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Katyanochek1 [597]
3 years ago
9

How does a dynamo work

Physics
1 answer:
Vladimir [108]3 years ago
8 0
The dynamo is made up of stationary magnets which creates a powerful magnetic field, and a rotating magnet which distorts and cuts through the magnetic lines of flux of the stater. When the rotor cuts through lines of magnetic flux it makes electricity 


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an audio CD has a diameter of 120 mm and spins at up to 540 rpm. When a CD is spinning at its maximum rate, how much time is req
Andru [333]

Answer:

 t = 0.1111 s

Explanation:

Let's reduce the magnitudes to the SI system

    d = 120 mm (1m / 1000 mm)

    d= 0.120 m

    w = 540 rpm (2pi rad / 1 rev) (1 min / 60s)  

     w= 56.55 rad / s

When at maximum speed we can use angular kinematic relationships to find the time for a sperm revolution with zero angular acceleration

     W = θ / t

     t = θ / w

     t = 2π / 56.55

     t = 0.1111 s

6 0
2 years ago
The mean diameters of Mars and Earth are 6.9 ✕ 103 km and 1.3 ✕ 104 km, respectively. The mass of Mars is 0.11 times Earth's mas
Roman55 [17]

Answer:

(a) Ratio of mean density is 0.735

(b) Value of g on mars 0.920 m,/sec^2

(c) Escape velocity on earth is 3.563\times 10^4m/sec

Explanation:

We have given radius of mars R_{mars}=6.9\times 10^3km=6.9\times 10^6m and radius of earth R_{E}=1.3\times 10^4km=1.3\times 10^7m

Mass of earth M_E=5.972\times 10^{24}kg

So mass of mars M_m=5.972\times\times 0.11 \times 10^{24}=0.657\times 10^{24}kg

Volume of mars V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (6.9\times 10^6)^3=1375.357\times 10^{18}m^3

So density of mars d_{mars}=\frac{mass}{volume}=\frac{0.657\times 10^{24}}{1375.357\times 10^{18}}=477.69kg/m^3

Volume of earth  V=\frac{4}{3}\pi R^3=\frac{4}{3}\times 3.14\times (1.3\times 10^7)^3=9.198\times 10^{21}m^3

So density of earth d_{E}=\frac{mass}{volume}=\frac{5.972\times 10^{24}}{9.198\times 10^{21}}=649.271kg/m^3

(A) So the ratio of mean density \frac{d_{mars}}{d_E}=\frac{477.69}{649.27}=0.735

(B) Value of g on mars

g is given by g=\frac{GM}{R^2}=\frac{6.67\times 10^{-11}\times0.657\times 10^{24}}{(6.9\times 10^6)^2}=0.920m/sec^2

(c) Escape velocity is given by

v=\sqrt{\frac{2GM}{R}}=\sqrt{\frac{2\times 6.67\times 10^{-11}\times 0.657\times 10^{24}}{6.9\times 10^6}}=3.563\times 10^4m/sec

5 0
3 years ago
Read 2 more answers
While standing on a balcony a child drops a penny. The penny lands on the ground floor 1.5 s later. How fast was the penny trave
sveticcg [70]

Answer:

14.7 m/s.

Explanation:

From the question given above, the following data were obtained:

Time (t) = 1.5 s

Acceleration due to gravity (g) = 9.8 m/s².

Height = 11.025 m

Final velocity (v) = 0 m/s

Initial velocity (u) =?

We, can obtain the initial velocity of the penny as follow:

H = ½(v + u) t

11.025 = ½ (0 + u) × 1.5

11.025 = ½ × u × 1.5

11.025 = u × 0.75

Divide both side by 0.75

u = 11.025/0.75

u = 14.7 m/s

Therefore, the penny was travelling at 14.7 m/s before hitting the ground.

8 0
3 years ago
What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t
Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

\Delta P =  \frac{8 \mu L Q}{\pi r^4}

where:

\Delta P is the pressure difference between the two ends

\mu is viscosity of the fluid

L is the length of the pipe

Q=Av is the volumetric flow rate, with A=\pi r^2 being the section of the tube and v the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

\mu=0.001 Pa/s is the dynamic water viscosity at 20^{\circ}

L=4.0 cm=0.04 m

Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s

and r=1 mm=0.001 m

Using these data in the formula, we get:

\Delta P = 3200 Pa

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0
3 years ago
Here is a "formula" for building a model airplane: 1 body + 2 wings + 4 jet engines + 1 tailpiece → 1 model airplane if a hobbyi
WINSTONCH [101]
Your answer is 44 wings

3 0
3 years ago
Read 2 more answers
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