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4 years ago

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A light string is wrapped around a solid cylindrical spool of radius 0.565 m and mass 1.76 kg. a 5.04 kg mass is hung from the s

tring, causing the spool to rotate and the string to unwind. assume that the system starts from rest and no slippage takes place between the string and the spool. use conservation of energy to determine the angular speed of the spool after the mass has dropped 4.19 m. hint: use the relation v=Ïr, and use conservation of energy.

1 answer:

Molodets [167]4 years ago

4 0

OOF OOF OOF OOF oof OOF OOF oof OOF oof

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after one hour, 1/16 of the initial amount of certain radioactive isotope remains undecayed. the half life of the isotope is:

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A spring has a constant k= 10.32 N/m and is hung vertically. A 0.400kg mass is suspended from the spring. What is the displaceme

alukav5142 [94]

Answer:

displacement (x) = 0.003798 meters

Explanation:

from the fact that the string is hung vertically we can deduce that:

Total force acting on the mass = Fs (by spring) + Fg (by gravity)

<em>where</em>

Fs = k*x , x is the displacement..

Fg = m*g

then:

Ftot = m*a, <em>but a = 0 m/(s^2) because the mass becames stationary.</em>

Ftot = 0

Fs + Fg = 0

<em>by direction, take down as negative.</em>

Fs - Fg = 0

k*x = m*g

x = m*g/k = [(0.400)(9.8)]/(10.32)

= 0.3798 meters

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3 years ago

A typical human lens has an index of refraction of 1.430 . The lens has a double convex shape, but its curvature can be varied b

rjkz [21]

Answer:

Maximum Power = 144.3 D

The associated focal length of the lens = $6.92*10^{-3} m$

Explanation:

According to the Lens maker's Formula:

$\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{R_2} )$

where;

$n_1$ = the refractive index of the medium

$R_1$ and $R_2$ = radius of curvature on each surface

For a convex lens, The radius of curvature in the front surface will be positive and that of the second surface will be negative . Therefore;

$\frac{1}{f} = (n-1) (\frac{1}{R_1}-\frac{1}{-R_2} ) \\ \\ \frac{1}{f} = (n-1) (\frac{1}{R_1}+\frac{1}{R_2} )$

At maximum power

$\frac{1}{f} = (1.430-1) (\frac{1}{6.50 \ mm}-\frac{1}{5.50 \ mm} )$

= $0.144 \ mm^{-1}$

This Implies

$f = 6.92 mm\\f = 6.92*10^{-3} \ m$

Therefore; the power is given by the formula:

$P_{max} = \frac{1}{f}$

$P_{max}= \frac{1}{6.92*10^{-3}}$

= 144.3 D

3 0

4 years ago

A 100-kg block being released from rest from a height of 1.0 m. It then takes it 1.40 s to reach the floor. What is the mass m o

dimulka [17.4K]

Complete Question

The complete question is shown on the first uploaded image

Answer:

The mass of the other block is $m_1 = 81.14 \ kg$

Explanation:

From the question we are told that

Mass of the first block is $m_1 = 100 \ kg$

The height is $s = 1.0 \ m$

The time it takes it is $t = 1.40 \ s$

Generally from kinematic equation

$s = ut + \frac{1}{2} at^2$

Here u is the initial velocity which zero given that it was at rest initially

So

$s = 0 * t + \frac{1}{2} at^2$

=> $s = \frac{1}{2} at^2$

=> $1 = \frac{1}{2}* a * (1.40 )^2$

=> $a = 1.0204 \ m/s^2$

Generally from the diagram the resultant force due to the weight of the first object and the tension on the string is mathematically represented as

$mg - T = ma$

=> $T = m g - ma$

=> $T = m(g - a)$

=> $T = 877.96 \ N$

Generally from the diagram the resultant force due to the weight of the second object and the tension on the string is mathematically represented as

$T - m_1g = m_1 a$

=> $877.96 = m_1 (a + g)$

=> $877.96 = m_1 (1.0204 + 9.8 )$

=> $m_1 = 81.14 \ kg$

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3 years ago