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a_sh-v [17]
3 years ago
13

What happens to the open circuit if a small fan is connected at point f and the circuit is closed

Physics
1 answer:
liraira [26]3 years ago
4 0
There's no way to tell.  Without seeing a diagram of the circuit,
I'll need to know much more about it than you've told me.
I don't know anything about the components or power supply
that are in the circuit, and I don't know where point ' f ' is in it.

Right now, even with the copious volume of all the available
information, no answer to your question is possible. 
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Weather maps use isobar lines to show weather conditions. Which of the following information is indicated by these lines?
marishachu [46]
A its is high and low pressure areas
4 0
4 years ago
Read 2 more answers
Having landed on a newly discovered planet, an astronaut sets up a simple pendulum of length 1.38 m and finds that it makes 441
Tasya [4]
The period of a simple pendulum is given by:
T=2 \pi  \sqrt{ \frac{L}{g} }
where L is the pendulum length, and g is the gravitational acceleration of the planet. Re-arranging the formula, we get:
g= \frac{4 \pi^2}{T^2}L (1)

We already know the length of the pendulum, L=1.38 m, however we need to find its period of oscillation.

We know it makes N=441 oscillations in t=1090 s, therefore its frequency is
f= \frac{N}{t}= \frac{441}{1090 s}=0.40 Hz
And its period is the reciprocal of its frequency:
T= \frac{1}{f}= \frac{1}{0.40 Hz}=2.47 s

So now we can use eq.(1) to find the gravitational acceleration of the planet:
g= \frac{4 \pi^2}{T^2}L =  \frac{4 \pi^2}{(2.47 s)^2} (1.38 m) =8.92 m/s^2
3 0
3 years ago
Two astronauts are floating close to each other in space. Can they talk to each other without using any special device? plsss he
storchak [24]

Answer:

no they can't talk to each other bcoz of the lack of atmosphere.

Explanation:

l hope it helps you

5 0
3 years ago
A 50.0 kg object rests on a horizontal surface. The coefficient of static friction between the box and the surface is 0.300 and
MariettaO [177]

Answer:

f=140\ N

Explanation:

Given:

  • mass of the object on a horizontal surface, m=50\ kg
  • coefficient of static friction, \mu_s=0.3
  • coefficient of kinetic friction, \mu_k=0.2
  • horizontal force on the object, F=140\ N

<u>Now the value of limiting frictional force offered by the contact surface tending to have a relative motion under the effect of force:</u>

F_s=\mu_s.N

where:

N= normal force of reaction acting on the body= weight of the body

F_s=0.3\times (50\times 9.8)

F_s=147\ N

As we know that the frictional force acting on the body is always in the opposite direction:

So, the frictional force will not be at its maximum and will be equal in magnitude to the applied external force and hence the body will not move.

so, the frictional force will be:

f=140\ N

8 0
3 years ago
A 14.0 gauge copper wire of diameter 1.628 mmmm carries a current of 12.0 mAmA . Part A What is the potential difference across
NARA [144]

Answer:

a) 2.063*10^-4

b) 1.75*10^-4

Explanation:

Given that: d= 1.628 mm = 1.628 x 10-3 I= 12 mA = 12.0 x 10-8 A The Cross-sectional area of the wire is:  

A=\frac{\pi }{4}d^{2}  \\=\frac{\pi }{4}*(1.628*10^-3 m)^2\\=2.082*10^-6 m^2\\

a) <em>The Potential difference across a 2.00 in length of a 14-gauge copper  </em>

<em>    wire: </em>

  L= 2.00 m

From Table  Copper Resistivity p= 1.72 x 10-8 S1 • m The Resistance of the Copper wire is:

R=\frac{pL}{A}

   =0.0165Ω

The Potential difference across the copper wire is:  

V=IR

 =2.063*10^-4

b) The Potential difference if the wire were made of Silver: From Table: Silver Resistivity p= 1.47 x 10-8 S1 • m

The Resistance of the Silver wire is:  

R=\frac{pL}{A}

   =0.014Ω

The Potential difference across the Silver wire is:  

V=IR

 =1.75*10^-4

4 0
3 years ago
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