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Alla [95]
3 years ago
14

Please show work with the answer and thank you

Physics
1 answer:
Alika [10]3 years ago
7 0

Answer:

its your homework

Explanation:

you aren't getting any free answers out of me because it sounds like your trying to be smart your welcome

(Not smarter than me)

You might be interested in
I dont know how to do this at all please help
worty [1.4K]
Wow !  I understand your shock.  I shook and vibrated a little
when I looked at this one too.

The reason for our shock is all the extra junk in the question,
put there just to shock and distract us.

"Neutron star", "5.5 solar masses", "condensed burned-out star".
That's all very picturesque, and it excites cosmic fantasies in
out brains when we read it, but it's just malicious decoration.
It only gets in the way, and doesn't help a bit.

The real question is:

What is the acceleration of gravity 2000 m from
the center of a mass of 1.1 x 10³¹ kg ?

Acceleration of gravity is

                           G  ·  M / R²

      =  (6.67 x 10⁻¹¹ N·m²/kg²) · (1.1 x 10³¹ kg) / (2000 m)²

      =  (6.67 x 10⁻¹¹  ·  1.1 x 10³¹ / 4 x 10⁶)      (N) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴           (kg · m / s²) · m² · kg / kg² · m²

      =             1.83 x 10¹⁴            m / s²      

That's about  1.87 x 10¹³  times the acceleration of gravity on
Earth's surface.

In other words, if I  were standing on the surface of that neutron star,
I would weigh  1.82 x 10¹² tons, give or take.     
3 0
3 years ago
A generator is designed to produce a maximum emf of 190 V while rotating with an angular speed of 3800 rpm. Each coil of the gen
Zinaida [17]

Answer:

The number of turns of wire needed is 573.8 turns

Explanation:

Given;

maximum emf of the generator, = 190 V

angular speed of the generator, ω = 3800 rev/min =

area of the coil, A = 0.016 m²

magnetic field, B = 0.052 T

The number of turns of the generator is calculated as;

emf = NABω

where;

N is the number of turns

\omega = 3800 \frac{rev}{min} \times \frac{2\pi}{1 \ rev} \times \frac{1 \min}{60 \ s } = 397.99 \ rad/s

N = \frac{emf}{AB\omega } \\\\N = \frac{190}{0.016 \times 0.052\times 397.99} \\\\N = 573.8 \ turns

Therefore, the number of turns of wire needed is 573.8 turns

4 0
3 years ago
you ride a bike for 2 hours at 25 km/hr then 3 hours at 34 km/hr. what is your average velocity (in km/hr)?
mamaluj [8]
<h3>Answer:</h3>

30.4 km/hr

<h3>Explanation:</h3>

<u>We are given</u>;

  • Speed in the first 2 hours as 25 km/hr
  • Speed in the next 3 hours as 34 km/hr

We are required to determine the average velocity in km/hr

  • To get the average velocity we divide total distance by total time.
  • Thus, we need to determine the total distance

Distance = Speed × time

Distance covered in the first 2 hours;

               = 25 km/hr × 2 hours

               = 50 km

Distance in the next 3 hours

                = 34 km/hr × 3 hours

                = 102 km

Therefore, total distance = 50 km + 102 km

                                        = 152 km

Total time = 2 hrs + 3 hrs

                = 5 hours

Therefore;

Average speed = 152 km ÷ 5 hours

                          = 30.4 km/hr

Thus, the average speed is 30.4 km/hr

4 0
3 years ago
What force is needed to accelerate a 1,000-kilogram car from a stop to 5 m/s/s?
solong [7]
That depends on how soon you want it to reach 5 m/s/s. Without friction, ANY force will accelerate the car, like a mosquito pushing on it, but a Space Shuttle booster will accelerate it at a greater rate.
3 0
3 years ago
Two equal charges with magnitude Q and Q experience a force of 12.3442 when held at a distance r. What is the force between two
andre [41]

Answer:

197.5072.

Explanation:

According to the Coulomb's law, the magnitude of the electrostatic force of interaction between two charges \rm q_1 and \rm q_2 which are separated by the distance \rm d is given by

\rm F = \dfrac{kq_1q_2}{d^2}.

<em>where,</em> k is the Coulomb's constant.

For the case, when,

  • \rm q_1 = Q.
  • \rm q_2 = Q.
  • \rm d=r.
  • \rm F=12.3442.

Then, using Coulomb's law,

\rm 12.3442 = \dfrac{kQQ}{r^2}=\dfrac{kQ^2}{r^2}\ \ \ \ .......\ (1).

For the case, when,

  • \rm q_1 = 2Q.
  • \rm q_2 = 2Q.
  • \rm d=\dfrac r2.

Then, using Coulomb's law, the new electric force between the charges is given by,

\rm F' = \dfrac{k(2Q)(2Q)}{\left (\dfrac r2\right )^2}\\=\dfrac{k\ 4Q^2}{\dfrac{r^2}{4}}\\=4\times 4 \times \dfrac{kQ^2}{r^2}\\=16\ \dfrac{kQ^2}{r^2}\\=16\times 12.3442\ \ \ \ \ \ \ \ (Using\ (1))\\=197.5072.

8 0
3 years ago
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