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zalisa [80]
3 years ago
8

El coeficiente de variación de la resistencia con la temperatura del carbón es -0.0005/°c.Si la resistencia de una resistencia d

e carbón es de 1000 a0°c , determine su resistencia a 120°c
Physics
1 answer:
Doss [256]3 years ago
3 0

Answer:

 R (120) = 940Ω

Explanation:

The variation in resistance with temperature is linear in metals

           ΔR (T) = R₀ α ΔT

where α is the coefficient of variation of resistance with temperature, in this case α = -0,0005 / ºC

let's calculate

            ΔR = 1000 (-0,0005) (120-0)

            ΔR = -60 Ω

            ΔR = R (120) + R (0) = -60

            R (120) = -60 + R (0)

            R (120) = -60 + 1000

            R (120) = 940Ω

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Virty [35]

Answer:

Well, I think you're talking about kinematics, especially uniform rectilinear motion. We know that there is a specific equation for that:

S = Vt + S0

With S being the distance, V the velocity, t the time and S0 the initial distance (initial displacement).

From this you can calculate t, if that's what you want.

8 0
2 years ago
Which statement is true about an object that sinks in water?
Rina8888 [55]
A > Is the correct answer :)
5 0
3 years ago
Read 2 more answers
An airplane accelerates from rest down a runway at 3.20 m/s2 for 32.8 seconds until it lifts off the ground. Determine the dista
vagabundo [1.1K]

Answer:

s=1721.344m  ,v=104.96m/s.

Explanation:

using thr equation of motion;

s=ut+\frac{1}{2}at^{2}

u=0, plane starts from rest,

s=\frac{1}{2}at^{2}

a=3.2m/s^{2}, t=32.8s \\ s=\frac{1}{2}*3.2*32.8^{2}

s=1721.344m

v=u+at

v=0 +3.2*32.8

v=104.96m/s

8 0
3 years ago
Two identical masses are connected to two different flywheels that are initially stationary. Flywheel A is larger and has more m
inysia [295]

Answer:

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

c) True. Information is missing to perform the calculation

Explanation:

Let's consider solving this exercise before seeing the final statements.

We use Newton's second law Rotational

      τ = I α

     T r = I α

     T gR = I α

     Alf = T R / I (1)

     T = α I / R

Now let's use Newton's second law in the mass that descends

     W- T = m a

     a = (m g -T) / m

The two accelerations need related

     a = R α

    α = a / R

    a = (m g - α I / R) / m

    R α = g - α I /m R

    α (R + I / mR) = g

    α = g / R (1 + I / mR²)

We can see that the angular acceleration depends on the radius and the moments of inertia of the steering wheels, the mass is constant

Let's review the claims

a) True. There is dependence on the radius and moment of inertia, no data is given to calculate the moment of inertia

b) False. Missing data for calculation

c) True. Information is missing to perform the calculation

d) False. There is a dependency if the radius and moment of inertia increases angular acceleration decreases

4 0
3 years ago
At a certain point in space, there is a potential of 800 V relative to zero. What is the potential energy of the system when a +
sammy [17]

To solve this problem we will apply the concepts related to electric potential and electric potential energy. By definition we know that the electric potential is determined under the function:

V = \frac{k_e q}{r}

k_e = Coulomb's constant

q = Charge

r = Radius

At the same time

U = \frac{k_e q_1q_2}{r}

The values of variables are the same, then if we replace in a single equation we have this expression,

U  = Vq

If we replace the values, we have finally that the charge is,

V = 800V

q = 1\mu C

U = (800V)(1*10^{-6}C)

U = 8*10^{-4}J

Therefore the potential energy of the system is 8*10^{-4} J

7 0
3 years ago
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