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alekssr [168]
4 years ago
15

A rectangular steel plate expands as it is heated. Find the rate of change of area with respect to temperature T when the width

is 1.6 cm and the length is 2.8 cm if d l divided by dt equals 1.9 times 10 Superscript negative 5 Baseline cm divided by degrees Upper C and dw divided by dt equals 8.5 times 10 Superscript negative 6 Baseline cm divided by degrees C. Round to one decimal place.
Physics
1 answer:
tensa zangetsu [6.8K]4 years ago
8 0

Answer:

The variation rate is 5.42 10⁻⁵ cm²/ºC

Explanation:

When we have a thermal expansion problem we must have the relationship of the change in length as a function of the temperature, which are given in this problem, so we can write the expression for the area of ​​a rectangle

      a = L W

They ask us to find the rate of variation of this area depending on the temperature, so we can derive this expression with respect to the temperature

    da / dT = d(LW) / dt

We use the derivative of a product since the two magnitudes change

    da / dT = W dL/dT + L dW/dT

The values ​​they give us are

\frac{dL}{dT} = 1.9 10⁻⁵ cm/ºC

\frac{dW}{dT} = 8.5 10⁻⁶ cm/ºC

W = 1.6  cm

L= 2.8 cm

 Substituting the values ​​and calculating

\frac{da}{dT} = 1.6 1.9 10⁻⁵ + 2.8 8.5 10⁻⁶

\frac{da}{dT} = 3.04 10⁻⁵ + 2.38 10⁻⁵

\frac{da}{dTy}=  5.42 10⁻⁵ cm²/ºC

The variation rate is 5.42 10⁻⁵ cm²/ºC

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A proton and an electron are fixed in space with a separation of 859 nm. Calculate the electric potential at the midpoint betwee
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Answer:

The electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

Explanation:

Electric potential is given as;

V = E*r

where;

E is the electric field strength, = kq/r²

V = ( kq/r²)*r

V = kq/r

k is coulomb's constant = 8.99 X 10⁹ Nm²/C²

q is the charge of the particles = 1.6 X 10⁻¹⁹ C

r is the distance between the particles = 859 nm

At midpoint, the distance = r/2 = 859nm/2 = 429.5 nm

V = (8.99 X 10⁹  * 1.6 X 10⁻¹⁹)/ (429.5 X 10⁻⁹)

V = 3.349 X 10⁻³ Volts

Therefore, the electric potential at the midpoint between the two particles is 3.349 X 10⁻³ Volts

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Answer:

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Explanation:

An oxygen sensor is being tested with a digital multimeter using the MIN/MAX function. The readings are: minimum = 78 mV; maximum = 932 mV; average = 442 mV. Technician A says that the engine is operating normally. Technician B says that the oxygen sensor is skewed too rich. Which technician is correct?

an oxygen sensor detects te amount of oyen in an engine at the exhaust pipe.

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3 years ago
(4. A bus is moving at 25 m/s when the driver steps on the brakes and brings the bus to a stop in
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Answer:

(a)  the average acceleration of the bus while braking is 8.333 m/s

(b) if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.  [¹/₂ (8.333 m/s) = 4.16 s]

Explanation:

Given;

initial velocity of the bus, v = 25 m/s

time of the motion, t = 3 s

(a)  the average acceleration of the bus while braking

a = dv/dt

where;

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dt is change in time

a = 25 / 3

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(b) If the bus took twice as long to stop, the duration = 2 x 3s

a = 25 / (2 x 3s)

a = ¹/₂ x (25 / 3)

a = ¹/₂ (8.333 m/s) = 4.16 s

Thus, if the bus took twice as long to stop, the acceleration will be half of the value obtained in part a.

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