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Elden [556K]
3 years ago
6

When operated on a household 110.0-V line, typical hair dryers draw about 1650 W of power. We can model the current as a long st

raight wire in the handle. During use, the current is about 2.25 cm from the user\'s hand. What is the current in the dryer
Physics
1 answer:
defon3 years ago
4 0

Answer:

Current in the hair dryer will be equal to 15 A

Explanation:

We have given that household is operated at 110 volt

So potential difference V =110 volt

Power drawn by hairdryer is P = 1650 watt

We have to find the current in the hair dryer

We know that power is given as P = VI, here V is potential difference and I is current

So 1650=110\times I

I = 15 A

So current in the hair dryer will be equal to 15 A

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If velocity is 0.2m/s, what is the kinetic energy?
Natalija [7]

\huge\boxed{♧ \: \mathfrak{ \underline{Answer} \: ♧}}

we know,

\boxed{kinetic  \:  \: energy =  \frac{1}{2} m {v}^{2} }

So,

\longmapsto \dfrac{1}{2}  \times 10 \times 0.2 \times 0.2

\longmapsto0.2 \: joules

3 0
3 years ago
What an image of a distant object is brought into focus in front of a persons retina the fact is called
beks73 [17]
When the image of a distant object is brought into focus of front of a person's retina, the defect is called: nearsightedness.
7 0
3 years ago
A 250 kg flatcar 25 m long is moving with a speed of 3.0 m/s along horizontal frictionless rails. A 61 kg worker starts walking
Anna11 [10]

Answer:

x=31.09m

Explanation:

p1=p2

The momentum of flatcar and the momentum of the worker so

The velocity of the worker is:

m_{f}*v_{f}=m_{w}*v_{w}\\\\v_{f}=\frac{m_{f}*v_{f}}{m_{w}}\\v_{f}=\frac{61kg*3.0\frac{m}{s}}{250kg}\\v_{f}=0.732\frac{m}{s}

The total motion has a total velocity and is

Vt=v_{w}+v_{f}\\Vt=0.732\frac{m}{s}+3.0\frac{m}{s}\\Vt=3.732\frac{m}{s}

The time the worker take walking is

t=\frac{x}{v_{w}}\\t=\frac{25m}{3\frac{m}{s}}=8.33s

Now the total time and the total velocity determinate the motion of tha flatcar how far has moved

x=t*Vt\\x=8.33s*3.732\frac{m}{s} \\x=31.09m

5 0
3 years ago
Two sources of light of wavelength 700 nm are 9 m away from a pinhole of diameter 1.2 mm. How far apart must the sources be for
Lelechka [254]

Answer:

The distance is  D  =  0.000712 \ m

Explanation:

From the question we are told that

    The wavelength of  the  light source is  \lambda  =  700 \ nm = 700 *10^{-9} \  m

     The distance from a pin hole is  x  =  9\ m

       The  diameter of the pin  hole is  d =  1.2 \ mm  =  0.0012 \ m

     

Generally the distance which the light source need to be in order for their diffraction patterns to be resolved by Rayleigh's criterion is

mathematically represented as

              D  =  \frac{1.22 \lambda }{d }

substituting values

             D  =  \frac{1.22 * 700 *10^{-9} }{ 0.0012 }

             D  =  0.000712 \ m

5 0
3 years ago
Find the current in the 12 ohm resistor.
disa [49]

Answer:

1.5 A

Explanation:

V =I.R

18 = I × 12

I = 18/12

= 3/2 = 1.5 A

3 0
3 years ago
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