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Law Incorporation [45]
3 years ago
7

A liquid with a mass of 75 g is at its boiling point. It completely boils off when AQ = 15000000) of heat energy is added to it.

Calculate the substance's latent heat of vaporization.
Temperature
Specific Heat
Latent Heat of
la
C - Celsius Temp (°C)
Physics
1 answer:
aev [14]3 years ago
8 0

Answer:

2 × 10⁸J/kg

Explanation:

Specific latent heat of vaporization (steam)  is the quantity of heat required or needed to change a unit of mass of liquid to gaseous state at constant temperature (boiling point) and pressure.

It can be calculated mathematically using,

Q = mL

Where,  Q = quantity of heat

m= mass in kg

L = specific latent heat (J/kg)

Therefore,from the question

Q= 15000000J

mass (m)  = 75g = 75/1000 kg = 0.075kg

Q = mL

L = Q/m

L = 15000000  ÷ 0.075

L= 200000000J/kg

L = 2 × 10⁸J/kg

I hope this was helpful, please mark as brainliest

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A 26 foot ladder is lowered down a vertical wall at a rate of 3 feet per minute. The base of the ladder is sliding away from the
lakkis [162]

Answer:

(i) 7.2 feet per minute.

(ii) No, the rate would be different.

(iii) The rate would be always positive.

(iv) the resultant change would be constant.

(v) 0 feet per min

Explanation:

Let the length of ladder is l, x be the height of the top of the ladder from the ground and y be the length of the bottom of the ladder from the wall,

By making the diagram of this situation,

Applying Pythagoras theorem,

l^2 = x^2 + y^2-----(1)

Differentiating with respect to t ( time ),

0=2x\frac{dx}{dt} + 2y\frac{dy}{dt}  ( l = 26 feet = constant )

\implies 2y\frac{dy}{dt} = -2x\frac{dx}{dt}

\implies \frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}

We have,

y = 10, \frac{dx}{dt}= -3\text{ feet per min}

\frac{dy}{dt}=\frac{3x}{10}-----(X)

(i) From equation (1),

26^2 = x^2 + 10^2

676=x^2 + 100

576 = x^2

\implies x = 24\text{ feet}

From equation (X),

\frac{dy}{dt}=\frac{3\times 24}{10}=7.2\text{ feet per min}

(ii) From equation (X),

\frac{dy}{dt}\propto x

Thus, for different value of x the value of \frac{dy}{dt} would be different.

(iii) Since, distance = Positive number,

So, the value of y will always a positive number.

Thus, from equation (X),

The rate would always be a positive.

(iv) The length of the ladder is constant, so, the resultant change would be constant.

i.e. x = increases ⇒ y = decreases

y = decreases ⇒ y = increases

(v) if ladder hit the ground x = 0,

So, from equation (X),

\frac{dy}{dt}=0\text{ feet per min}

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3 years ago
The midpoint M of a guitar string is pulled a distance d = 1.7 mm from equilibrium and released. Point M is observed to undergo
nlexa [21]

Answer:

ω = 380π rad/s

Explanation:

The formula for the angular frequency is the oscillation frequency f (hertz) multiplied by 2π

ω = 2πf

then

ω = 2π(190)

ω = 380π rad/s

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To measure the height of the cloud cover at an airport, a worker shines a spotlight upward at an angle 75° from the horizontal.
Yuri [45]

Answer: The height of the cloud = 394.55 m

Explanation:

The observer is 500m away from the spotlight.

Let x be the distance from the observer to the interception of the segment of the height, h with the floor. The equations are thus:

Tan 45° = h/x ... eq1

Tan 75° = h/(500- x ) ... eq2

From eq 1, Tan 45° = 1, therefore eq1 becomes:

h = x ... eq3

Put eq3 into eq2

Tan 75° = h/(500- h)

h = ( 500 - h ) Tan 75°

h = 500Tan 75° - hTan75°

h + h Tan 75° = 500 Tan 75°

h ( 1 + Tan 75° ) = 500 Tan75°

h = 500Tan75°/ (1 + Tan 75°)

h= 1866.02 / 4.73

h = 394.55m

4 0
3 years ago
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