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nirvana33 [79]
3 years ago
9

Which of the following would have the strongest electric field? MULTIPLE CHOICE.

Physics
1 answer:
irina [24]3 years ago
7 0
Under a thundercloud or lightning bolt
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The earth travels around the sun once a year in an approximately circular orbit whose radius is 1.50x10^11 m. From this data det
seraphim [82]
(a) Determine the circumference of the Earth through the equation,
            C = 2πr
Substituting the known values, 
           C = 2π(1.50 x 10¹¹ m)
             C = 9.424 x 10¹¹ m

Then, divide the answer by time which is given to a year which is equal to 31536000 s. 
          orbital speed = (9.424 x 10¹¹ m)/31536000 s

               orbital speed = 29883.307 m/s

Hence, the orbital speed of the Earth is ~29883.307 m/s.

(b) The mass of the sun is ~1.9891 x 10³⁰ kg. 
8 0
3 years ago
Crickets Chirpy and Milada jump from the top of a vertical cliff. Chirpy drops downward and reaches the ground in 2.70 s, while
Vinvika [58]

Answer:

Explanation:

Given

Time taken to reach ground is t=2.7\ s

Malda initial velocity u=95\ cm/s

Let h be the height of Cliff

using h=ut+\frac{1}{2}at^2

where, u=initial velocity

t=time

In first case chirpy drop downward thus u=0

h=0+\frac{1}{2}(9.8)(2.7)^2

h=35.72\ m

For Milada there is horizontal velocity u=95 cm/s=0.95 m/s[/tex]

time taken to reach the ground will be same so distance traveled in this time with 0.95 m/s horizontal velocity is given by

R=u\times t

R=0.95\times 2.7=2.43\ m    

7 0
3 years ago
An energy storage system based on a flywheel (a rotating disk) can store a maximum of 3.7 MJ when the flywheel is rotating at 16
Likurg_2 [28]

The moment of inertia of the flywheel is 2.63 kg-m^{2}

It is given that,

The maximum energy stored on the flywheel is given as

E=3.7MJ= 3.7×10^{6} J

Angular velocity of the flywheel is 16000\frac{rev}{min} = 1675.51\frac{rad}{sec}

So to find the moment of inertia of the flywheel. The energy of a flywheel in rotational kinematics is given by :

E = \frac{1}{2}Iw^{2}

By rearranging the equation:

I = \frac{2E}{w_{2} }

I = 2.63 kg-m^{2}

Thus the moment of inertia of the flywheel is 2.63 kg-m^{2}.

Learn more about moment of inertia here;

brainly.com/question/13449336

#SPJ4

7 0
1 year ago
the nucleus and the subatomic particles in the nucleus, what is the charge of the nucleus of an atom?
pychu [463]
The Nucleus contains Protons and Neutrons.

The Neutrons does not have a charge.

The Protons are positively charge.

Hence the charge on the Nucleus, would be the charge of the proton, which is positive.

Hence Nucleus is Positively Charged.
4 0
3 years ago
A charge of 25 nC is uniformly distributed along a straight rod of length 3.0 m that is bent into a circular arc with a radius o
Greeley [361]

Answer:

E = 31.329 N/C.

Explanation:

The differential electric field dE at the center of curvature of the arc is

dE = k\dfrac{dQ}{r^2}cos(\theta ) <em>(we have a cosine because vertical components cancel, leaving only horizontal cosine components of E. )</em>

where r is the radius of curvature.

Now

dQ = \lambda rd\theta,

where \lambda is the charge per unit length, and it has the value

\lambda = \dfrac{25*10^{-9}C}{3.0m} = 8.3*10^{-9}C/m.

Thus, the electric field at the center of the curvature of the arc is:

E = \int_{\theta_1}^{\theta_2} k\dfrac{\lambda rd\theta  }{r^2} cos(\theta)

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2}cos(\theta) d\theta.

Now, we find \theta_1 and \theta_2. To do this we ask ourselves what fraction is the arc length  3.0 of the circumference of the circle:

fraction = \dfrac{3.0m}{2\pi (2.3m)}  = 0.2076

and this is  

0.2076*2\pi =1.304 radians.

Therefore,

E = \dfrac{\lambda k}{r} \int_{\theta_1}^{\theta_2} cos(\theta)d\theta= \dfrac{\lambda k}{r} \int_{0}^{1.304}cos(\theta) d\theta.

evaluating the integral, and putting in the numerical values  we get:

E = \dfrac{8.3*10^{-9} *9*10^9}{2.3} *(sin(1.304)-sin(0))\\

\boxed{ E = 31.329N/C.}

4 0
3 years ago
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