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2 years ago

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A brass lid screws tightly onto a glass jar at 20 degrees C. To help open the jar, it can be placed into a bath of hot water. Af

ter this treatment, the temperature of the lid and the jar are both 60 degrees C. The inside diameter of the lid is 8.0 cm at 20 degrees C. Find the size of the gap (difference in radius) that develops by this procedure.

1 answer:

omeli [17]2 years ago

8 0

Answer:

0.0016 cm

Explanation:

$\alpha_b$ = Thermal coefficient of expansion of brass = $19\times 10^{-6}\ /^{\circ}C$

$\alpha_g$ = Thermal coefficient of expansion of glass = $9\times 10^{-6}\ /^{\circ}C$

$\Delta T$ = Change in temperature = $(60-20)^{\circ}C$

$R_0$ = Initial radius = 4 cm

Change in radius of material is given by

$R=R_0(1+\alpha\Delta T)$

Difference in radii of the lid and jar

$\Delta R=R_b-R_g\\\Rightarrow \Delta R=R_0(1+\alpha_b\Delta T)-R_0(1+\alpha_g\Delta T)\\\Rightarrow \Delta R=R_0(\alpha_b-\alpha_g)\Delta T\\\Rightarrow \Delta R=4\times (19\times 10^{-6}-9\times 10^{-6})\times (60-20)\\\Rightarrow \Delta R=0.0016\ cm$

The size of the gap is 0.0016 cm or 0.000016 m

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Answer:

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Explanation:

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Let's use trigonometry to find the breast

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We replace

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λ = 189.43 10⁻⁹ m

Part B

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λ = 332.33 10⁻⁹ 12.15 10⁻² / 15 10⁻²

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When a test charge q0 = 2 nC is placed at the origin, it experiences a force of 8 times 10-4 N in the positive y direction. What

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Answer:

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Explanation:

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$E=4\times 10^5\ N/C$

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1 year ago

Read 2 more answers

An electron in a vacuum chamber is fired with a speed of 9800 km/s toward a large, uniformly charged plate 75 cm away. The elect

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Answer:

The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

Explanation:

Given that,

Speed = 9800 km/s

Distance d= 75 cm

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Suppose we determine the plate's surface charge density?

We need to calculate the surface charge density

Using work energy theorem

$W=\Delta K.E$

$W=\dfrac{1}{2}mv_{f}^2-\dfrac{1}{2}mv_{i}^2$

Here, final velocity is zero

$W=0-\dfrac{1}{2}mv_{i}^2$ ...(I)

We know that,

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$W=-E\times e\times d$

$W=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$ ...(II)

From equation (I) and (II)

$-\dfrac{1}{2}mv_{i}^2=-\dfrac{\lambda}{2\epsilon_{0}}\times e\times d$

Charge is negative for electron

$\lambda=\dfrac{mv^2\epsilon_{0}}{(-e)d}$

Put the value into the formula

$\lambda=-\dfrac{9.1\times10^{-31}\times(9800\times10^{3})^2\times8.85\times10^{-12}}{1.6\times10^{-19}\times(75-15)\times10^{-2}}$

$\lambda=-8.056\times10^{-9}\ C/m^2$

Hence, The plate's surface charge density is $-8.056\times10^{-9}\ C/m^2$

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Answer:

C

Explanation:

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d₀ = 3.994 cm

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