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hodyreva [135]
3 years ago
8

Have you ever written a bio-data or an application letter? Share your experience in the

Physics
1 answer:
Georgia [21]3 years ago
8 0

Answer:

I found the experience tasking

Explanation:

I wouldn't say it was hard, neither was it easy. I'd rather go for something like it being tasking. It's worthy of note that it was my first time, and I think it's very normal especially when one hasn't been doing something of that nature previously. Of course I did my draft, which unsurprisingly happened to be not good enough, and I had to look for templates to guide me through the acceptable way.

I still did it in my own way, but in the right way. Ever since then though, I have never stuttered when writing application letters, as it had since then seem inborn

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How did the angular acceleration change with the new moment of inertia? was your prediction correct?
elena55 [62]

In Newtonian physics, the acceleration of a body is inversely proportional to mass. In Newtonian rotational physics, angular acceleration is inversely proportional to the moment of inertia of a frame.

The moment of Inertia is frequently given the image I. it's miles the rotational analog of mass. The moment of inertia of an object is a measure of its resistance to angular acceleration. because of its rotational inertia, you want torque to change the angular pace of an object. If there may be no net torque acting on an object, its angular speed will no longer change.

In linear momentum, the momentum p is the same as the mass m instances of the velocity v; whereas for angular momentum, the angular momentum L is the same as the instant of inertia I times the angular pace ω.

Learn more about angular acceleration here:-brainly.com/question/21278452

#SPJ4

6 0
1 year ago
Which climate condition is typically found in the tropics due to the interaction of the atmosphere and hydrosphere?
Ilya [14]
The answer is A! Hope I helped! 

7 0
3 years ago
Read 2 more answers
Two spherical conductors are separated by a distance much larger than either of their radii. Sphere A has a radius of 11.5 cm an
bonufazy [111]

Explanation:

As the given spheres are connected by a thin wire so, the potential on the spheres are the same.

          \frac{q_{1}}{r_{1}} = \frac{q_{2}}{r_{2}} ......... (1)

Hence, total charge will be as follows.

              q_{1} + q_{2} = Q = -95.5 nC .......... (2)

Using the above two equations, the final equation will be as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

and,    q_{1} = \frac{Qr_{1}}{r_{1} + r_{2}}

Hence, we will calculate the charge on sphere B after the equilibrium is reached as follows.

          q_{2} = \frac{Qr_{2}}{r_{1} + r_{2}}

                     = \frac{-95.5 \times 74.4 cm}{(11.5 + 74.4) cm}

                     = 82.714 nC

Thus, we can conclude that the charge on sphere B after equilibrium has been reached is 82.714 nC.

                       

5 0
3 years ago
The field is strongest at which point?
Licemer1 [7]

Answer:The poles

Explanation:

The field is strongest at the poles

4 0
3 years ago
Calculate the kinetic energy of a 64.g bullet moving at a speed of 411.ms . Round your answer to 2 significant digits.
Mandarinka [93]

Answer:

The kinetic energy of the bullet is 5.4 × 10³ J

Explanation:

Hi there!

The equation of kinetic energy is the following:

KE = 1/2 · m · v²

Where:

KE = kinetic energy.

m = mass of the bullet.

v = speed of the bullet.

Let´s convert the mass unit to kg so that our result is in Joules:

64 g · ( 1 kg / 1000 g) = 0.064 kg

Then, the kinetic energy will be the following:

KE = 1/2 · 0.064 kg · (411 m/s)²

KE = 5.4 × 10³ J

8 0
3 years ago
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