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const2013 [10]
3 years ago
13

A closed container is filled with oxygen. The pressure in the container is 245 kPa . What is the pressure in millimeters of merc

ury? Express the pressure numerically in millimeters
Physics
1 answer:
Makovka662 [10]3 years ago
6 0

Answer:

Answer to the question is: 1837.65 millimeters of mercury are equal to 245 kPa.

Explanation:

1 kPa are equal to 7.50062 millimeters of mercury.

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In----- theory, the actual wishes that are threatening to the dreamers conscious awareness are disguised
Mandarinka [93]

Answer:

Unconscious wish fulfillment

Explanation:

Unconscious wish fulfillment

(In the unconscious wish fulfillment theory, the actual wishes of the dreamer that are threatening to the dreamer's conscious awareness are disguised.)

7 0
3 years ago
A 1460-kg submarine rises straight up towards the surface. Seawater exerts both an upward buoyant force of 16670 N on the submar
Amiraneli [1.4K]

Answer:

a = 0.63 m/s²

Explanation:

given,

mass of submarine = 1460-kg

upward buoyant force = 16670 N

downward resistive force = 1150 N

submarine acceleration = ?

assuming g = 10 m/s²

now,

B - (R + mg) = ma

16670 - 1150 - 1460 × 10 = 1460 × a

1460× a = 920

a = 0.63 m/s²

hence, the acceleration of submarine is equal to a = 0.63 m/s²

5 0
3 years ago
While seeds are forming, another plant part forms around them. Animals eat this part and carry the seeds to a new spot.
Eddi Din [679]

Answer:

<em>The correct option is D) fruit</em>

Explanation:

Both the gymnosperm and the angiosperms plants produce seed. Seeds are an essential part of reproduction in plants.

After fertilization has occurred, the ripened ovule is the part which turns into a seed. The size of the seeds depends on the type if plant.

After fertilization, the ovary forms the fruit. As the ovule is present in the ovule hence after fertilization, the seed which was made from the ovule gets enclosed in the fruit which was made from the ovary.

6 0
3 years ago
4.77 Augment the rectifier circuit of Problem 4.70 with a capacitor chosen to provide a peak-to-peak ripple voltage of (i) 10% o
goblinko [34]

The question incomplete! The complete question along with answer and explanation is provided below.

Question:

Augment the rectifier circuit of Problem 4.68 with a  capacitor chosen to provide a peak-to-peak ripple voltage of  (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Problem 4.68:

A half-wave rectifier circuit with a 1-kΩ load operates from a 120-V (rms) 60-Hz household supply through  a 10-to-1 step-down transformer. It uses a silicon diode  that can be modeled to have a 0.7-V drop for any current.

Given Information:

Input voltage = 120 Vrms

10 to 1 step-down transformer

Voltage drop at diode = 0.7 V

Load resistance = R = 1 kΩ

Required Information:

 (i) 10% of the peak output and (ii) 1% of the peak output. In  each case:

(a) What average output voltage results?

(b) What fraction of the cycle does the diode conduct?

(c) What is the average diode current?

(d) What is the peak diode current?

Answer:

Case (i)

Vavg = 15.45 V

Conduction of diode = 7.11 %

Iavg = 0.232 A

Ip = 0.449 A

Case (ii)

Vavg = 16.18 V

Conduction of diode = 2.25 %

Iavg = 0.735 A

Ip = 1.453 A

Explanation:

Voltage at the secondary side of the transformer is

Vrms = Vpri/turn ratio

Vrms = 120/10 = 12 V

The relation between rms voltage and peak voltage is

Vp = Vrms/√2

Vp = 12√2 = 16.97 V

Vd = 0.7 V

First we will calculate all the required parameters for the 10% ripple voltage and then for 1% ripple voltage.

case (i) 10% of the peak output:

(a) What average output voltage results?

Average output voltage = Vavg = Vp - Vd - 0.5Vr

Where Vp is the peak output voltage Vd is the voltage drop of diode and Vr is the ripple voltage which is given as a percentage of Vp

Vavg = Vp - Vd - 0.5Vr

Vavg = 16.97 - 0.7 - 0.5[0.1(16.97 - 0.7)]

Vavg = 15.45 V

(b) What fraction of the cycle does the diode conduct?

ω = √2Vr/Vp - Vd

ω = √2*0.1(Vp-Vd)/Vp - Vd

ω = √2*0.1(16.97-0.7)/16.97 - 0.7

ω = 0.447 rad

Conduction of diode = (ω/2π)*100

Conduction of diode = (0.447/2π)*100

Conduction of diode = 7.11 %

(c) What is the average diode current?

Average current = Iavg = Vavg/R[ 1 + π( √2(Vp - Vd)/0.1(Vp-Vd))]

Average current = Iavg = 15.45/1000[ 1 + π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Average current = Iavg = 0.232 A

(d) What is the peak diode current?

Peak current = Ip = Vavg/R[ 1 + 2π( √2(Vp - Vd)/0.1(Vp-Vd))]

Peak current = Ip = 15.45/1000[ 1 + 2π( √2(16.97 - 0.7)/0.1(16.97-0.7))]

Peak current = Ip = 0.449 A

case (ii) 1% of the peak output:

(a) What average output voltage results?

Vavg = 16.97 - 0.7 - 0.5[0.01(16.97 - 0.7)]

Vavg = 16.18 V

(b) What fraction of the cycle does the diode conduct?

ω = √2*0.01(Vp-Vd)/Vp - Vd

ω = √2*0.01(16.97-0.7)/16.97 - 0.7

ω = 0.1417 rad

Conduction of diode = (0.1417/2π)*100

Conduction of diode = 2.25 %

(c) What is the average diode current?

Average current = Iavg = 16.18/1000[ 1 + π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Average current = Iavg = 0.735 A

(d) What is the peak diode current?

Peak current = Ip = 16.18/1000[ 1 + 2π( √2(16.97 - 0.7)/0.01(16.97-0.7))]

Peak current = Ip = 1.453 A

3 0
3 years ago
hhTwo cups of the same size are filled to the brim with clear liquids. Cup A holds water. Cup B contains alcohol. Your teacher c
Taya2010 [7]
You used density, because water/ice has a density of 1, and ice will sink in anything with a lesser density 
4 0
3 years ago
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