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Dmitry_Shevchenko [17]
3 years ago
13

عند زيادة تردد الدينامو . فماذا يحدث لشده التيار المار بالدائرة ​

Physics
1 answer:
saveliy_v [14]3 years ago
4 0

Answer:

تقل شدة التيار

Explanation:

عند زيادة التردد تزداد سرعة الدوران وهذا يعني زيادة في الفولتية.

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Consider water flowing through a cylindrical pipe with a variable cross-section. The velocity is v at a point where the pipe dia
Harrizon [31]

Answer:

one ninth

Explanation:

d = 1 cm , v = v

D = 3d, V = ?

By the equation of continuity,

A V = a v

3.14 x D^2 / 4 x V = 3.14 x d^2 / 4 x v

9d^2 x V = d^2 x v

V = v / 9

Thus, the velocity becomes one ninth the initial velocity

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Mandarinka [93]

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Explanation:

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3 years ago
Find the work done in pumping gasoline that weighs 6600 newtons per cubic meter. A cylindrical gasoline tank 3 meters in diamete
Sonbull [250]

Answer:

<em>work done in pumping the entire fuel is 466587 J</em>

<em></em>

Explanation:

weight of the gasoline per volume = 6600 N/m^3

diameter of the tank = 3 m

length of the tank = 2 m

height of the tractor tank above the top of the tank = 5 m

work done in pumping fuel to this height = ?

First, we find the volume of the fuel

since the tank is cylindrical,<em> we assume that the fuel within also takes the cylindrical shape.</em>

<em>Also, we assume that the fuel completely fills the tank.</em>

volume of a cylinder = \pi r^{2}l

where r = radius = diameter ÷ 2 = 3/2 = 1.5 m

volume of the cylinder = 3.142 x 1.5^{2} x 2 = 14.139 m^3

we then find the total weight of the fuel in Newton

total weight = (weight per volume) x volume

total weight = 6600 x 14.139 = 93317.4 N

work done = (total weight of the fuel) x (height through which the fuel is pumped)

work done in pumping = 93317.4 x 5 = <em>466587 J</em>

8 0
3 years ago
A high-speed railway car goes around a flat, horizontal circle of radius 480 m at a constant speed. The magnitudes of the horizo
kvv77 [185]

Answer:

Net force ( F ) = 208 N.

Speed of car ( v ) = 68.58 m/s.

Explanation:-

The radius ( r ) = 480 m.

Mass of passenger ( m ) = 65 kg.

Horizontal force ( f_{x} ) = 208 N.

Vertical force ( f_{n} ) = 637 N.

Net vertical component of force ( f_{x} ) = m*g - 637   = 65 * 9.8 - 637 =  0 N.

a ) magnitude of net force .

F = \sqrt{f^{2} _{x} + f^{2} _{y}  }   = \sqrt{208^{2} + 0^{2} }   = 208 N.

b) The speed of car.

f_{n}  = \frac{m*v^{2} }{r}    where v is the velocity.

637 = \frac{65 *v^{2} }{480}

v² = 4704

v = 68.58 m/s.

6 0
3 years ago
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