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Ksenya-84 [330]
3 years ago
6

Jadeen says that you can increase the resistance of a copper wire by hammering the wire to make it narrower and longer. Arnell s

ays that you can increase its resistance by heating the wire. Which one, if either, is correct, and why?
1) Arnell, because the conductivity of the wire increases when it is heated.
2) Arnell, because the conductivity of the wire decreases when it is heated.
3) Jadeen, because the conductivity of a wire is directly proportional to its area and inversely proportional to its length.
4) Jadeen, because the conductivity of a copper wire does not increase and might decrease when it is hammered.
5) Both are correct because (b) and (d) are both correct.
Physics
1 answer:
katrin [286]3 years ago
5 0

Answer:

The answer is "Option 5".

Explanation:

Jadeen claims that only by hitting the wire to make it thinner and wider, it can improve a copper wire's strength, which is why a copper wire's permeability doesn't quite improve and can reduce once it is pounded. Arnell says so by heating its wire, it can improve its strength, and when it is heated, the wire's permeability reduces.

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Classify each of the following hydro carbons as alkane,alkene or alkyne
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Answer:

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3 years ago
A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing
Damm [24]

Given:

Shaft Power, P = 7.46 kW = 7460 W

Speed, N = 1200 rpm

Shearing stress of shaft, \tau _{shaft} = 30 MPa

Shearing stress of key, \tau _{key} = 240 MPa

width of key, w = \frac{d}{4}

d is shaft diameter

Solution:

Torque, T = \frac{P}{\omega }

where,

\omega = \frac{2\pi  N}{60}

T = \frac{7460}{\frac{2\pi  (1200 )}{60}} = 59.365 N-m

Now,

\tau _{shaft} = \tau _{max} = \frac{2T}{\pi (\frac{d}{2})^{3}}

30\times 10^{6} = \frac{2\times 59.365}{\pi (\frac{d}{2})^{3}}

d = 0.0216 m

Now,

w =  \frac{d}{4} =  \frac{0.02116}{4} = 5.4 mm

Now, for shear stress in key

\tau _{key} = \frac{F}{wl}

we know that

T = F \times r =  F. \frac{d}{2}

⇒ \tau _{key} = \frac{\frac{T}{\frac{d}{2}}}{wl}

⇒ 240\times 10^{6} = \frac{\frac{59.365}{\frac{0.0216}{2}}}{0.054l}

length of the rectangular key, l = 4.078 mm

7 0
3 years ago
Read 2 more answers
Si aplicamos una fuerza constante de 30 N sobre un cuerpo de 25 Kg, este se mueve de tal manera que en 5 s adquiere la velocidad
Ganezh [65]

Answer:

<em>Si hay rozamiento y el valor de la fuerza de roce es 10 N</em>

Explanation:

<u>Fuerza Neta</u>

La fuerza neta sobre un cuerpo es la suma vectorial de todas las fuerzas actuantes sobre el mismo.

Si conocemos el módulo de la fuerza neta F y la masa m del cuerpo, aplicamos la segunda ley de Newton para relacionarlas con la aceleración a:

F=m.a

Tenemos los datos cinemáticos de la situación, según la cual el cuerpo adquiere una velocidad (desde el reposo) de 4 m/s en 5 s.

Utilizamos la fórmula:

v_f=v_o+a.t

Y despejamos la aceleración:

\displaystyle a=\frac{v_f-v_o}{t}

\displaystyle a=\frac{4-0}{5}

a=0.8 \ m/s^2

Podemos calcular la aceleración real que el cuerpo adquiere, producto de una fuerza efectiva igual a:

F_e=25\ Kg\cdot 0.8 \ m/s^2

F_e=20\ N

Si se está aplicando una fuerza de F_a= 30 N y solo 20 N producen movimiento, entonces se está perdiendo en rozamiento una fuerza:

F_r=F_a-F_e=30 - 20=10

F_r=10\ N

Si hay rozamiento y el valor de la fuerza de roce es 10 N

8 0
3 years ago
A bicycle wheel has a diameter of 63.0 cm and a mass of 1.75 kg. Assume that the wheel is a hoop with all of the mass concentrat
Masteriza [31]

Answer:

F2 = 834 N

Explanation:

We are given the following for the bicycle;

Diameter; d1 = 63 cm = 0.63 m

Mass; m = 1.75 kg

Resistive force; F1 = 121 N

For the sprocket, we are given;

Diameter; d2 = 8.96 cm = 0.0896 m

Radius; r2 = 0.0896/2 = 0.0448 m

Radial acceleration; α = 4.4 rad/s²

Now moment of inertia of the wheel which is assumed to be a hoop is given by; I = m(r1)²

Where r1 = (d1)/2 = 0.63/2

r1 = 0.315 m

Thus, I = 1.75 × 0.315²

I = 0.1736 Kg.m²

The torque is given by the relation;

I•α = F1•r1 - F2•r2

Where F2 is the force that must be applied by the chain to give the wheel an acceleration of 4.40 rad/s².

Thus;

0.1736 × 4.4 = (121 × 0.315) - (0.0448F2)

>> 0.76384 = 38.115 - (0.0448F2)

>> 0.0448F2 = 38.115 - 0.76384

>> F2 = (38.115 - 0.76384)/0.0448

>> F2 = 833.73 N

Approximately; F2 = 834 N

7 0
3 years ago
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