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svetoff [14.1K]
3 years ago
14

How does an ice cube cause hot coffee to become cool and through what heat transfer

Physics
1 answer:
Trava [24]3 years ago
3 0
An ice cube causes hot coffee to become cool because the amount of coldness contrasts the hot coffee to make it a little cooler
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The fact that there are seasons on the earth is largely due to
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The main reason Earth have Seasons it's because of the Earth tilt. One side would be tilted to the sun which causes summer and the other side would be tilted away whitch causes Winter.

The S side of Earth for example is face to the sun which is summer

The N side of Earth of pointed away which is winter.

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3 years ago
Marcus is on a train that travels 20 and takes 5 seconds to slow to 10 .
Vera_Pavlovna [14]

Answer:

-2

Explanation:

v0=20

v=10

t=5

a=(v-v0)/t=(10-20)/5=-2

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Why is declaring faith the first of the five pillars of Islam?
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The philosophy is simple that no one can start the journey towards being a Muslim unless he professes the basic principles mentioned in the  "Declaration of Faith", which is "<span>“There is none worthy of worship but Allah and Muhammad is His Messenger”.
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3 years ago
A torque of 50 N*m acts on a wheel of moment of inertia 25 kg*m^2 for 4 s and then is removed.
motikmotik
You have to find the calculate<span> the circumference first then you can just multiply the diameter by π, which is about 3.142. That gives you the distance for each </span>revolution<span>. Then you can multiply by the </span>number of revolutions<span> per minute.

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3 0
2 years ago
A 6.5 kg rock thrown down from a 120m high cliff with initial velocity 18 m/s down. Calculate
Olegator [25]

Answer:

See the answers below.

Explanation:

In order to solve this problem we must use the principle of energy conservation. Which tells us that the energy of a body will always be the same regardless of where it is located. For this case we have two points, point A and point B. Point A is located at the top at 120 [m] and point B is in the middle of the cliff at 60 [m].

E_{A}=E_{B}

The important thing about this problem is to identify the types of energy at each point. Let's take the reference level of potential energy at a height of zero meters. That is, at this point the potential energy is zero.

So at point A we have potential energy and since a velocity of 18 [m/s] is printed, we additionally have kinetic energy.

E_{A}=E_{pot}+E_{kin}\\E_{A}=m*g*h+\frac{1}{2}*m*v^{2}

At Point B the rock is still moving downward, therefore we have kinetic energy and since it is 60 [m] with respect to the reference level we have potential energy.

E_{B}=m*g*h+\frac{1}{2}*m*v^{2}

Therefore we will have the following equation:

(6.5*9.81*120)+(0.5*6.5*18^{2} )=(6.5*9.81*60)+(0.5*6.5*v_{B}^{2} )\\3.25*v_{B}^{2} =4878.9\\v_{B}=\sqrt{1501.2}\\v_{B}=38.75[m/s]

The kinetic energy can be easily calculated by means of the kinetic energy equation.

KE_{B}=\frac{1}{2} *m*v_{B}^{2}\\KE_{B}=0.5*6.5*(38.75)^{2}\\KE_{B}=4878.9[J]

In order to calculate the velocity at the bottom of the cliff where the reference level of potential energy (potential energy equal to zero) is located, we must pose the same equation, with the exception that at the new point there is only kinetic energy.

E_{A}=E_{C}\\6.5*9.81*120+(0.5*9.81*18^{2} )=0.5*6.5*v_{C}^{2} \\v_{c}^{2} =\sqrt{2843.39}\\v_{c}=53.32[m/s]

5 0
3 years ago
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