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Colt1911 [192]
3 years ago
9

what is the relationship between the number of turns of the wire and electrical output of the generator?

Physics
1 answer:
xxTIMURxx [149]3 years ago
4 0

The number of turns of wire is directly proportional to voltage.

The higher the number of turns of wire the higher the voltage

Since .

Power (output) of a generator = voltage x current. Therefore the higher the voltage the higher the output of the generator.

So, the higher the number of turns of wire the higher the output of the generator.

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Biologists think that some spiders "tune" strands of their web to give enhanced response at frequencies corresponding to those a
Nat2105 [25]

Answer:

Explanation:

Given

diameter d=20 \mu m

density \rho =1300 kg/m^3

frequency \nu =150 Hz

Length of silk strand L=14 cm

Velocity in the string is as follows

\nu =\sqrt{\frac{T}{\mu }}

The expression for Fundamental Frequency

f=\frac{\nu }{2l}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\mu }}

f=\frac{1}{2l}\times \sqrt{\frac{T}{\frac{m}{l}}}

f=\frac{1}{2l}\times \sqrt{\frac{Tl}{\rho V}}

Squaring

f^2=\frac{1}{4l^2}\times \frac{Tl}{\rho V}

T=4\rho \cdot A\cdot l^2\cdot f^2

T=4\times 1300\times \frac{\pi }{4}(20\times 10^{-6})^2\times (0.14)^2\times 150^2

T=7.2\times 10^{-4} N

8 0
3 years ago
You are driving at the speed of 33.4 m/s (74.7296 mph) when suddenly the car in front of you (previously traveling at the same s
romanna [79]

Answer:

Part a)

a = - 8.45 m/s/s

Part b)

d = 66 m

Part c)

d = 14.16 m

Explanation:

Part a)

when car apply brakes then the friction force on the car in front of us is given as

F_f = \mu mg

here we need to find deceleration due to friction

a = \frac{F_f}{m}

a = -\mu g

a = -8.45 m/s^2

Part b)

Braking distance of the car is the distance that it move till it stops

so we will have

v_f^2 - v_i^2 = 2ad

0 - 33.4^2 = 2(-8.45)d

d = 66 m

Part c)

Since we know that average reaction time for human is 0.424 s

now we know that during reaction time our car will travel at uniform speed

so we will have

d = vt

d = (33.4) (0.424)

d = 14.16 m

4 0
2 years ago
What holds the moon in place, orbiting around earth?
pshichka [43]

Answer:

Gravity

Explanation:

5 0
1 year ago
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i'm begging ya'll i posted this about 3 times already and i'm really stuck. can someone please give me the answers to the cart?
Vedmedyk [2.9K]

Answer:

just put the molecules and atoms where they belong

Explanation:

4 0
2 years ago
A space vehicle approaches a space station in orbit. The intent of the engineers is to have the vehicle slowly approach, reducin
N76 [4]

Answer: The total momentum before the docking maneuver is mV_{1}+MV_{2} and after the docking maneuver is (m+M) U

Explanation:

Linear momentum p (generally just called momentum) is defined as mass in motion and is given by the following equation:  

p=m.v  

Where m is the mass of the object and v its velocity.

According to the conservation of momentum law:

<em>"If two objects or bodies are in a closed system and both collide, the total momentum of these two objects before the collision </em>p_{i} <em>will be the same as the total momentum of these same two objects after the collision </em>p_{f}<em>". </em>

<em />

p_{i}=p_{f}

This means, that although the momentum of each object may change after the collision, the total momentum of the system does not change.

Now, the docking of a space vehicle with the space station is an inelastic collision, which means both objects remain together after the collision.

Hence, the<u> initial momentum</u> is:

p_{i}=mV_{1}+MV_{2}

Where:

m is the mass of the vehicle

V_{1} is the velocity of th vehicle

M is the mass of the space station

V_{2} is the velocity of the space station

And the <u>final momentum</u> is:

p_{f}=(m+M)U

Where:

U is the velocity of the vehicle and space station docked

6 0
3 years ago
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