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2 years ago

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A 30kg uniform solid cylinder has a radius of 0.18m. if the cylinder accelerates at 0.023 rad/s^2 as it rotates about an axis th

rough its center, how large is the torque acting on the cylinder? With work please

1 answer:

Nuetrik [128]2 years ago

7 0

Answer:

0.011 N-m

Explanation:

Given that

The mass of a solid cylinder, m = 30 kg

The radius of the cylinder, r = 0.18 m

The acceleration of the cylinder, $\alpha =0.023\ rad/s^2$

It rotates about an axis through its center. We need to find the torque acting on the cylinder. The formula for the torque is given by :

$\tau=I\alpha$

Where

I is the moment of inertia of the cylinder,

For cylinder,

$I=\dfrac{mr^2}{2}$

So,

$\tau=\dfrac{mr^2\alpha }{2}\\\\\tau=\dfrac{30\times (0.18)^2\times 0.023 }{2}\\\\\tau=0.011\ N-m$

So, the required torque on the cylinder is 0.011 N-m.

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In 1.71 minutes, a ski lift raises four skiers at constant speed to a height of 148 m. The average mass of each skier is 62.9 kg

Vitek1552 [10]

Answer:

3560.36 Watts

Explanation:

Power, $P=\frac {nΔW}{Δt)$ where P is power, n is the number of skiers, t is time in seconds and Δt is change in time, ΔW is given by mgh where m is mass, g is gravitational constant, h is height

Substituting n for 4 skiers, m for 62.9 Kg, g for 9.81, h for 148 m and t for 1.71*60=102.6 seconds

P= $\frac {4*62.9*9.81*148}{1.71*60}=3560.360702 Watts$

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3 years ago

In a Hydrogen atom an electron rotates around a stationary proton in a circular orbit with an approximate radius of r =0.053nm.

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Answer:

(a): $F_e = 8.202\times 10^{-8}\ \rm N.$

(b): $F_g = 3.6125\times 10^{-47}\ \rm N.$

(c): $\dfrac{F_e}{F_g}=2.27\times 10^{39}.$

Explanation:

Given that an electron revolves around the hydrogen atom in a circular orbit of radius r = 0.053 nm = 0.053 $\times 10^{-9}$ m.

Part (a):

According to Coulomb's law, the magnitude of the electrostatic force of interaction between two charged particles of charges $q_1$ and $q_2$ respectively is given by

$F_e = \dfrac{k|q_1||q_2|}{r^2}$

where,

$k$ = Coulomb's constant = $9\times 10^9\ \rm Nm^2/C^2.$
$r$ = distance of separation between the charges.

For the given system,

The Hydrogen atom consists of a single proton, therefore, the charge on the Hydrogen atom, $q_1 = +1.6\times 10^{-19}\ C.$

The charge on the electron, $q_2 = -1.6\times 10^{-19}\ C.$

These two are separated by the distance, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the electrostatic force of attraction between the electron and the proton is given by

$F_e = \dfrac{(9\times 10^9)\times |+1.6\times 10^{-19}|\times |-1.6\times 10^{-19}|}{(0.053\times 10^{-9})^2}=8.202\times 10^{-8}\ \rm N.$

Part (b):

The gravitational force of attraction between two objects of masses $m_1$ and $m_1$ respectively is given by

$F_g = \dfrac{Gm_1m_2}{r^2}.$

where,

$G$ = Universal Gravitational constant = $6.67\times 10^{-11}\ \rm Nm^2/kg^2.$
$r$ = distance of separation between the masses.

For the given system,

The mass of proton, $m_1 = 1.67\times 10^{-27}\ kg.$

The mass of the electron, $m_2 = 9.11\times 10^{-31}\ kg.$

Distance between the two, $r = 0.053\times 10^{-9}\ m.$

Thus, the magnitude of the gravitational force of attraction between the electron and the proton is given by

$F_g = \dfrac{(6.67\times 10^{-11})\times (1.67\times 10^{-27})\times (9.11\times 10^{-31})}{(0.053\times 10^{-9})^2}=3.6125\times 10^{-47}\ \rm N.$

The ratio $\dfrac{F_e}{F_g}$ :

$\dfrac{F_e}{F_g}=\dfrac{8.202\times 10^{-8}}{3.6125\times 10^{-47}}=2.27\times 10^{39}.$

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Since conservation of momentum states that the momentum will remain the same. We have:

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