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kondor19780726 [428]
3 years ago
11

An underwater air bubble has an excess inside pressure of 13 pa. what is the excess pressure inside an air bubble with twice the

radius?
Physics
1 answer:
Evgesh-ka [11]3 years ago
6 0
Pressure = 2 * pi * sigma * radius where sigma is the pressure constant of the medium or the material.

Thus, Pressure/Radius = constant

Based on this:
(pressure*radius) in case 1 = (pressure*radius) in case 2
(13 * R) = (P * 2R)
P = 13/2 = 6.5 pa
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a = 2 m/s2

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a = \frac{kx}{m}

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3 years ago
The critical angle for water is 49°. If a ray of light
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Answer:

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3 0
3 years ago
Heat can be transferred in several ways. Which of the following is an example of conduction? (DOK 2) A The sun is shining on a m
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8 0
3 years ago
What is the smallest radius of an unbanked (flat) track around which a bicyclist can travel if her speed is 22 km/h and the coef
Aleks [24]
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Now the radial force required to keep an object of mass m, moving in circular motion around a radius R, is given by

F_{rad}=m \frac{v^2}{R}

The force of friction is given by the normal force (here, just the weight, mg) times the static coefficient of friction:

F_{fric}= mg \mu_{s}

Notice we don't use the kinetic coefficient even though the bike is moving.  This is because when the tires meet the road they are momentarily stationary with the road surface.  Otherwise the bike is skidding.

Now set these equal, since friction is the only thing providing the ability to accelerate (turn) without skidding off the road in a line tangent to the curve:

m\frac{v^2}{R} = mg \mu_{s} \\ \\ \frac{v^2}{R} = g \mu_{s} \\ \\R= \frac{v^2}{g \mu_{s}} \\ \\ R= \frac{6.11}{9.8 \times 0.37}=1.685m

3 0
3 years ago
Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d
Step2247 [10]
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

9 0
3 years ago
Read 2 more answers
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