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fenix001 [56]
3 years ago
14

Suppose that at room temperature, a certain aluminum bar is 1.0000 m long. The bar gets longer when its temperature is raised. T

he length l of the bar obeys the following relation: l=1.0000+2.4×10−5T, where T is the number of degrees Celsius above room temperature. What is the change of the bar's length if the temperature is raised to 14.1 ∘C above room temperature?
Physics
1 answer:
Deffense [45]3 years ago
4 0

Answer:

0.00034 m

Explanation:

Since the length of the aluminium bar, L is given by , L = 1.0000 + 2.4 × 10⁻⁵T and T = 14.1°C, we substitute the value of T into L. So, we have L = 1.0000 + 2.4 × 10⁻⁵ × 14.1°C = 1.0000 + 0.0003384 = 1.0003384 m. The change in length is thus 1.0003384 - 1.0000 = 0.0003384 m ≅ 0.00034 m

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Leokris [45]
Equation for ke = 1/2mv^2
1) ke = 1/2 x 5000 x (2x2)
= 10,000J
2) ke = 1/2 x 4000 x (3x3)
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3 0
3 years ago
A 500 N force accelerates an object at 20 m s-2. What is its mass?
avanturin [10]

<u>Answer</u>: The mass of the object is 25kg.

The given question deals with Newton's second law of motion and its applications.

<u>Explanation:</u> Given force, F=500N

                                 acceleration, a=20 m/s^{2}

  From Newton's 2nd law of motion , we have

                             F=ma where m=mass of the object

                         ⇒500=m×20

                         ⇒m=500/20=25

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<u> </u><u>Reference Link: </u>brainly.com/question/1141170

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8 0
2 years ago
A tuning fork generates sound waves with a frequency of 240 Hz. The waves travel in opposite directions along a hallway, are ref
Bingel [31]

Answer:

The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

Explanation:

Given that,

Frequency of sound wave = 240 Hz

Distance = 46.0 m

Distance of fork = 14 .0 m

We need to calculate the path difference

Using formula of path difference

\Delta x=2(L_{2}-L_{1})

Put the value into the formula

\Delta x =2((46.0-14.0)-14.0)

\Delta x=36\ m

We need to calculate the wavelength

Using formula of wavelength

\lambda=\dfrac{v}{f}

Put the value into the formula

\lambda=\dfrac{343}{240}

\lambda=1.42\ m

We need to calculate the phase difference

Using formula of the phase difference

\phi=\dfrac{2\pi}{\lambda}\times \delta x

Put the value into the formula

\phi=\dfrac{2\pi}{1.42}\times36

\phi=159.29\ rad

\phi\approx 68.2^{\circ}

Hence, The phase difference between the reflected waves when they meet at the tuning fork is 159.29 rad.

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