change in heat

thus,

W₁ = Q₁ - Q₂

W₂ = Q₂ - Q₃

Thus,

$\eta=\frac{(Q_1-Q_2)\ +\ (Q_2-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_1-Q_3)}{Q_1}}$

or

$\eta=1-\frac{(Q_3)}{Q_1}}$

also,

$\frac{Q_1}{T_1}=\frac{Q_2}{T_2}=\frac{Q_3}{T_3}$

or

$\frac{T_3}{T_1}=\frac{Q_3}{Q_1}$

thus,

$\eta=1-\frac{(T_3)}{T_1}}$

thus,

$\eta=1-\frac{(240\ K)}{500\ K}}$

or

$\eta=0.52$

or

Efficiency = 52%

8 0

3 years ago

A 0.5 kg cheeseburger is lobbed at a particularly unhappy customer with a force of 10 N.

aleksley [76]

The acceleration that the cheeseburger experienced is 20 m/s^2.

6 0

2 years ago

Estimate the mass of the Great Pyramid of Giza, in tons. You make may use of the following information: the Great Pyramid is in

postnew [5]

Answer:

6005803.83105 short tons

Explanation:

The definition of density is $\rho = \frac{m}{V}$ , and the volume of a pyramid is (confusingly written on the proposal) $V=\frac{1}{3} Ah$ , so we can write:

$m=\rho V=\rho V \frac{1}{3} Ah=\rho V \frac{1}{3} s^2h$

Where s is the side of the base, being $s^2$ the area of that square.

We will write everything in S.I., and the best way to convert units is using conversion factors, for example, since 1m=100cm, we know that $\frac{1m}{100cm}=1$ , and we can use this factor to convert anything written in cm to anything written in m. Example:

$500cm=500cm\frac{1m}{100cm}=5m$

Here we just multiplied 500cm by something that is equal to 1 (as every conversion factor must), so it's not doing anything but changing the units.

We can use this tool like this:

$2.1\frac{g}{cm^3}=2.1\frac{g}{cm^3}(\frac{1Kg}{1000g})(\frac{100cm}{1m})^3=2100Kg/m^3$

Where we have used the fact that $1^3=1$ (we can elevate any conversion factor to any number and they still will be 1) and where we have placed strategically what is the numerator and what in the denominator so the units we don't want cancel out and the units we want appear.

Substituting then our values:

$m=\rho V \frac{1}{3} s^2h=(2100Kg/m^3)\frac{1}{3} (230.34m)^2(146.7m)=5448373586.96Kg$

And now we will convert to short tons using two conversion factors at the same time:

$m=5448373586.96\ Kg(\frac{1\ lb}{0.45359237\ Kg})(\frac{1\ short\ ton}{2000\ lb} )=6005803.83105\ short \ tons$

Remember, their value is 1, and we place the units to cancel the ones we don't want and keep the ones we want, here Kg cancel out, and lb cancel out, leaving the short tones.

8 0

3 years ago

Consider two diffraction gratings. One grating has 3000 lines per cm, and the other one has 6000 lines per cm. Both gratings are

Usimov [2.4K]

Answer:

The 6000 lines per cm grating, will produces the greater dispersion .

Explanation:

A diffraction grating is an optical component with a periodic (usually one that has ridges or rulings on their surface rather than dark lines) structure that splits and diffracts light into several beams travelling in different directions.

The directions of the light beam produced from a diffraction grating depend on the spacing of the grating, and also on the wavelength of the light.

For a plane diffraction grating, the angular positions of principle maxima is given by

(a + b) sin ∅n = nλ

where

a+b is the distance between two consecutive slits

n is the order of principal maxima

λ is the wavelength of the light

From the equation, we can see that without sin ∅ exceeding 1, increasing the number of lines per cm will lead to a decrease between the spacing between consecutive slits.

In this case, light of the same wavelength is used. If λ and n is held constant, then we'll see that reducing the distance between two consecutive slits (a + b) will lead to an increase in the angle of dispersion sin ∅. So long as the limit of sin ∅ not greater that one is maintained.

7 0

3 years ago