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vivado [14]
3 years ago
15

A Nichrome wire 44 cm long and 0.30 mm in diameter is connected to a 3.1 V flashlight battery. What is the electric field inside

the wire? (Enter the magnitude.)
Physics
1 answer:
Alexeev081 [22]3 years ago
4 0

Answer:

7.05 Volts/m

Explanation:

L = length of the Nichrome wire = 44 cm = 0.44 m

V = Potential difference across the end of the wire = battery voltage = 3.1 Volts

E = magnitude of electric field inside the wire

Magnitude of electric field inside the wire is given as

E = \frac{V}{L}

Inserting the values

E = \frac{3.1}{0.44}

E = 7.05 Volts/m

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If freshwater consumption was greater than freshwater renewal.

Explanation:

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3 years ago
A camera lens with focal length f = 50 mm and maximum aperture f>2
Brut [27]

Answer:

The minimum distance between two points on the  object that are barely resolved is 0.26 mm

The corresponding distance between the  image points = 0.0015 m

Explanation:

Given  

focal length f = 50 mm and maximum aperture f>2

s =  9.0 m

aperture = 25 mm = 25 *10^-3 m

Sin a = 1.22 *wavelength /D  

Substituting the given values, we get –  

Sin a = 1.22 *600 *10^-9 m /25 *10^-3 m

Sin a = 2.93 * 10 ^-5 rad

Now  

Y/9.0 m = 2.93 * 10 ^-5

Y = 2.64 *10^-4 m = 0.26 mm

Y’/50 *10^-3 = 2.93 * 10 ^-5  

Y’ = 0.0015 m

8 0
2 years ago
Match the following items.
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Je ne Sachez que Qu’est-ce que le

8 0
2 years ago
If i ask wat r the application of simple machine some of u will say-----> Simple machines that are widely used include the wh
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3 0
3 years ago
In February 1955, a paratrooper fell 370 m from an airplane without being able to open his chute but happened to land in snow, s
nevsk [136]

a) 0.94 m

The work done by the snow to decelerate the paratrooper is equal to the change in kinetic energy of the man:

W=\Delta K\\-F d = \frac{1}{2}mv^2 - \frac{1}{2}mu^2

where:

F=1.1 \cdot 10^5 N is the force applied by the snow

d is the displacement of the man in the snow, so it is the depth of the snow that stopped him

m = 68 kg is the man's mass

v = 0 is the final speed of the man

u = 55 m/s is the initial speed of the man (when it touches the ground)

and where the negative sign in the work is due to the fact that the force exerted by the snow on the man (upward) is opposite to the displacement of the man (downward)

Solving the equation for d, we find:

d=\frac{1}{2F}mu^2 = \frac{(68 kg)(55 m/s)^2}{2(1.1\cdot 10^5 N)}=0.94 m

b) -3740 kg m/s

The magnitude of the impulse exerted by the snow on the man is equal to the variation of momentum of the man:

I=\Delta p = m \Delta v

where

m = 68 kg is the mass of the man

\Delta v = 0-55 m/s = -55 m/s is the change in velocity of the man

Substituting,

I=(68 kg)(-55 m/s)=-3740 kg m/s

7 0
3 years ago
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