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3 years ago

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A certain quantity of a liquid has a volume of 10cm at 20°C. Calculate its volume at 50°C, if it’s cubic expansivity is 10^(-3)

K^(-1)

1 answer:

Pavel [41]3 years ago

6 0

Answer:

10.3 cm³

Explanation:

From the question given above, the following data were obtained:

Original volume (V₁) = 10 cm³

Initial temperature (θ₁) = 20 °C

Final temperature (θ₂) = 50 °C

Cubic expansivity (γ) = 10¯³ K¯¹

Final volume (V₂) =?

γ = V₂ – V₁ / V₁(θ₂ – θ₁)

10¯³ = V₂ – 10 / 10( 50 – 20)

10¯³ = V₂ – 10 / 10(30)

10¯³ = V₂ – 10 / 300

Cross multiply

10¯³ × 300 = V₂ – 10

0.3 = V₂ – 10

Collect like terms

0.3 + 10 = V₂

10.3 = V₂

V₂ = 10.3 cm³

Thus, the volume at 50 °C is 10.3 cm³

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How do i find stretch? The problem in questioning has already given me the elastic energy and k-value, but I have no idea how to

finlep [7]

Answer:

Stretch can be obtained using the Elastic potential energy formula.

The expression to find the stretch (x) is $x=\sqrt{\frac{2\times EPE}{k}}$

Explanation:

Given:

Elastic potential energy (EPE) of the spring mass system and the spring constant (k) are given.

To find: Elongation in the spring (x).

We can find the elongation or stretch of the spring using the formula for Elastic Potential Energy (EPE).

The formula to find EPE is given as:

$EPE=\frac{1}{2}kx^2$

Rewriting the above expression in terms of 'x', we get:

$x=\sqrt{\frac{2\times EPE}{k}}$

Example:

If EPE = 100 J and spring constant, k = 2 N/m.

Elongation or stretch is given as:

$x=\sqrt{\frac{2\times EPE}{k}}\\\\x=\sqrt{\frac{2\times 100}{2}}\\\\x=\sqrt{100}=10\ m$

Therefore, the stretch in the spring is 10 m.

So, stretch in the spring can be calculated using the formula for Elastic Potential Energy.

6 0

3 years ago

If you put a drop of food coloring in water and watch the drop disperse, is entropy increasing or decreasing.

den301095 [7]

Answer:

Entropy is increasing. Entropy is decreasing.

Explanation:

The Entropy doesn't change.

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3 years ago

A vinyl record is played by rotating the record so that an approximately circular groove in the vinyl slides under a stylus. Bum

AleksandrR [38]

Answer:

Hits per second=199 hit/s

Explanation:

#Given the angular velocity, $\omega=33\frac{1}{3} rev/min$ , radius of the record $r=0.1m$ and the distance between any two successive bumps on the groove as $d=1.75mm$ .

The linear speed of the record in meters per second is:

$v=\omega r=33\frac{1}{3}\times\frac{2\pi}{60}\times 10\times 10^{_2}\\\\=0.3843m/s\\$

#From $v$ above, if the bumps are uniformly separated by 1m, then the rate at which they hit the stylus is:

$Hits/second=v/d \ \ \ \ d=1.75mm\\\\=0.3483/0.000175\\\\=199.0385714\approx 199$

Hence the bumps hit the stylus at around 199hit/s

8 0

3 years ago

50 grams of ice cubes at -15°C are used to chill a water at 30°C with mass mH20 = 200 g. Assume that the water is kept in a foam

Arada [10]

Answer : The final temperature is, $25.0^oC$

Explanation :

In this problem we assumed that heat given by the hot body is equal to the heat taken by the cold body.

$q_1=-q_2$

$m_1\times c_1\times (T_f-T_1)=-m_2\times c_2\times (T_f-T_2)$

where,

$c_1$ = specific heat of ice = $2.09J/g^oC$

$c_2$ = specific heat of water = $4.18J/g^oC$

$m_1$ = mass of ice = 50 g

$m_2$ = mass of water = 200 g

$T_f$ = final temperature = ?

$T_1$ = initial temperature of ice = $-15^oC$

$T_2$ = initial temperature of water = $30^oC$

Now put all the given values in the above formula, we get:

$50g\times 2.09J/g^oC\times (T_f-(-15))^oC=-200g\times 4.184J/g^oC\times (T_f-30)^oC$

$T_f=25.0^oC$

Therefore, the final temperature is, $25.0^oC$

5 0

3 years ago

A flat, circular, steel loop of radius 75 cm is at rest in a uniform magnetic field, as shown in an edge-on view in the figure (

SIZIF [17.4K]

The solution to the questions are given as

$t=40.39 \mathrm{sec}$
$\varepsilon &=(0.12v)e^{0.057t}$
the direction of induced current will be Counterclock vise.

<h3>What is the direction of the current induced in the loop, as viewed from above the loop.?</h3>

Given, $B(t)=(1.4 T) e^{-0.057 t}$

$$\varepsilon m f(\varepsilon)=-\frac{d \phi_{B}}{d t}$

$\quad$ and, $\phi_{B}=\int B \cdot d A=\int B \cdot d A \cdot \cos \theta$$

$\begin{aligned}\text { Here, } \theta &=30^{\circ} ; \\A &=\pi r^{2} \\a n \delta, R &=0.75 \mathrm{~m} \\\therefore \varepsilon &=-\frac{d}{d t}(B A \cdot \cos \theta)=-A \cdot \cos \theta \cdot \frac{d}{d t}(B(t)) \\\therefore \varepsilon &=-\pi R^{2} \cdot \cos \theta \cdot \frac{d}{d t}\left(e^{-0.057 t}\right)(1.4 T) \\\therefore \varepsilon &=+\pi(0.75)^{2} \cdot \cos 30 \cdot(0.057)(1.4) \cdot e^{-0.057 t}\left\{\because \frac{d}{d t} e^{-x}=-x \cdot e^{-x} .\right.\end{aligned}$

$\varepsilon &=(0.12v)e^{0.057t}$

(b) $Here, $\varepsilon_{0}=0.12 \mathrm{~V} \quad\left(a t_{2} t=0 \mathrm{sec}\right)$$

$\begin{aligned}&\therefore 1 . \varepsilon_{0}=\varepsilon_{0} \cdot e^{-e .057 t} \\&\therefore e^{0.057 t}=10 \quad \text { (taking log both thesides) } \\&\therefore 0.057 t=\ln (10)=2.303 \\&\therefore t=40.39 \mathrm{sec}\end{aligned}$

c)

In conclusion, the direction of the induced current will be Counterclockwise.

Read more about current

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4 0

2 years ago