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Leto [7]
3 years ago
5

calculate the force required to take away a flat corcular plate of radius 0.01m from the surface of water. the surface tention o

f water is 0.075N/m.​
Physics
1 answer:
leonid [27]3 years ago
8 0

Answer:

Force = 0.0047175\ N

Explanation:

Given

T = 0.075N/m --- Surface Tension

r = 0.01m --- Radius

Required

Determine the required force

First, we calculate the circumference (C) of the circular plate

C= 2\pi r

C= 2 * \frac{22}{7} * 0.01m

C= \frac{2 * 22 * 0.01}{7}m

C= \frac{0.44}{7}m

C= 0.0629 m

The applied force is then calculated using;

Force = C * T

Force = 0.0629m * 0.075N/m

Force = 0.0047175\ N

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A swimmer bounces straight up from a diving board and falls feet first into a pool. She starts with a velocity of 4.00 m/s, and
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Answer:

a = 1.152s

b = 0.817 m

c = 7.29m/s

Explanation: let the following

From the first equation of linear motion

V = u+at..........1

parameters be represented as :

t = Time taken

v = Final velocity

a = Acceleration due to gravity = 9.8m/s²

u = Initial velocity = 4 m/s

s = Displacement

V = 0

Substitute the values into equation 1

0 = 4-9.8(t)

-4 = -9.8t

t = 4/9.8

t = 0.408s

From : s = ut+1/2at^2.........2

S = 4×0.408+0.5(-9.8)×0.408^2

S= 1.632-4.9(0.166)

S = 1.632-0.815

S = 0.817m

Her highest height above the board is 0.817 m

Total height she would fall is 0.817+1.90 = 2.717 m

From equation 2

s = ut+1/2at^2

2.717 m = 0t+0.5(9.8)t^2

2.717 m = 0+4.9t^2

2.717 m = 4.9t^2

2.717/4.9 = t^2

0.554 =t^2

t =√0.554

t = 0.744s

Hence, her feet were in the air for 0.744+0.408seconds

= 1.152s

Also recall from equation 1

V= u+at

V = 0+9.8(0.744)

V = 7.29m/s

Hence, the velocity when she hits the water is 7.29m/s

Finally,

a = 1.152s

b = 0.817 m

c = 7.29m/s

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