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gavmur [86]
3 years ago
7

What is the absolute pressure at a depth of 9.05 m below the surface of a deep lake? Assume atmospheric pressure is 1.01 × 10 5

Pa .
Physics
1 answer:
Arte-miy333 [17]3 years ago
8 0

Answer:

1.89 x 10⁵ Pa

Explanation:

given,

depth of the lake = 9.05 m

atmospheric pressure = 1.01 × 10⁵ Pa

absolute pressure = ?

we know,

Absolute pressure = atmospheric pressure + gauge pressure

Absolute pressure = 1.01 × 10⁵ Pa + ρgh

Absolute pressure = 1.01 × 10⁵ Pa + 1000 x 9.8 x 9.05

Absolute pressure = 1.89 x 10⁵ Pa

hence, absolute pressure is equal to 1.89 x 10⁵ Pa

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A Ping-Pong ball with mass 2.5 g is attached by a thread to the bottom of a beaker. When the beaker is filled with water so that
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Answer:

0.022m or 2.2cm

Expxlanation:

Step 1:

Data obtained from the question. This includes:

Mass (m) = 2.5g = 2.5/1000 = 2.5x10^-3Kg

Tension (T) = 0.029 N

Density (ρ) = 1000 kg/m3

Acceleration due to gravity (g) = 9.81 m/s2

Diameter (d) =?

Step 2:

Finding an expression to calculate the diameter of the ball. This is illustrated below:

Tension = weight displaced - weight of the ball

Weight displaced = Mass of water x acceleration due to gravity

Mass of water = Density x volume

Mass of water = ρxV

Weight displaced = ρxVxg = ρVg

Weight of the ball = Mass of the ball x acceleration due to gravity

Weight of the ball = mg

Therefore,

Tension = weight displaced - weight of the ball

T = ρVg - mg

Make V the subject of the formula

T = ρVg - mg

T + mg = ρVg

Divide both side by ρg

V = ( T + mg) /ρg. (1)

Recall that the ball is spherical in shape and the Volume of a sphere is given by

V = 4/3πr^3

Radius (r) = diameter (d) /2

V = 4/3π(d/2)^3

V = 4/3πd^3/8

V = πd^3 /6

Substituting the value of V into equation 1, we have

V = ( T + mg) /ρg

πd^3 /6 = ( T + mg) /ρg.

Making d the subject of the formula, we have:

πd^3 /6 = (T + mg) /ρg.

d^3 = 6(T + mg) /πρg.

Taking the cube root of both sides

d = [6(T + mg) /πρg]^1/3

Step 3:

Determination of the diameter of the ball. This is illustrated below:

T = 0.029 N

m = 2.5x10^-3Kg

g = 9.81 m/s2

ρ = 1000 kg/m3

d =?

d = [6(T + mg) /πρg]^1/3

d = [6(0.029 + 2.5x10^-3x9.81)/ πx1000x9.81]^1/3

d = 0.022m

Therefore, the diameter of the ball is 0.022m or 2.2cm

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3 years ago
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