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Orlov [11]
2 years ago
15

Water is boiled in a pan on a stove at sea level. During 10 min of boiling, it is observed that 200 g of water has evaporated. D

etermine the rate of heat transfer to the water. Solve this using appropriate software. Use data from the tables.
Physics
1 answer:
Ymorist [56]2 years ago
4 0

Answer : The rate of heat transfer to the water is, 37.92 kJ/min

Explanation : Given,

Time = 10 min

Mass of water = 200 g

Latent heat of fusion of water = 334 J/g

Latent heat of vaporization of water = 2230 J/g

Now we have to calculate the rate of heat transfer to the water.

Q=\frac{m\times (L_v-L_f)}{t}

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

Q=\frac{200g\times (2230-334)J/g}{10min}

Q=37920J/min=37.92kJ/min

Thus, the rate of heat transfer to the water is, 37.92 kJ/min

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Darya [45]

Answer:

y = -8.37 cm

Explanation:

As we know that the equation of SHM is given as

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here we know that

\omega = \frac{2\pi}{T}

here we have

T = 0.66 s

now we have

\omega = \frac{2\pi}{0.66}

\omega = 3\pi

now we have

y = (8.8 cm) cos(3\pi t)

now at t = 2.3 s we have

y = (8.8 cm) cos(3\pi \times 2.3)

y = -8.37 cm

6 0
1 year ago
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Quinn is testing the motion of two projectiles x and y by shooting them from a sling shot. What can we say best describes the mo
Studentka2010 [4]

Explanation:

            A projectile motion may be defined as that form of a motion that is experienced by an object or a particle which is projected near the surface of the Earth and the particle moves along the curved path  subjected to gravity force only.

           Thus a projectile motion is always acted upon by a constant acceleration due to gravity in the down ward direction.

             In the context, Quinn shoots two particle x and y from his sling shot and he observes that both his projectiles travels in a parabola curve in the air. Both the object x and y touches the ground a distance apart from him which is known as the range and it depends upon the velocity of the projectile. Both the projectile reaches a maximum height and then drop on the ground in a parabola shape.

3 0
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Cass is walking her dog (Oreo) around the neighborhood. Upon arriving at Calina's house (a friend of Oreo's), Oreo turns part mu
MArishka [77]

Answer:

Horizontal component: F_x = 58\ N

Vertical component: F_y = 33.5\ N

Explanation:

To find the horizontal and vertical components of the force, we just need to multiply the magnitude of the force by the cosine and sine of the angle with the horizontal, respectively.

Therefore, for the horizontal component, we have:

F_x = F * cos(angle)

F_x = 67 * cos(30)

F_x = 58\ N

For the vertical component, we have:

F_y = F * sin(angle)

F_y = 67 * sin(30)

F_y = 33.5\ N

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4 0
2 years ago
Two objects are placed in thermal contact and are allowed to come to equilibrium in isolation. the heat capacity of object a is
Harman [31]
Given:
Ca = 3Cb                      (1)
where
Ca =  heat capacity of object A
Cb =  heat capacity f object B

Also,
Ta = 2Tb                     (2)
where
Ta = initial temperature of object A
Tb = initial temperature of object B.

Let
Tf =  final equilibrium temperature of both objects,
Ma = mass of object A,
Mb = mass of object B.

Assuming that all heat exchange occurs exclusively between the two objects, then energy balance requires that
Ma*Ca*(Ta - Tf) = Mb*Cb*(Tf - Tb)           (3)

Substitute (1) and (2) into (3).
Ma*(3Cb)*(2Tb - Tf) = Mb*Cb*(Tf - Tb)
3(Ma/Mb)*(2Tb - Tf) = Tf - Tb

Define k = Ma/Mb, the ratio f the masses.
Then
3k(2Tb - Tf) = Tf - Tb
Tf(1+3k) = Tb(1+6k)
Tf = [(1+6k)/(1+3k)]*Tb

Answer:
T_{f} =( \frac{1+6k}{1+3k} )T_{b}= \frac{1}{2}( \frac{1+6k}{1+3k})T_{a}
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k= \frac{M_{a}}{M_{b}} 
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2 years ago
When a vertical beam of light passes through a transparent medium, the rate at which its intensity I decreases is proportional t
antoniya [11.8K]

Answer:

Intensity of beam 18 feet below the surface is about 0.02%

Explanation:

Using Lambert's law

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then dI / I = kdt

taking log,

ln(I) = kt + ln C

I = Ce^kt

t=0=>I=I(0)=>C=I(0)

I = I(0)e^kt

t=3 & I=0.25I(0)=>0.25=e^3k

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k = -0.4621

I = I(0)e^(-0.4621t)

I(18) = I(0)e^(-0.4621*18)

I(18) = 0.00024413I(0)

Intensity of beam 18 feet below the surface is about 0.2%

3 0
2 years ago
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