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Alik [6]
3 years ago
8

A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th

e rim of the disk (a) grows with the time t as ________ for (αt2) << 1.
Physics
1 answer:
solniwko [45]3 years ago
4 0

Answer:

a = R\alpha\sqrt{1 + \alpha^2t^4}

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

a_t = R\alpha

2) Centripetal acceleration

a_c = \omega^2 R

here we know that

\omega = \alpha t

a_c = (\alpha t)^2 R

now we know that net linear acceleration is given as

a = \sqrt{a_c^2 + a_t^2}

so we have

a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}

a = R\alpha\sqrt{1 + \alpha^2t^4}

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What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor
const2013 [10]

Complete Question

An athlete at the gym holds a 3.0 kg steel ball in his hand. His arm is 70 cm long and has a mass of 4.0 kg. Assume, a bit unrealistically, that the athlete's arm is uniform.

What is the magnitude of the torque about his shoulder if he holds his arm straight out to his side, parallel to the floor? Include the torque due to the steel ball, as well as the torque due to the arm's weight.

Answer:

The torque is  \tau = 34.3 \  N\cdot m

Explanation:

From the question we are told that

   The mass of the steel ball is  m  =  3.0 \  kg

    The length of arm is  l =  70 \ cm  = 0.7 \  m

    The mass of the arm is m_a  = 4.0 \  kg

Given that the arm of the athlete is uniform them the distance from the shoulder to the center of gravity of the arm is mathematically represented as

       r = \frac{l}{2}

=>    r = \frac{ 0.7}{2}  

=>    r = 0.35 \ m  

Generally the magnitude of torque about the athlete shoulder is mathematically represented as

      \tau =  m_a * g * r  + m * g *  L

=>    \tau =  4 * 9.8 * 0.35 + 3 * 9.8 *  0.70

=>    \tau = 34.3 \  N\cdot m

5 0
3 years ago
The term ampacity is defined as the _____ current, in amperes, that a conductor can carry continuously under conditions of use w
White raven [17]

Answer:

The maximum current, in amperes, that a conductor can carry continuously under the conditions of abuse without exceeding its temperature rating.

5 0
3 years ago
The bending of light as it passes into a transparent material of different optical intensity is known as A. conversion. B. aberr
stiks02 [169]

Answer:

<em>D. refraction</em>

Explanation:

Refraction: Refraction is change in direction of light rays. Refraction occurs whenever light rays travels from a transparent medium to another transparent medium of different density. The abrupt change in direction at the surface of the surface of the two media is referred to as <em>refraction</em><em>.</em>

<em>Refraction occurs when light travels from air to glass or from air to liquid.</em>

<em>Laws Of Refraction:</em>

(i) The incident ray, the refracted ray and the normal, all at the point of incident lies in the same plane.

(ii) The ratio of the sine of the angle of incident to the sine of the angle of refraction is a constant for a given pair of media.

<em>Thus the right option is D. refraction</em>

6 0
3 years ago
Determine qual é a carga elétrica de um corpo que possui 1 milhão de partículas?
MaRussiya [10]

Answer:

This can be translated to:

"find the electrical charge of a body that has 1 million of particles".

First, it will depend on the charge of the particles.

If all the particles have 1 electron more than protons, we will have that the charge of each particle is q = -e = -1.6*10^-19 C

Then the total charge of the body will be:

Q = 1,000,000*-1.6*10^-19 C = -1.6*10^-13 C

If we have the inverse case, where we in each particle we have one more proton than the number of electrons, the total charge will be the opposite of the one of before (because the charge of a proton is equal in magnitude but different in sign than the charge of an electron)

Q = 1.6*10^-13 C

But commonly, we will have a spectrum with the particles, where some of them have a positive charge and some of them will have a negative charge, so we will have a probability of charge that is peaked at Q = 0, this means that, in average, the charge of the particles is canceled by the interaction between them.

7 0
3 years ago
If the mass of a 1.8 g paperclip was able to be completely converted to energy, how much energy would you obtain?
Anton [14]

Answer:

E=1.62\times 10^{14}\ J

Explanation:

Given that,

The mass of the paperclip, m = 1.8 g = 0.0018 kg

We need to find the energy obtained. The relation between mass and energy is given by :

E=mc^2

Where

c is the speed of light

So,

E=0.0018\times (3\times 10^8)^2\\\\E=1.62\times 10^{14}\ J

So, the energy obtained is 1.62\times 10^{14}\ J.

7 0
3 years ago
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