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Ket [755]
3 years ago
10

The largest graduated cylinder in my lab holds 2 L and has an inner diamter (the part that holds the water) of 8 cm. When it is

full of 2 L of water, what is the pressure at the bottom of the cylinder in Pa?
Physics
1 answer:
mestny [16]3 years ago
3 0

Answer:

<em>3924 Pa</em>

<em></em>

Explanation:

Volume of cylinder = 2 L = 0.002 m^3  (1000 L = 1 m^3)

diameter of the inner cylinder = 8 cm = 0.08 m  (100 cm = 1 m)

radius of the inner cylinder = diameter/2 = 0.08/2 = 0.04 m

area of the inner cylinder = \pi r^{2}

where \pi = 3.142,

and r = radius = 0.04 m

area of inner cylinder = 3.142 x 0.04^{2} = 0.005 m^2

<em>height h of the water in this cylinder = volume/area</em>

h = 0.002/0.005 = 0.4 m

<em>pressure at the bottom of the cylinder due to the height of water = pgh</em>

where

p = density of water = 1000 kg/m^3

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

h = height of water within this cylinder = 0.4 m

pressure = 1000 x 9.81 x 0.4 = <em>3924 Pa</em>

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Answer:

(a) a_s=0.62\frac{m}{s^2}

(b) a_s=0.13\frac{m}{s^2}

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Explanation:

(a) According to Newton's second law, the acceleration of a body is directly proportional to the force exerted on it and inversely proportional to it's mass.

a_s=\frac{F}{m_s}\\a_s=\frac{5.2N}{8.4kg}\\a_s=0.62\frac{m}{s^2}

(b) According to Newton's third law, the force that the sled exerts on the girl is equal in magnitude but opposite in the direction of the force that the girl exerts on the sled:

a_g=\frac{F}{m_g}\\a_g=\frac{5.2N}{40kg}\\a_g=0.13\frac{m}{s^2}

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x_f=x_0+v_0t \pm  \frac{at^2}{2}

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x_f_g=\frac{a_gt^2}{2}(1)

For the sled, we have v_0=0. So:

x_f_s=x_0_s-\frac{a_st^2}{2}(2)

When they meet, the final positions are the same. So, equaling (1) and (2) and solving for t:

x_0_s-\frac{a_st^2}{2}=\frac{a_st^2}{2}\\t^2(a_g+a_s)=2x_0_s\\t=\sqrt{\frac{2x_s_0}{a_g+a_s}}\\t=\sqrt{\frac{2(15m)}{0.13\frac{m}{s^2}+0.62\frac{m}{s^2}}}\\t=6.32s

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