During which change of state would the volume of a aubstance increase the most ?

Tungsten has a temperature coefficient of resistivity of 0.0045 (c°)-1. a tungsten wire is connected to a source of constant vol

Murljashka [212]

Answer:

$131.1^{\circ}C$

Explanation:

The power delivered in the wire is given by:

$P=\frac{V^2}{R}$

where V is the voltage of the battery and R is the resistance of the wire.

Since the voltage of the battery is constant, we can rewrite this equation as follows:

$V^2 = PR=const.$ (1)

At the beginning, the initial resistance is $R_0$ , and the power delivered is $P_0$ . Later, when the temperature increases, the power becomes $P_1 = \frac{2}{3}P_0$ , and the new resistance is $R_1$ . Using (1), we can write

$P_0 R_0 = \frac{2}{3}P_0 R_1\\R_1 = \frac{3}{2}\frac{P_0 R_0}{P_0}=\frac{3}{2}R_0$ (2)

So, the new resistance must be 3/2 of the initial resistance.

We know that the resistance increases linearly with the temperature, as

$R_1 = R_0 (1+\alpha \Delta T)$

where

$\alpha = 0.0045 ^{\circ}C^{-1}$ is the temperature coefficient

$\Delta T$ is the change in temperature

Using (2), we can rewrite this equation as

$\frac{3}{2}R_0 = R_0(1+ \alpha \Delta T)$

and we find:

$\frac{3}{2}=1+\alpha \Delta T\\\Delta T=\frac{\frac{3}{2}-1}{\alpha}=111.1 ^{\circ}$

So, the new temperature of the wire must be

$T_f = 21^{\circ}+111.1^{\circ}=132.1^{\circ}$

6 0

3 years ago

The cross-section of a pipe has an area of 2.0 cm2. Calculate the flow rate (in grams per second) of a liquid with p=1.0 g/cm3 t

Maurinko [17]

flow rate = velocity of fluid x area of cross section x density
= 42 x 2 x 1 = 84 g/s

4 0

4 years ago

Shakina and Juliette set the car's initial velocity to zero and set the acceleration to +1.2 m/s2, then clicked "start." Answer

Katarina [22]

Let's start by differentiating the terms distance and displacement. They both refer to the length of paths. Distance only accounts for the total length regardless of the path taken. Displacement measures the linear path from the starting point to the end point. So, it does not necessarily follow the actual path. However, for this problem, assuming that the path is just in one direction, displacement and distance would just be equal. The equation would be:

Distance = Displacement = v₀t + 0.5at² = 0(10 s) + 0.5(+1.2 m/s²)(10 s)²
Distance = Displacement = 60 meters

6 0

3 years ago

A jogger whose mass is 65 kg is moving at a velocity of 4 m/s. What is the

mina [271]

Answer:

KE = 1/2 m v^2

= 520 joules

4 0

3 years ago

A road with a radius of 75.0 m is banked so that a car can navigate the curve at a speed of 15.0 m/s without any friction. When

sesenic [268]

Find the angle θ made by the road. When rounding the curve at 15.0 m/s, the car has a radial acceleration of

a = (15.0 m/s)² / (75.0 m) = 3.00 m/s²

There are two forces acting on the car in this situation:

• the normal force of the road pushing upward on the car, perpendicular to the surface of the road, with magnitude n

• the car's weight, pointing directly downward; its magnitude is mg (where m is the mass of the car and g is the acceleration due to gravity), and hence its perpendicular and parallel components are, respectively, -mg cos(θ) and mg sin(θ)

By Newton's second law, the net forces in the perpendicular and parallel directions are

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) = ma

==> sin(θ) = a/g ==> θ = arcsin(a/g) ≈ 17.8°

(Notice that in the paralell case, the positive direction points toward the center of the curve.)

When rounding the curve at 31.8 m/s, the car's radial acceleration changes to

a = (31.8 m/s)² / (75.0 m) ≈ 13.5 m/s²

and there is now static friction (mag. f = µn, where µ is the coefficient of static friction) acting on the car and keeping from sliding off the road, hence pointing toward the center of the curve and acting in the parallel direction. Newton's second law gives the same equations, with an additional term in the parallel case:

(perp.) ∑ F = n - mg cos(θ) = 0

(para.) ∑ F = mg sin(θ) + f = ma

The first equation gives

n = mg cos(θ)

and substituting into the second equation, we get

mg sin(θ) + µmg cos(θ) = ma

==> µ = (a - g sin(θ)) / (g cos(θ)) = a/g sec(θ) - tan(θ) ≈ 1.12

5 0

3 years ago

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