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3 years ago

How can flowers or flower products be sold or distributed ?

Physics

2 answers:

Dmitrij [34]3 years ago

6 0

Answer:

by marketing the product

kondaur [170]3 years ago

3 0

Answer:

bfhdjfndjckdkfmkfkckfj

Explanation:

hdjdkckfkvkfkfkxkfkck ex XD FB fact trees

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Coherent light with wavelength 599 nm passes through two very narrow slits with separation of 20 μm, and the interference patter

goblinko [34]

Answer:

134.77 mm

Explanation:

Wave length of light λ = 599 x 10⁻⁹ m

Slit separation d = 20 x 10⁻⁶ m

Screen distance D = 3 m

Distance of second dark fringe from centre

= 1.5 x λ D / d

Putting the values given above

distance = $\frac{1.5\times599\times10^{-9}\times 3}{20\times10^{-6}}$

= 134.77 x 10⁻³ m

= 134.77 mm.

7 0

4 years ago

1. The volume of a given mass of gas is 20cm when its

Pavel [41]

Answer:

Explanation:

The way to show a cubed substance is either like this³ or like this x^3. The small three is found at the bottom toolbar at the bottom of the question space marked by the Ω symbol.

100 mmHg

Givens

V1 = 20 cm^3

V2 = 80 cm^3

P1 = 400 mmHg

P2 = ?

Formula

V1 * P1 = V2 * P2

Solution

20 * 400 = 80 * P2 Divide by 80

20 * 400/80 = P2

P2 = 8000 / 80

P2 = 100 mmHg

5 0

3 years ago

A dog running at 10 m/s is 30m behind a rabbit moving at 5 m/s. when will the dog catch up with the rabbit assuming both their v

dedylja [7]

The will dog catch up with the rabbit in 6 minutes assuming both their velocities remain constant during the chase.

<h3>What time will the dog catch the rabbit?</h3>

The time that the dog will catch up with the rabbit is given as follows:

Let the distance covered by the rabbit be x.

Distance covered by dog = x + 30

Time taken = distance/speed

The time taken will be the same T

Time taken by dog, T = (x + 30)/10
Time taken by rabbit, T = x/5

Equating both times.

(x + 30)/10 = x/5

x = 30 m

Solving for T in equation (ii);

T = 30/5 = 6 minutes

In conclusion, time is obtained as a ratio of distance and speed.

Learn more about time and speed at: brainly.com/question/26046491

#SPJ1

7 0

2 years ago

Find the quantity of heat needed

krok68 [10]

Answer:

Approximately $3.99\times 10^{4}\; \rm J$ (assuming that the melting point of ice is $0\; \rm ^\circ C$ .)

Explanation:

Convert the unit of mass to kilograms, so as to match the unit of the specific heat capacity of ice and of water.

$\begin{aligned}m&= 100\; \rm g \times \frac{1\; \rm kg}{1000\; \rm g} \\ &= 0.100\; \rm kg\end{aligned}$

The energy required comes in three parts:

Energy required to raise the temperature of that $0.100\; \rm kg$ of ice from $(-10\; \rm ^\circ C)$ to $0\; \rm ^\circ C$ (the melting point of ice.)
Energy required to turn $0.100\; \rm kg$ of ice into water while temperature stayed constant.
Energy required to raise the temperature of that newly-formed $0.100\; \rm kg$ of water from $0\; \rm ^\circ C$ to $10\;\ rm ^\circ C$ .

The following equation gives the amount of energy $Q$ required to raise the temperature of a sample of mass $m$ and specific heat capacity $c$ by $\Delta T$ :

$Q = c \cdot m \cdot \Delta T$ ,

where

$c$ is the specific heat capacity of the material,
$m$ is the mass of the sample, and
$\Delta T$ is the change in the temperature of this sample.

For the first part of energy input, $c(\text{ice}) = 2100\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (0\; \rm ^\circ C) - (-10\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_1 &= c(\text{ice}) \cdot m(\text{ice}) \cdot \Delta T\\ &= 2100\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 2.10\times 10^{3}\; \rm J\end{aligned}$ .

Similarly, for the third part of energy input, $c(\text{water}) = 4200\; \rm J \cdot kg \cdot K^{-1}$ whereas $m = 0.100\; \rm kg$ . Calculate the change in the temperature:

$\begin{aligned}\Delta T &= T(\text{final}) - T(\text{initial}) \\ &= (10\; \rm ^\circ C) - (0\; \rm ^\circ C) \\ &= 10\; \rm K\end{aligned}$ .

Calculate the energy required to achieve that temperature change:

$\begin{aligned}Q_3&= c(\text{water}) \cdot m(\text{water}) \cdot \Delta T\\ &= 4200\; \rm J \cdot kg \cdot K^{-1} \\ &\quad\quad \times 0.100\; \rm kg \times 10\; \rm K\\ &= 4.20\times 10^{3}\; \rm J\end{aligned}$ .

The second part of energy input requires a different equation. The energy $Q$ required to melt a sample of mass $m$ and latent heat of fusion $L_\text{f}$ is:

$Q = m \cdot L_\text{f}$ .

Apply this equation to find the size of the second part of energy input:

$\begin{aligned}Q_2&= m \cdot L_\text{f}\\&= 0.100\; \rm kg \times 3.36\times 10^{5}\; \rm J\cdot kg^{-1} \\ &= 3.36\times 10^{4}\; \rm J\end{aligned}$ .

Find the sum of these three parts of energy:

$\begin{aligned}Q &= Q_1 + Q_2 + Q_3 = 3.99\times 10^{4}\; \rm J\end{aligned}$ .

3 0

3 years ago

The Kentucky Derby, the first leg of horse racing’s Triple Crown, was won by a time of 122.2 s. If the race covers 2011.25 m, wh

kirza4 [7]

The answer is 789.25 which you’d subtract 2011.25-122.2 I think sry if I’m wrong

3 0

3 years ago

Read 2 more answers

How can flowers or flower products be sold or distributed ?​

How can flowers or flower products be sold or distributed ?