n Section 12.3 it was mentioned that temperatures are often measured with electrical resistance thermometers made of platinum wire. Suppose that the resistance of a platinum resistance thermometer is 125 Ω when its temperature is 20.0°C. The wire is then immersed in boiling chlorine, and the resistance drops to 99.6 Ω. The temperature coefficient of resistivity of platinum is α = 3.72 × 10−3(C°)−1. What is the temperature of the boiling chlorine?

Answers

Answer 1

Answer:

Answer:

- 48.55 degree

Explanation:

Resistance increases when temperature increases. This increase is linear and which is defined by the following relation between resistance and temperature

$R_t=R_0(1-\alpha* t)$

$R_t$ is resistance at temperature t , R₀ is resistance at temperature t₀ , t is rise in temperature and α is temperature coefficient of resistance .

Putting the values we get

125 = 99.6 ( 1 + 3.72x 10⁻³ x t )

t = 68.55

Therefore temperature of boiling chlorine is 20-68.55 = - 48.55 degree celsius .

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Answers

This is an example of an elastic collision. The two objects collide and return to their original shapes and move separately. In such a collision, kinetic energy is conserved. I think we can agree that this represents Newton's third law by demonstrating conservation of momentum.

Answer:

Yes, the law of conservation of momentum is satisfied. The total momentum before the collision is 1.5 kg • m/s and the total momentum after the collision is 1.5 kg • m/s. The momentum before and after the collision is the same.

Explanation:

A pitcher is in 85° of abduction, holding a 1.4 N baseball at point C, 65 cm from the joint axis at point O • The center of gravity of his arm is 25 cm from the joint axis of shoulder abduction at point O • The weight of the arm W is 0.06 of the pitcher’s weight of 100 N • Deltoids muscles are at an angle θ of 15° with respect to the humerus and insert 15 cm from the joint axis at point A • Determine the force applied by the Deltoid muscles and the joint reaction force at the shoulder joint and its orientation β

Answers

I attached a Diagram for this problem.

We star considering the system is in equlibrium, so

Fm makes $90-(\theta+5)$ with vertical

Fm makes 70 with vertical

Applying summatory in X we have,

$\sum F_x = 0$

$W+1.4-Fm cos(70)$

We know that W is equal to

$W= 0.06*100N = 6N$

Substituting,

$Fm cos (70) = W+1.4N$

$Fm cos (70) = 6N + 1.4N$

$Fm = (7.4)/(cos(70))$

$Fm = 21.636N$

For the second part we know that the reaction force Fj on deltoid Muscle is equal to Fm, We can assume also that $\beta = \theta$

To estimate the width of a valley, a climber starts a stopwatch as he shouts. He hears an echo from the opposite side of the valley after 1.0s. The sound travels at 340m/ s. What is the width of the valley?

Answers

$\tt w/d=(v.t)/(2)$

w = width/distance

v = velocity

t = time

input the value:

$\tt w=(340* 1)/(2)=170~m$

A 6.0-cm-diameter horizontal pipe gradually narrows to 4.0 cm. When water flows through this pipe at a certain rate, the gauge pressure in these two sections is 32.0 kPa and 24.0 kPa, respectively. What is the volume rate of flow?

Answers

Answer:

a n c

Explanation:

Final answer:

The volume rate of flow can be determined using the equation Q = Av, where Q is the volume rate of flow, A is the cross-sectional area of the pipe, and v is the average speed of the water. Given the diameter of the wider section of the pipe is 6.0 cm and the gauge pressure is 32.0 kPa, we can calculate the volume rate of flow using the provided information. The volume rate of flow is found to be 0.0018 m³/s.

Explanation:

The volume rate of flow can be determined using the equation Q = Av, where Q is the volume rate of flow, A is the cross-sectional area of the pipe, and v is the average speed of the water.

Given that the diameter of the wider section of the pipe is 6.0 cm, the radius is 3.0 cm and the gauge pressure is 32.0 kPa. Similarly, for the narrower section with a diameter of 4.0 cm, the radius is 2.0 cm and the gauge pressure is 24.0 kPa.

Using the equation Q = Av and the fact that the flow rate must be the same at all points along the pipe, we can set up the equation A₁v₁ = A₂v₂. Solving for v₂, we have v₂ = A₁v₁ / A₂ = πr₁²v₁ / πr₂², where r₁ is the radius of the wider section and r₂ is the radius of the narrower section.

Substituting the values, we get v₂ = (3.14)(3.0 cm)²(32.0 kPa) / [(3.14)(2.0 cm)²] = 18.0 cm/s. Since v = d/t, we can convert cm/s to m³/s by multiplying by 0.0001, so the volume rate of flow is 0.0018 m³/s.

Learn more about volume rate of flow here:

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