Plz help will give brainliest and ten points

You carry a fire hose up a ladder to a height of 10 m above ground level and aim the nozzle at a burning roof that is 9 m high.

schepotkina [342]

Answer:

The value is $P_1 = 314645 \ Pa$

Explanation:

From the question we are told that

The height is $h_2 = 10 m$

The height of the burning roof is $k = 9 m$

The horizontal distance is $d = 7 \ m$

The height of the truck is $h_1 = 0.5 \ m$

Generally the time for the water to hit the roof from the hose is mathematically represented as

$t = \sqrt{\frac{2 * (h_2 - k)}{g} }$

=> $t = \sqrt{\frac{2 * (10 - 9)}{9.8} }$

=> $t = 0.4518 \ s$

Generally the velocity of the water is mathematically evaluated as

$v_2 = \frac{d}{t}$

$v_2 = \frac{ 7}{0.4518}$

$v_2 = 15.5 \ m/s$

Generally from Bernoulli's Equation we have that

$P_1 + \frac{1}{2} v_1^2 * \rho + \rho *g *h_1 = P_2 + \frac{1}{2} v_2^2 * \rho + \rho *g *h_2$

Here $P_1 [\tex] is pressure in the chamber which we are to calculate , [tex]P_2 [\tex] is the atmospheric pressure with value [tex]P_2 = 101325 \ Pa [\tex] , [tex]v_1 [\tex] is the velocity of the water before it starts flowing with value [tex]v_1 = 0 m/s [\tex] , [tex]\rho [\tex] is the density of water with value [tex]\rho = 1000 \ kg/m^3 [\tex] So [tex]P_1 + \frac{1}{2} 0^2 * 1000 + 1000 *9.81 *0.5 = 101325 + \frac{1}{2}* 15.5^2* 1000 + 1000 *9.81 *10$

$P_1 = 314645 \ Pa$

8 0

3 years ago

What is the difference between Drift current & Diffusion current ?

dsp73

The main difference between the drift current and diffusion current is the corresponding reaction of the semiconductor materials in the electric fields. In electric fields, when semiconductor material's movement go along with the direction of the electric field it is called drift current, but when it tends to go from high concentrated area to lower concentrated area then it is diffusion current.

8 0

3 years ago

How are frequency and were peroid related?

hodyreva [135]

Answer:

Frequency and period are distinctly different, yet related, quantities. Frequency refers to how often something happens. Period refers to the time it takes something to happen. Frequency is a rate quantity.

7 0

2 years ago

All else equal, the payback period for a project will decrease whenever the_______.

telo118 [61]

Answer:

(B) cash inflows are moved earlier in time.

Explanation:

The payback period stated time-frame during which the initial amount of investment should be recovered. It is expressed in the year form

The formula to compute the payback period is shown below:

Payback period = Initial investment ÷ Net cash flow

where,

The net cash flow = annual net operating income + depreciation expenses

The payback period of the project decreases when the accumulated starting year cash flows increases that results the movement of the cash inflows earlier in time

3 0

3 years ago

Read 2 more answers

Consider the classic problem with holiday lights, one little bulb goes out and the whole string goes out. First consider a strin

Flura [38]

Answer:

a. 0.33 Amp

b. 2.4 Volt

c. 0

d. 2.45 Amp

e. Infinite

f. Series is safer

Explanation:

Series Connection of Resistors

When two or more resistors are connected in series, the current through each one of them is the same, and the voltage divides depending on the particular value of each resistance. If all the resistances are equal, then the voltage is equally divided.

a. The string of 50 bulbs is connected to a 120 VAC outlet and consumes 40 W. The power of a circuit is given by

$P=V.I$

Solving for I

$\displaystyle I=\frac{P}{V}=\frac{40}{120}=0.33\ Amp$

Since all the bulbs are connected in series the current is the same for all of them.

b. The voltage is equally divided, so each bulb has 120/50= 2.4 V

c. If one of the bulbs burns out and its resistance becomes infinite, then the series circuit is open and no current flows through it, neither through the rest of the bulbs. The typical case of the whole string going out.

d. If one of the bulbs short circuits, the resistance of that bulb is zero and the voltage is distributed by the 49 remaining bulbs. Thus the new current is

$\displaystyle I=\frac{V}{R}=\frac{120}{49}=2.45\ A$

e. If the bulbs were connected in parallel, all of them would have the same voltage, and the total current will be equally divided among them. In that case, a short circuit in one of the bulbs will cause a parallel short, theoretically producing an infinite current and making the short circuit protection blow up.

f. The condition described above makes the strings be made of series-connected bulbs which is safer than the parallel circuit. If a single bulb shorts, the entire string goes out in a series connection, but the breaker would trigger disconnection of the house circuit if it's a parallel connection. That is why we must deal with unusable strings instead of burning cables.

6 0

3 years ago