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MrRa [10]
3 years ago
15

A ledge on a building is 18 m above the ground. A taut rope attached to a 4.0-kg can of paint sitting on the ledge passes up ove

r a pulley and straight down to a 3.0-kg can of nails on the ground.If the can of paint is accidently knocked off the ledge, what time interval does a carpenter have to catch the can before it smashes on the floor?

Physics
1 answer:
True [87]3 years ago
5 0

Answer:t=5.07 s

Explanation:

Given

height of Building h=18 m

mass of Paint can m_1=4 kg

mass of second can m_2=3 kg

let T be the Tension in the rope

For  4 kg can

4g-T=4a

T=4(g-a)----1

For 3 kg can

T-3g=3a

T=3(g+a)----2

From 1 and 2

4(g-a)=3(g+a)

g=7a

a=\frac{g}{7}

So time taken to cover 18 m is

h=ut+\frac{at^2}{2}

18=0+\frac{g\cdot t^2}{7\times 2}

t^2=\frac{18\times 2\times 7}{g}

t=5.07 s

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A rocket sled accelerates at a rate of 49.0 m/s2 . Its passenger has a mass of 75.0 kg. (a) Calculate the horizontal component o
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Explanation:

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F = m a

F=75\ kg\times 49\ m/s^2

F = 3675 N

Ratio, R=\dfrac{F}{W}

R=\dfrac{3675}{75\times 9.8}=5

So, the ratio between the horizontal force and the weight is 5 : 1.

(b) The magnitude of total force the seat exerts against his body is F' i.e.

F'=\sqrt{F^2+W^2}

F'=\sqrt{(3675)^2+(75\times 9.8)^2}

F' = 3747.7 N

The direction of force is calculated as :

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For small amplitudes of oscillation the motion of a pendulum is simple harmonic. Consider a pendulum with a period of 0.550 s Fi
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To solve this problem we will use the concepts related to the expression of energy for harmonic oscillator. From our given values we have that the period is equivalent to

T = 0.55s

Therefore the frequency will be the inverse of the period and would be given as

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f = \frac{1}{0.55}

f = 1.82s^{-1}

The ground state energy of the pendulum is,

E = \frac{1}{2} hv

E = \frac{1}{2}(6.626*10^{-34}J\cdot s)(1.82s^{-1})

E = 6.03*10^{-34}J

The ground state energy in eV,

E = 6.03 * 10^{-34}J(\frac{1eV}{1.6*10^{-19}J})

E = 3.8*10^{-15}eV

The energy difference between adjacent energy levels,

\Delta E = hv

\Delta E = (6.626*10^{-34}J\cdot s)(1.82s)

\Delta E = 12.1*10^{-34}J

4 0
3 years ago
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