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Advocard [28]
2 years ago
14

A mass M tied to a light string is moving in a vertical circle. The tension in the string at the top is TT and TB at the bottom

of the circle. The work done by the tension on M as it moves from bottom to top is:
Physics
1 answer:
natita [175]2 years ago
5 0

Answer:

Explanation:

Tension provides centripetal force in the circular motion . In circular motion work done by force = torque x angle

torque is zero as , centripetal force passes through axis of rotation that is center.

So work done by centripetal force  = 0

So work done by tension on M = 0

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Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
When the starter motor on a car is engaged, there is a 310 A current in the wires between the battery and the motor. Suppose the
Blababa [14]

Answer:

Diameter will be 351.42 mm

Explanation:

We have given current flowing in the copper wire i = 310 A

Voltage drop across the wire V = 0.55 volt

We know that resistance is given by R=\frac{V}{i}=\frac{0.55}{310}=0.00177\Omega

Length of the copper wire l = 1 m

Resisitivity of the copper wire \rho =1.72\times 10^{-8}\Omega m

We know that resistance R=\frac{\rho l}{A}

0.00177=\frac{1.72\times 10^{-8}\times  1}{A}

A=969.4545\times 10^{-8}m^2

As area A=\pi r^2

3.14\times r^2=969.4545\times 10^{-8}

r=17.57\times 10^{-4}m

So diameter d=17.57\times 10^{-4}\times 2=35.142\times \times 10^{-4}m = 351.42 mm

3 0
3 years ago
On the kelvin scale what is the freezing point of water
Alla [95]
Freezing point of the water is known as 273 K

Hope this helps!
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You forgot to add a photo.
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Which measurements are equal to 321 decimeters
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