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valentinak56 [21]
3 years ago
11

Find the resistance of a 0.03 m long copper wire with a radius of .005 m (Area of a circle: π r²).

Physics
1 answer:
Bas_tet [7]3 years ago
5 0

Answer:

578feet ygti

bkyjuuhmjkungi9

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An object accelerates at 32 m/s² when a force of 71 N is applied to it. What is the object’s mass? show your work
amid [387]

Answer:

Its  answer is 2.21 kg.

Explanation:

F =m × a

71 = m × 32

71 ÷ 32 = m

2.21 kg = m

6 0
2 years ago
Does the voltage of a battery affect the strength of an electromagnet?
swat32

I'm trying to make an electromagnet that's strength is constantly getting incremented by small amounts every second. I need to know, which would have a greater effect on the electromagnet's strength, amps or volts? (I know increasing the turns and/or density of the magnet wire will increase the strength, but I am looking for answers other than that particular one.)

7 0
2 years ago
Bob is threatening Tom’s life with a giant laser with wavelength (650 nm), a distance (D = 10 m) from the wall James is shackled
Fittoniya [83]

Answer:

He should stand from the center of laser pointed on the wall at 1.3 m.

Explanation:

Given that,

Wave length = 650 nm

Distance =10 m

Double slit separation d = 5 μm

We need to find the position of fringe

Using formula of distance

d\sin\theta=n\lambda

d\dfrac{y}{D}=n\lambda

y=\dfrac{\lambda D}{d}

Put the value into the formula

y=\dfrac{650\times10^{-9}\times10}{5\times10^{-6}}

y=1.3\ m

Hence, He should stand from the center of laser pointed on the wall at 1.3 m.

8 0
3 years ago
If it takes you 10 seconds to move a chair 5 meters across the floor, using a force of 2 Newtons, how much power did you put out
Maru [420]

Answer:

power=work done÷time taken

2×5=10

10÷10=1

ans 1J per second

5 0
2 years ago
g The electric power needs of a community are to be met by windmills with 40-m-diameter rotors. The windmills are to be located
Ksenya-84 [330]

Answer:

Explanation:

Given Data

The diameter of the wind mills is d = 40m

Velocity of the air is V = 6 m / s

Required power output is:  P ₀ = 2100 k W

Expression to calculate the exergy of the air is

E = V ² / 2

Substitute the value in above expression

E = ( 6 m / s ) ² / 2

E = 18 m ² / s ² x (1kJ/kg / 1000m²/s²)

E = 0.018 k J / k g

Expression to calculate the density of the air is

P v =m R T

m /v = P  /RT ⋯ ⋯( I )

Here  

m  is the mass of the air,  

v  is the volume of the air,  

P  is the atmospheric pressure,  

T  is the standard temperature at the atmospheric pressure and  

R  is the gas constant

As the density is

ρ = m /V

Substitute the value in expression (I)

ρ = 101  kP a /( 0.287 k J / k g ⋅ K ) ( 298 K )

ρ = 1.180 k g / m ²

Expression to calculate the mass flow rate is

m = ρ A V ⋯ ⋯ ( I I )

Here  A  is the area of the windmill

Expression to calculate the  A  is

A = π /4  d ²

Substitute the value in above expression

A = π /4 ( 40 m ²)

A = 1256.63 m ²

Substitute the value in expression (II)

m = ( 1.180 k g / m ³) ( 1256.63 m ²) ( 6 m / s )

m = 8896.94  k g / s

Expression to calculate the maximum power available to the windmill is

P w = m ( V ² /2 )

Substitute the value in above expression

P w = 8896.94  k g / s ( (6m/s)²/2 )

P w = 160144.92 W  × ( 1 W /1000 k W )

P w = 160.144 k W

Expression to calculate the number of windmills required is

n = P o /P w

Substitute the value in above expression

n=2100kw/160.144kw

n=13.11

8 0
3 years ago
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