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3 years ago

13

Think about a good game story that made you feel a mix of positive and negative emotions. What was the story, what emotions did

you feel, and how did it make you feel them?

1 answer:

Margarita [4]3 years ago

3 0

Answer:

dead in the inside

Explanation:

it made me feel sad and mad at the same time I am sleepy

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Adam has 5 barrels of different volumes. The barrel of which volume can contain the most liquid?

Natalija [7]

Answer:

D

Explanation:

5 0

2 years ago

Steep safety ramps are built beside mountain highways to enable vehicles with defective brakes to stop safely. A truck enters a

Virty [35]

Answer:

a) The additional time required for the truck to stop is <u>8.5 seconds</u>

b) The additional distance traveled by the truck is <u>230.05 ft</u>

Explanation:

Since the acceleration is constant, the average speed is:

(final speed - initial speed) / 2 = 0.75 v0

Since travelling at this speed for 8.5 seconds causes the vehicle to travel 690 ft, we can solve for v0:

0.75v0 * 8.5 = 690

v0 = 108.24 ft/s

The speed after 8.5 seconds is: 108.24 / 2 = 54.12 ft/s

We can now use the following equation to solve for acceleration:

$v^2 - u^2 = 2*a*s$

$54.12^2 - 108.24^2 = 2*a*690$

a = -6.367 m/s^2

Additional time taken to decelerate: 54.12/6.367 = 8.5 seconds

Total distance traveled:

$v^2 - u^2 = 2*a*s$

0 - 108.24^2 = 2 * (-6.367) * s

solving for s we get total distance traveled = 920.05 ft

Additional Distance Traveled: 920.05 - 690 = 230.05 ft

5 0

3 years ago

A power of 100 kW (105 W) is delivered to the other side of a city by a pair of power lines, between which the voltage is 12,000

OLga [1]

Answer:

I = 8.3 Amp

potential drop = 83 V

Explanation:

Power = 100 KW

V = 12,000 V

R = 10 ohms

a)

Calculate current I in each wire:

P = I*V

I = P / V

I = 100 / 12 = 8.333 A

b)

Calculate potential drop in each wire:

V = I*R

V = (8.3) * (10)

V = 83 V

4 0

2 years ago

Air in a large tank at 300C and 400kPa, flows through a converging diverging nozzle with throat diameter 2cm. It exits smoothly

-Dominant- [34]

Answer:

The answer is " $3.74 \ cm\ \ and \ \ 0.186 \frac{kg}{s}$ "

Explanation:

Given data:

Initial temperature of tank $T_1 = 300^{\circ}\ C= 573 K$

Initial pressure of tank $P_1= 400 \ kPa$

Diameter of throat $d* = 2 \ cm$

Mach number at exit $M = 2.8$

In point a:

calculating the throat area:

$A*=\frac{\pi}{4} \times d^2$

$=\frac{\pi}{4} \times 2^2\\\\=\frac{\pi}{4} \times 4\\\\=3.14 \ cm^2$

Since, the Mach number at throat is approximately half the Mach number at exit.

Calculate the Mach number at throat.

$M*=\frac{M}{2}\\\\=\frac{2.8}{2}\\\\=1.4$

Calculate the exit area using isentropic flow equation.

$\frac{A}{A*}= (\frac{\gamma -1}{2})^{\frac{\gamma +1}{2(\gamma -1)}} (\frac{1+\frac{\gamma -1}{2} M*^2}{M*})^{\frac{\gamma +1}{2(\gamma -1)}}$

Here: $\gamma$ is the specific heat ratio. Substitute the values in above equation.

$\frac{A}{3.14}= (\frac{1.4-1}{2})^{-\frac{1.4+1}{2(1.4 -1)}} (\frac{1+\frac{1.4-1}{2} (1.4)^2}{1.4})^{\frac{1.4+1}{2(1.4-1)}} \\\\A=\frac{\pi}{4}d^2 \\\\10.99=\frac{\pi}{4}d^2 \\\\d = 3.74 \ cm$

exit diameter is 3.74 cm

In point b:

Calculate the temperature at throat.

$\frac{T*}{T}=(1+\frac{\Gamma-1}{2} M*^2)^{-1}\\\\\frac{T*}{573}=(1+\frac{1.4-1}{2} (1.4)^2)^{-1}\\\\T*=411.41 \ K$

Calculate the velocity at exit.

$V*=M*\sqrt{ \gamma R T*}$

Here: R is the gas constant.

$V*=1.4 \times \sqrt{1.4 \times 287 \times 411.41}\\\\=569.21 \ \frac{m}{s}$

Calculate the density of air at inlet

$\rho_1 =\frac{P_1}{RT_1}\\\\=\frac{400}{ 0.287 \times 573}\\\\=2.43\ \frac{kg}{m^3}$

Calculate the density of air at throat using isentropic flow equation.

$\frac{\rho}{\rho_1}=(1+\frac{\Gamma -1}{2} M*^2)^{-\frac{1}{\Gamma -1}} \\\\\frac{\rho *}{2.43}=(1+\frac{1.4-1}{2} (1.4)*^2)^{-\frac{1}{1.4-1}} \\\\\rho*= 1.045 \ \frac{kg}{m^3}$

Calculate the mass flow rate.

$m= \rho* \times A* \times V*\\\\= 1.045 \times 3.14 times 10^{-4} \times 569.21\\\\= 0.186 \frac{kg}{s}$

5 0

2 years ago

Explain the following statement: In a windowless proportional counter, the output pulse height from an alpha particle source wil

-BARSIC- [3]

Answer / Explanation:

Regarding α and β Particles in Windowless Counter, the range of α particles is lower than β particles. Alpha particles typically have range less than the dimensions of the gas chamber so that proportional counters are able to easily record. Hence, with almost 100% efficiency, each particle which enters the so called active volume.

Then, since the pulse height spectra depends on the number of ion pairs which have formed, an aplha particle with higher energy creates more ion pairs in the chamber. However, beta particle range usually exceeds the dimensions of the chamber and therefore most of the betas hit the walls where they deposit energy. Then, fewer ion pairs are formed because very few β’s give their energy to the bulk gas.

5 0

3 years ago