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Veseljchak [2.6K]

3 years ago

11

A body of mass 2 kilograms moves on a circle of radius 3 meters, making one revolution every 5 seconds. Find the magnitude of th

e centripetal force acting on the body. (Be sure to correctly state the units of the force.

1 answer:

harkovskaia [24]3 years ago

7 0

Answer: $F_c=9.375 N$

Explanation:

Given

mass of body $m=2 kg$

radius of circle $r=3 m$

Time Period $T=5 s$

suppose $\omega$ is the angular velocity of revolution

therefore $\omega T=2\pi$

$\omega =\frac{2\pi }{5}=1.25 rad/s$

Centripetal acceleration $a_c=\omega ^r$

$a_c=1.25^2\times 3$

$a_c=4.68 m/s^2$

therefore centripetal $F_c=m\omega ^2\times r$

$F_c=9.375 N$

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Pages 64 to 65 of your reading material trace an extended problem that involes lifting a bag of sugar up to a shelf at first it

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The strength of an electromagnet can be altered by

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3 0

3 years ago

Read 2 more answers

A rock is thrown upward from the top of a 30 m building with a velocity of 5 m/s. Determine its velocity (a) When it falls back

Kobotan [32]

Answer:

a) 5 m/s

b) 17.8542 m/s

c) 24.7212 m/s

0.229

Explanation:

t = Time taken

u = Initial velocity = 5 m/s

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 9.81 m/s²

$v=u+at\\\Rightarrow 0=5-9.81\times t\\\Rightarrow \frac{-5}{-9.81}=t\\\Rightarrow t=0.51 \s$

$s=ut+\frac{1}{2}at^2\\\Rightarrow s=5\times 0.51+\frac{1}{2}\times -9.81\times 0.51^2\\\Rightarrow s=1.27\ m$

So, the stone would travel 1.27 m up

$v^2-u^2=2as\\\Rightarrow v=\sqrt{2as+u^2}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.27+0^2}\\\Rightarrow v=5\ m/s$

Velocity as the rock passes through the original point is 5 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 1.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{1.27\times 2}{9.81}}\\\Rightarrow t=0.51\ s$

Time taken to reach the original point is 0.51+0.51 = 1.02 seconds

So, total height of the rock would fall is 30+1.27 = 31.27 m

$s=ut+\frac{1}{2}at^2\\\Rightarrow 16.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{16.27\times 2}{9.81}}\\\Rightarrow t=1.82\ s$

Time taken by the stone to reach 15 m above the ground is 1.82+0.51 = 2.33 seconds

$v=u+at\\\Rightarrow v=0+9.81\times 1.82\\\Rightarrow v=17.8542\ m/s$

Speed of the ball at 15 m above the ground is 17.8542 m/s

$s=ut+\frac{1}{2}at^2\\\Rightarrow 31.27=0t+\frac{1}{2}\times 9.81\times t^2\\\Rightarrow t=\sqrt{\frac{31.27\times 2}{9.81}}\\\Rightarrow t=2.52\ s$

$v=u+at\\\Rightarrow v=0+9.81\times 2.52\\\Rightarrow v=24.7212\ m/s$

Speed of the stone just before it hits the street is 24.7212 m/s

F = Force

m = Mass = 100 kg

g = Acceleration due to gravity = 9.81 m/s²

s = Displacement = 4 km = 4000 m

P = Power = 1 hp = 745.7 Watt

t = Time taken = 20 minutes = 1200 seconds

μ = Coefficient of sliding friction

F = μ×m×g

⇒F = μ×100×9.81

W = Work done = F×s

P = Work done / Time

⇒P = F×s / t

$745.7=\frac{\mu \times 981\times 4000}{1200}\\\Rightarrow \mu=\frac{747.5\times 1200}{981\times 4000}\\\Rightarrow \mu=0.229$

Coefficient of sliding friction is 0.229

8 0

3 years ago