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Cerrena [4.2K]
3 years ago
15

What conditions must be present for a translational and rotational equilibrium of a rigid body?

Physics
1 answer:
zheka24 [161]3 years ago
5 0
If a rigid body is in translational and rotational equilibrium, then
1. The sum of all applied forces is zero.
2. The sum of all applied moments is zero.
3. There is no linear acceleration.
4. There is no rotational acceleration.

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Formula for calcium chloride​
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Formula : CaCl2

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The yearly rainfall in the rainforest is between 50- 260 inches ?id weather or climate ​
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3 0
4 years ago
A car moves with constant velocity along a straight road. Its position is x1 = 0 m at t1 = 0 s and is x2 = 66 m at t2 = 6.0 s .
UNO [17]

Answer: 1. 33, 2. 264

Explanation: 66m= 6s so, to find the position at 3s you just need to take 66/2 = 33m cause 3 is half of 6. & for 2 you will take 66x4= 264m cause it took 4s multiply by the original 6s to get 24s. Answer: 1 is 33m and 2 is 264m

7 0
3 years ago
Two very small +3.00-μC charges are at the ends of a meter stick. Find the electric potential (relative to infinity) at the cent
PtichkaEL [24]

Answer:

The electric potential at the center of the meter stick is 54 KV.

Explanation:

Electric potential (V) is given as:

i.e V = \frac{kq}{r}

Where: k is the Coulomb constant, q is the charge and r is the distance.

Given: q = 3.0 μC = 3.0 x 10^{-6} C, r = 0.5 m

So that,

V = \frac{9*10^{9}*3.0*10^{-6}  }{0.5}

   = \frac{2.7*10^{4} }{0.5}

V = 54000

  = 54 000 volts

The electric potential at the center of the meter stick is 54 KV.

4 0
3 years ago
Electromagnetic radiation is emitted by accelerating charges. The rate at which energy is emitted from an accelerating charge th
statuscvo [17]

Answer:

 P /K = 1,997 10⁻³⁶  s⁻¹

Explanation:

For this exercise let's start by finding the radiation emitted from the accelerator

       \frac{dE}{dt} = \frac{q^{2} a^{2} }{6\pi  \epsilon_{o} c^{2}    }

the radius of the orbit is the radius of the accelerator a = r = 0.530 m

let's calculate

       \frac{dE}{dt} = [(1.6 10⁻¹⁹)² 0.530²] / [6π 8.85 10⁻¹² (3 108)³]

      P= \frac{dE}{dt}= 1.597 10⁻⁵⁴ W

Now let's reduce the kinetic energy to SI units

       K = 5.0 10⁶ eV (1.6 10⁻¹⁹ J / 1 eV) = 8.0 10⁻¹⁹ J

the fraction of energy emitted is

      P / K = 1.597 10⁻⁵⁴ / 8.0 10⁻¹⁹

      P /K = 1,997 10⁻³⁶  s⁻¹

3 0
3 years ago
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