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dimaraw [331]
4 years ago
8

Which is true about the resistivities of a metal and a semiconductor as the temperature is increased?

Physics
2 answers:
allochka39001 [22]4 years ago
7 0

Answer:

The resistivity of metals decreases on increase in temperature whereas the resistivity of semiconductors increases with increase in temperature.

Explanation:

The majority charge carriers in metals are electrons which are explained by the electron-sea model. The electron sea model considers metal to have a sea of conducting electrons in between the nucleus arrangement.

The nucleus is bind firmly within the lattice structure and when the temperature has increased this accounts for the lattice vibrations as a result offering obstructions to the valence electrons which increases its bulk resistivity.

The variation in the resistivity of metal is given by:

\rho=\rho_0[1+\alpha_t(T-T_0)]

where:

\alpha_t=temperature coefficient of resistivity which is defined as the fractional change in resistivity per unit change in temperature.

\rm \rho\ and\ \rho_0\ are\ the\ resistivity\ at\ temperature\ T\ and\ T_0\ respectively

At temperatures 500 K above the room temperature we have the variation of resistivity of the metal in a linear manner.

  • For semiconductors, the value of coefficient of resistivity is negative which implies that the resistivity of semiconductors decreases as temperature increases.
MAVERICK [17]4 years ago
4 0

Answer:

Explanation:

According to the electrical conductivity, there are three types of materials:

Metals are the conductors which allows the electric current to flow through them. For example, iron , copper, silver, etc.

As the temperature of a metal increases the resistivity also increases.

Semiconductors are the material which are insulators at room temperature but they behaves like a conductor as the temperature increases.

The resistivity of a semiconductor decreases as the temperature increases.

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A pyrotechnical releases a 3 kg firecracker from rest. at t=0.4 s, the firecracker is moving downward with a speed 4 m/s. At the
olga2289 [7]

Answer:

a) F = 30 N, b)   I = 12 N s , c)  I = -12 N s , d) ΔI = 0 N s

Explanation:

This exercise is a case at the moment, let's define the system formed by the firecracker and its two parts, in this case the forces during the explosion are internal and the moment is conserved

Initial, before the explosion

     p₀ = m v

The speed can be found by kinematics

     v = v₀ - g t

     v = 0 - 10 0.4

     v = -4.0 m / s

Final after division

     pf = m₁ v₁f + m₂ v₂f

    p₀ = pf

    M v = m₁ v₁f + m₂ v₂f

Where M is the initial mass (M = 3 kg), m₁ is the mass mtop (m₁ = 1 kg) and m₂ in the mass m botton (m₂ = 2kg) and the piece that moves up (v₁f = 6m/s )

a) before the explosion the only force acting on the body is gravity

     F = mg

     F = 3 10 = 30 N

b) The expression for momentum is

     I = Ft

Before the explosion the only force that acts is the weight

    I = mg t

    I = 3 10 0.4

    I = 12 N s

c) To calculate this part we use the conservation of the moment and calculate the speed of the body that descends body 2

    M v = m₁ v₁f + m₂ v₂f

    v₂f = (M v - m₁ v₁f) / m₂

    v₂f = (3 (-4) - 1 6) / 2

   v₂f = - 9 m / 2

The negative sign indicates that body 2 (botton) is descending

Now we can use the momentum and momentum relationship for the body during the explosion

    I = F t = Dp

   F t = pf –po)

   F t= [m₁ v₁f + m₂ v₂f]

   

   I = [1 6 + 2 (-9) -0]

   I = -12 N s

This is the impulse during the explosion the negative sign indicates that it is headed down

d) impulse change

I₀ = Mv

I₀ = 3 *4

I₀ =-12 N s

 ΔI =If – I₀  

ΔI = - 12 – (-12)

ΔI = -0 N s

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3 years ago
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Why is energy required to get an object moving?
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Answer:

this is what popped up when I searched it up:In physics, the kinetic energy (KE) of an object is the energy that it possesses due to its motion. It is defined as the work needed to accelerate a body of a given mass from rest to its stated velocity. Having gained this energy during its acceleration, the body maintains this kinetic energy unless its speed changes.

Explanation:

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During a baseball game, a batter hits a high pop-up. If the ball remains in the air for 4.02 s, how high above the point where i
alukav5142 [94]

Answer:

d = 19.796m

Explanation:

Since the ball is in the air for 4.02 seconds, the ball should reach the maximum point from the ground in half the total time, therefore, t=2.01s to reach maximum height. At the maximum height, the velocity in the y-direction is 0.

So we know t=2.01, vi=0, g=a=9.8m/s and we are solving for d.

Next, you look for a kinematic equation that has these parameters and the one you should choose is:

d=vt+\frac{1}{2}at^2

Now by substituting values in, we get

d=(0\frac{m}{s})*(2.01s)+\frac{1}{2}(9.8\frac{m}{s^2})(2.01)^2

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