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postnew [5]
3 years ago
10

Pressure of gas is 13.6 cm of hg what does it means​

Physics
1 answer:
vlada-n [284]3 years ago
5 0

Explanation:

13 cmHg (centimeters of mercury) is the pressure at the bottom of a column of mercury 13 cm deep.  It is the equivalent of about 17.3 kPa or 2.5 psi.

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True or false: Balanced forces can change an object's direction?
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The statement is false.  Balanced forces can NOT change the speed OR direction of an object's motion.  (See Newton's #1 law of motion.)

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How can an animal regeneration produce two results?
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Joseph jogs from one end A to the other end B of a straight 300 m road in 2 minutes 30 seconds and then turns around and jogs 10
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Answer:

Explanation:

(a) It is given that Joseph jogs on a straight road of 300m in a time interval of 2 minutes and 30 seconds, which is equal to 150seconds. Therefore, when Joseph jogs from point A to point B, he covers a distance of 300m in time of 150seconds. Hence, his average speed is 300m/150s=2ms^−1. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m. Since it is a straight road and he jogs in a single direction in this case, his displacement is equal to 300m.

Hence, his average velocity is 300m/150s=2ms^−1

(b) Then it is given that he turns back and points B and jogs on the same road but in the opposite direction for a time interval for 1 minute and covers a distance of 100m.If we consider the whole motion of Joseph, i.e. from point A to point C, then he covers a total distance of 300m+100m=400m. And he covers this total distance in a time interval of 2.5min+1min=3.5min=210s.

Therefore, his average speed for this journey is 400m210s=1.9ms−1.

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7 0
3 years ago
A uniform electric field is produced due to the charge distribution inside the closed cylindrical surface. (a) What type of char
DiKsa [7]

Answer with Explanation:

a. Option d is true.

a negatively charged plane parallel to the end faces of the cylinder

b. Radius of cylinder, r=0.66m

Magnitude of electric field, E=300 N/C

We have to find the net flux through the closed surface.

Net electric flux,\phi=-2 EA=-2E(\pi r^2)

\phi=-2\times 300\times (3.14\times (0.66)^2)

\phi=-820.67 Nm^2/C

c.

Net charge,Q=\epsilon_0\times \phi

Where

\epsilon_0=8.85\times 10^{-12}

Q=-820.67\times 8.85\times 10^{-12}

Q=-7.26\times 10^{-9} C

Q=-7.26nC

Where 1nC=10^{-9}C

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