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Leviafan [203]
3 years ago
6

Your office has a 0.025 m^3 cylindrical container of drinking water. The radius of the container is about 13 cm. Required:a. Whe

n the container is full, what is the pressure that the water exerts on the sides of the container at the bottom of the container above the atmospheric pressure? b. When the container is full, what is the pressure that the water exerts on the sides of the container halfway down from the top above the atmospheric pressure?
Physics
1 answer:
uranmaximum [27]3 years ago
7 0

Answer:

<em>a) 105935.7 Pa</em>

<em>b) 103630.35 Pa</em>

Explanation:

The volume of the container = 0.025 m^3

The radius of the container = 13 cm = 0.13 m

We have to find the height of the tank

From the equation for finding the volume of the cylinder,

V = \pi r^2h

where

V is the volume of the cylinder

h is the height of the cylinder

substituting values, we have

0.025 = 3.142 x 0.13^2 x h

0.025 = 0.0531h

h = 0.025/0.0531 = 0.47 m

Pressure at the bottom of the tank P = ρgh

where

ρ is the density of water = 1000 kg/m^3

g is the acceleration due to gravity = 9.81 m/s^2

h is the depth of water which is equal to the height of the tank

substituting values, we have

P = 1000 x 9.81 x 0.47 = 4610.7 Pa

atmospheric pressure = 101325 Pa

therefore, the pressure in the tank bottom above atmospheric pressure = 101325 Pa + 4610.7 Pa = <em>105935.7 Pa</em>

b) For half way down the container, depth of water will be = 0.47/2 = 0.235 m

pressure P = 1000 x 9.81 x 0.235 = 2305.35 Pa

This pressure above atmospheric pressure = 101325 Pa + 2305.35 Pa = <em>103630.35 Pa</em>

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6 0
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A simple pendulum is made from a 0.54-m-long string and a small ball attached to its free end. The ball is pulled to one side th
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Answer:

0.37sec

Explanation:

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T = 2 x 3.14 x √[0.54/9.8]

T = 1.47sec

An oscillating pendulum, or anything else in nature that involves "simple harmonic" (sinusoidal) motion, spends 1/4 of its period going from zero speed to maximum speed, and another 1/4 going from maximum speed to zero speed again, etc. After four quarter-periods it is back where it started.

The ball will first have V(max) at T/4,

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3 0
3 years ago
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erma4kov [3.2K]

Answer:

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Answer:

option C

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given,

Q = +3.2 x 10⁻¹⁹ C

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B = 0.80 T

ion's acceleration is zero  

when acceleration is zero the magnitude of both the forces becomes equal.

q E = q V B

v = \dfrac{E}{B}

v= \dfrac{5 \times 10^5}{0.80}

v = 6.25 × 10⁵ m/s ≈ 6.3 × 10⁵ m/s

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