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Kipish [7]
2 years ago
9

Suppose a grower sprays (2.2x10^1) kg of water at 0 °C onto a fruit tree of mass 180 kg. How much heat is released by the water

when it freezes?
Physics
1 answer:
Bond [772]2 years ago
6 0

There is no temperature change which drives heat flow, thus no heat will be released by the water.

<h3>Heat released by the water when it freezes</h3>

The heat released by the water when it freezes is calculated as follows;

Q = mcΔФ

where;

  • m is mass of water
  • c is specific heat capacity of water
  • ΔФ is change in temperature = Фf - Фi

Initial temperature of water, Фi = 0 °C

when water freezes, the final temperature, Фf = 0 °C

Q = 22 x 4200 x (0 - 0)

Q = 0

Since there is no temperature change which drives heat flow, thus no heat will be released by the water.

Learn more about heat flow here: brainly.com/question/14437874

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Explanation:

The volume of the bread decreases, making the bread appear more compact, and smaller in size. The mass stays the same, it won't change unless part of the bread is removed. The density increases, the air bubbles inside of the bread get squished down, causing the bread to be smaller, and in turn, causing it to be more solid.

I hope this helped!

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Convert 26.4 mi to km
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What is the weight of a 4.5 kg mass on Earth?
swat32

Answer:

7.535×10^25 earth mass

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An 20-cm-long Bicycle Crank Arm. With A Pedal At One End. Is Attached To A 25-cm-diameter Sprocket, The Toothed Disk Around Whic
malfutka [58]

To solve the problem, it is necessary to apply the concepts related to the kinematic equations of the description of angular movement.

The angular velocity can be described as

\omega_f = \omega_0 + \alpha t

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\omega_f =Final Angular Velocity

\omega_0 =Initial Angular velocity

\alpha = Angular acceleration

t = time

The relation between the tangential acceleration is given as,

a = \alpha r

where,

r = radius.

PART A ) Using our values and replacing at the previous equation we have that

\omega_f = (94rpm)(\frac{2\pi rad}{60s})= 9.8436rad/s

\omega_0 = 63rpm(\frac{2\pi rad}{60s})= 6.5973rad/s

t = 11s

Replacing the previous equation with our values we have,

\omega_f = \omega_0 + \alpha t

9.8436 = 6.5973 + \alpha (11)

\alpha = \frac{9.8436- 6.5973}{11}

\alpha = 0.295rad/s^2

The tangential velocity then would be,

a = \alpha r

a = (0.295)(0.2)

a = 0.059m/s^2

Part B) To find the displacement as a function of angular velocity and angular acceleration regardless of time, we would use the equation

\omega_f^2=\omega_0^2+2\alpha\theta

Replacing with our values and re-arrange to find \theta,

\theta = \frac{\omega_f^2-\omega_0^2}{2\alpha}

\theta = \frac{9.8436^2-6.5973^2}{2*0.295}

\theta = 90.461rad

That is equal in revolution to

\theta = 90.461rad(\frac{1rev}{2\pi rad}) = 14.397rev

The linear displacement of the system is,

x = \theta*(2\pi*r)

x = 14.397*(2\pi*\frac{0.25}{2})

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Answer:

Explanation:

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