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love history [14]
3 years ago
14

La resistencia de un termómetro de platino es de 6Ω a30°C. Hallar su valor correspondiente a 100°C,sabiendo que el coeficiente d

e temperatura de resistividad del platino vale 0,00392°C^(-1).
Physics
1 answer:
Sonja [21]3 years ago
6 0

Respuesta:

7,6 Ω

Explicación:

Paso 1: Información dada

  • Resistencia a 30 °C (R₀): 6 Ω
  • Coeficiente de temperatura (α): 0,00392 °C⁻¹

Paso 2: Hallar la resistencia (R) a 100 °C

Podemos ver la relación entre la resistencia de un material y la temperatura usando la siguiente ecuación.

R = R₀ (1  + α × ΔT)

R = 6 Ω (1  + 0,00392 °C⁻¹ × (100 °C - 30 °C)) = 7,6 Ω

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The ideal mechanical advantage of a pulley system is equal to the?
marishachu [46]

Answer: The correct answer is "Number of rope segments supporting the load".

Explanation:

Mechanical advantage: It is defined as the ratio of the force produced by a machine to the force applied on the machine. The ideal mechanical advantage of a machines is mechanical advantage in the absence of friction.

The ideal mechanical advantage of a pulley system is equal to the number of rope segments which is supporting the load. More the rope segments, It is more helpful to do the lifting the work.

It means that less force is needed for this task to complete.

Therefore, the correct option is (C).

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3 years ago
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Which of the following is not an element of installment credit?
Alex787 [66]

Answer:

c line of credit I believe

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Why are the wheels on a roller coaster hot after a ride. (I know it’s due to friction but I need more of a detailed answer).
telo118 [61]
2 someone's glasses could fall of because of the inertia
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8 0
3 years ago
4. The blades on a fan have a frequency of 15 Hz.
vichka [17]

Answer:

a) 4500 cycles b) 0.0667s c) 6.67s

Explanation:

a) 15 Hz= 15 cycles/ s

   5 mins= 300s

   15 cycles/s * 300s= 4500 cycles

b) Period= 1/ frequency

   Period= 1/ 15 cycles/s

   Period= 0.0667s

c) Period * number of revolutions= time

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6 0
3 years ago
A stone with a mass of 0.100kg rests on a frictionless, horizontal surface. A bullet of mass 2.50g traveling horizontally at 500
jolli1 [7]

Answer:

Explanation:

Given that:

mass of stone (M) = 0.100 kg

mass of bullet (m) = 2.50 g = 2.5  ×10 ⁻³ kg

initial velocity of stone (u_{stone}) = 0 m/s

Initial velocity of bullet (u_{bullet}) = (500 m/s)i

Speed of the bullet after collision (v_{bullet}) = (300 m/s) j

Suppose we represent (v_{stone}) to be the velocity of the stone after the truck, then:

From linear momentum, the law of conservation can be applied which is expressed as:

m*u_{bullet} + M*{u_{stone}}= mv_{bullet}+Mv_{stone}

(2.50*10^{-3} \ kg) (500)i+0 = (2.50*10^{-3} \ kg)(300 \ m/s)j + (0.100 \ kg)v_{stone}

(2.50*10^{-3} \ kg) (500)i- (2.50*10^{-3} \ kg)(300 \ m/s)j=  (0.100 \ kg)v_{stone}

v_{stone}= (1.25\  kg.m/s)i-(0.75\ kg m/s)j

v_{stone}= (12.5\  m/s)i-(7.5\ m/s)j

∴

The magnitude now is:

v_{stone}=\sqrt{ (12.5\  m/s)^2-(7.5\ m/s)^2}

\mathbf{v_{stone}= 14.6 \ m/s}

Using the tangent of an angle to determine the direction of the velocity after the struck;

Let θ represent the direction:

\theta = tan^{-1} (\dfrac{-7.5}{12.5})

\mathbf{\theta = 30.96^0 \ below \ the \ horizontal\ level}

5 0
3 years ago
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