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Rama09 [41]
3 years ago
8

(a) Calculate the absolute pressure at the bottom of a fresh- water lake at a depth of 27.5 m. Assume the density of the water i

s 1.00 3 103 kg/m3 and the air above is at a pressure of 101.3 kPa. (b) What force is exerted by the water on the window of an underwater vehicle at this depth if the window is circular and has a diameter of 35.0 cm?

Physics
2 answers:
mart [117]3 years ago
6 0

Answer:

(a) The absolute pressure is 371.6 kPa

(b) The force exerted by the water is 26.0 kN

Explanation: Please see the attachments below

ddd [48]3 years ago
5 0

Answer:

a) P = 370.993\,kPa, b) F = 25.948\,kN

Explanation:

a) The absolute pressure at a depth of 27.5 meters is:

P = P_{atm} + P_{man}

P = 101.3\,kPa + \left(1000\,\frac{kg}{m^{3}}\right)\cdot \left(9.807\,\frac{m}{s^{2}} \right)\cdot (27.5\,m)\cdot \left(\frac{1\,kPa}{1000\,Pa} \right)

P = 370.993\,kPa

b) The force exerted by the water is:

F = (P - P_{atm})\cdot A

F = (370.993\,kPa-101.3\,kPa)\cdot \left(\frac{\pi}{4} \right)\cdot (0.35\,m)^{2}

F = 25.948\,kN

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Aluminum oxide can be produced during rocket launches. Show that the sum of positive and negative charges in a unit of Al2O3 equ
Triss [41]

Answer:

The sum of positive and negative charges in a unit of Al2O3 equals zero.

Aluminium has a charge of +3 while Oxygen has a charge of -2 on each ion.

Al203 has 2 Al atoms and 3 O atoms.

Charge on Al2O3 = 2(charge on Al ion) + 3(charge on O ion)

= 2(3) + 3(-2)

= 6 - 6

= 0

Explanation:

Aluminium has 3 electrons in the outermost shell and has the tendency to lose those 3 electrons to form a positive ion and have a complete outermost shell.

Whereas, Oxygen has 6 electrons in the outermost and has the tendency to accept two more electrons to form a negative ion and have a complete outermost shell.

4 0
3 years ago
3.1
Ilia_Sergeevich [38]

Answer:

The value is  C = 30729\  c

Explanation:

From the question we are told that

  The power rating of the stove is  P = 4.4 KW  = 4.4 *10^{3} \  W

   The duration of its use everyday is t_1  = 70 \ minutes  = 1.167 \ hours    

    The rating of the light bulbs is P_2'  = 150 W

   The number is n = 7

   The power rating of the total  bulb is  P_2 = 7 * 150 = 1050 \  W

    The duration of its use everyday is  t_2  = 7 hours

    The power rating of miscellaneous appliance P_3 = 1.8 \ KW = 1.8 *10^{3} \  W

    The duration of its use everyday is t_3 = 1 hour

     The power rating of hot water P = 4 KW  = 4 *10^{3} \  W

      The duration of its use everyday is t_4 = 120 \ minutes  = 2 hours

Generally the total electrical energy used in 1 month is mathematically represented as

    E = P_1 * t_1 * 30 + P_2 * t_2 * 30 + P_3 * t_3 * 30+ P_4 * t_4 * 30

=>  E = 4.4*10^{3} * 1.167  * 30 + 1050 * 7 * 30 + 1.8 *1 * 30+ 4*10^{3} * 2 * 30

=>  E = 614598  \ W \cdot h

=>   E = 614.6  \  K W \cdot h

Generally the monthly electricity  bill is mathematically represented as

    C = 614.6 * 50

=> C = 30729\  c

8 0
3 years ago
When the Glen Canyon hydroelectric power plant in Arizona is running at capacity, 690 m3 of water flows through the dam each sec
bixtya [17]

Answer:

1340.2MW

Explanation:

Hi!

To solve this problem follow the steps below!

1 finds the maximum maximum power, using the hydraulic power equation which is the product of the flow rate by height by the specific weight of fluid

W=αhQ

α=specific weight for water =9.81KN/m^3

h=height=220m

Q=flow=690m^3/s

W=(690)(220)(9.81)=1489158Kw=1489.16MW

2. Taking into account that the generator has a 90% efficiency, Find the real power by multiplying the ideal power by the efficiency of the electric generator

Wr=(0.9)(1489.16MW)=1340.2MW

the maximum possible electric power output is 1340.2MW

3 0
3 years ago
If the distance to a distant galaxy is recorded as 10,000 light years, then light must have been traveling for 5000 years to rea
satela [25.4K]
False false false false
5 0
3 years ago
Read 2 more answers
Tyrone, an 85-kg teenager, runs off the end of a horizontal pier and lands on a free-floating 130-kg raft that was initially at
Ghella [55]

Answer:u=4.04 m/s

Explanation:

Given

Mass m=85 kg

mass of Raft M=130 kg

velocity of raft and man  v=1.6 m/s

Let initial speed of Tyrone is u

Conserving Momentum as there is no external Force

mu=(M+m)v

85\times u=(85+130)\cdot 1.6

u=\frac{215}{85}\cdot 1.6

u=2.529\cdot 1.6=4.04 m/s  

4 0
3 years ago
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