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2 years ago

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At a certain depth in the ocean. the absolute pressure is p. If you 90 to twice that depth (treating the water as incompressible

) A) the absolute pressure will be 2p.B) the absolute pressure will be less than 2p. C) the absolute pressure will be greater than 2p. D) the gauge pressure will not change. E) Use gauge pressure will increase but will not double.

1 answer:

KatRina [158]2 years ago

5 0

Answer:

correct option is B

the absolute pressure will be less than 2p

Explanation:

given data

absolute pressure = p

depth = twice

to find out

what is correct option

solution

we know here that p absolute pressure so we consider atmospheric pressure p1

so we can say at some depth h pressure will be

pressure p = p1 + ρ g h

so if depth is double

pressure will be

2p = 2p1 +2 (ρ g h )

so

2p - p1 = p1 + 2 (ρ g h )

so correct option is B

the absolute pressure will be less than 2p

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Answer:

Net force acting on them is 16 N and it is acting to the right side.

Explanation:

It is given that,

Force acting by the dog, $F_1 = 32\ N$ (right side)

Force acting by Simone , $F_2 = -16\ N$ (backward)

Let backward direction is taken to be negative while right side is taken to be positive.

The net force will act in the direction where the magnitude of force is maximum. Net force is given by :

$F=F_1+F_2$

$F=32+(-16)$

F = 16 N

So, the net force is 16 N and it is acting to the right side.

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2 years ago

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Find the truth table for the circuit shown. Explain the working principle for all the inputs, briefly. Explain why D1 is used in

Ymorist [56]

Answer:

see below for the truth table

Explanation:

<u>Truth Table</u>

As we will see from the description of operation, any input low causes the output to be high. This is the logic of a NAND gate. The truth table is attached.

<u>Working Principle</u>

Pulling any of A, B, or C low will saturate transistor Q1, depriving Q2 of any base current, cutting it off. Then Q5 is also deprived of base current and is cut off. Meanwhile, the current through R2 supplies base current to Q4, allowing it to pull the output high.

If all of A, B, and C are high (or open), base current is supplied to Q2 through the base-collector junction of Q1. Then Q2 saturates, supplying base current to Q3. Diode D1 ensures that the voltage across Q2 will be insufficient to supply any base current to Q4, so it stays cut off.

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2 years ago

A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the c

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Explanation:

The given data is as follows.

Mass of oxygen present = 100 mg = $100 \times 10^{-3}$ g

So, moles of oxygen present are calculated as follows.

n = $\frac{100 \times 10^{-3}}{32}$

= $3.125 \times 10^{-3}$ moles

Diameter of cylinder = 6 cm = $6 \times 10^{-2}$ m

= 0.06 m

Now, we will calculate the cross sectional area (A) as follows.

A = $\pi \times \frac{(0.06)^{2}}{4}$

= $2.82 \times 10^{-3} m^{2}$

Length of tube = 11 cm = 0.11 m

Hence, volume (V) = $2.82 \times 10^{-3} \times 0.11$

= $3.11 \times 10^{-4} m^{3}$

Now, we assume that the inside pressure is P .

And, $P_{atm}$ = 100 kPa = 100000 Pa,

Pressure difference = 100000 - P

Hence, force required to open is as follows.

Force = Pressure difference × A

= $(100000 - P) \times 2.82 \times 10^{-3}$

We are given that force is 173 N.

Thus,

$(100000 - P) \times 2.82 \times 10^{-3}$ = 173

Solving we get,

P = $3.8650 \times 10^{4} Pa$

= 38.65 kPa

According to the ideal gas equation, PV = nRT

So, we will put the values into the above formula as follows.

PV = nRT

$38.65 \times 3.11 \times 10^{-4} = 3.125 \times 10^{-3} \times 8.314 \times T$

T = 462.66 K

Thus, we can conclude that temperature of the gas is 462.66 K.

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2 years ago

A coil of 1000 turns of wire has a radius of 12 cm and carries a counterclockwise current of 15A. If it is lying flat on the gro

grin007 [14]

Answer:

torque is 1.7 * $10^{-2}$ Nm

Explanation:

Given data

turns n = 1000 turns

radius r = 12 cm

current I = 15A

magnitude B = 5.8 x 10^-5 T

angle θ = 25°

to find out

the torque on the loop

solution

we know that torque on the loop is

torque = N* I* A*B* sinθ

here area A = πr² = π(0.12)²

put all value

torque = N* I* A*B* sinθ

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2 years ago

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Answer:

$V=14m/s$

Explanation:

From the Question we are told that

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Generally to avoid collision the speed of astronaut be should be less than or equal to that of astronaut A with the bag

Therefore for minimum requirement speed of astronaut A should be given by astronaut B's speed which is equal to 1

Therefore

$130=(60*1)=(5*v)$

$V=14m/s$

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1 year ago