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Margaret [11]
3 years ago
7

A high school student makes a large pot of hot chocolate for her friends. Once it is warm enough, she pours a little bit of the

hot chocolate into a mug so that she can taste it.
Which option correctly assesses the thermal and average kinetic energy in the scenario?
Select all that apply.

The hot chocolate in the large pot and the hot chocolate in the mug have the same amount of average kinetic energy.

The hot chocolate in the large pot has more thermal energy than the hot chocolate in the mug.

The hot chocolate in the large pot and the hot chocolate in the mug have the same amount of thermal energy.

The hot chocolate in the large pot has more average kinetic energy than the hot chocolate in the mug.
Physics
2 answers:
Harrizon [31]3 years ago
8 0

the hot chocolate in the large pot has more therma energy than the hot chocolate in the mug

mash [69]3 years ago
6 0

Answer:

The hot chocolate in the large pot has more thermal energy than the hot chocolate in the mug.

The hot chocolate in the large pot has more average kinetic energy than the hot chocolate in the mug.

Explanation:

Thermal energy is defined as the energy posses by an object because of kinetic energy of atoms present in it.

These atoms moves in a great pace or more rapidly when heat is transferred to them hence more the heat more will be the kinetic energy of the material.

In this case, hot chocolate is in large pot and thus has more temperature than the other mug.

Which explains it has more kinetic energy as well as more thermal energy.

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Andrews [41]
The first one is the light bends sheikh is known as refraction
8 0
3 years ago
Belly-flop Bernie dives from atop a tall flagpole into a swimming pool below. His potential energy at the top is 7000 J (relativ
elena55 [62]

Answer:

KE₂ = 6000 J

Explanation:

Given that

Potential energy at top U₁= 7000 J

Potential energy at bottom U₂= 1000 J

The kinetic energy at top ,KE₁= 0 J

Lets take kinetic energy at bottom level =  KE₂

Now from energy conservation

U₁+ KE₁= U₂+ KE₂

Now by putting the values

U₁+ KE₁= U₂+ KE₂

7000+ 0 = 1000+ KE₂

KE₂ = 7000 - 1000 J

KE₂ = 6000 J

Therefore the kinetic energy at bottom is 6000 J.

5 0
3 years ago
A fisherman is fishing from a bridge and is using a "44.0-N test line." In other words, the line will sustain a maximum force of
vitfil [10]

Answer:

a) 4.485 kg b) 3.94 kg

Explanation:

since the maximum tension the line can stand is 44 N and for question a the speed is constant (acceleration must be zero since the velocity or speed is not changing), F(tension) = mass * acceleration due to gravity (g) .

44 = m * 9.81m/s^2

m = 44/9.81 = 4.485kg

b) F(tension) = ma + mg ( where a is the acceleration of the body and g is the acceleration of the gravity)

44 = m (a +g)

44 = m (1.37 + 9.81)

44/11.18 = m

m = 3.94 kg

3 0
3 years ago
You push a desk with 245 N, but the desk doesn't move due to its friction with the ground. What is the magnitude of the friction
const2013 [10]

If the desk doesn't move, then it's not accelerating.

If it's not accelerating, then the net force on it is zero.

If the net force on it is zero, then any forces on it are balanced.

If there are only two forces on it and they're balanced, then they have equal strengths, and they point in opposite directions.

So the friction on the desk must be equal to your<em> 245N</em> .

7 0
3 years ago
A bolt is dropped from a bridge under construction, falling 94 m to the valley below the bridge. (a) how much time does it take
gregori [183]
Refer to the diagram shown below.

When the bolt is dropped at a height of 94 m, its initial velocity, V, is zero.
The last 26% of its fall is at a height of 0.26*94 = 24.4 m.
At that time, the bolt has fallen by 94 - 24.4 = 69.56 m.

The time, t, for the bolt to fall a known distance obeys the equation 
s = Vt + (1/2)gt²,
where
s = 69.56 m, vertical distance traveled, and
g = acceleration due to gravity.

Therefore
69.56 = 0 + (1/2)*9.8*t²
t² = (69.56*2)/9.8
t = 3.7677 s

The total time, T, to fall 94 m is given by 
94 = (1/2)*9.8*T^2
T² = 19.1837
T = 4.38 s

The time taken to pass through the last 26% of its fall is
T - t = 4.38 - 3.7677 = 0.6122 s

The speed after falling 69.56 m is given by
V₁ = 0 + g*t = 36.9235 m/s

The speed with which the bolt strikes the ground is given by
V₂ = 0 +g*T = 9.8*4.38 = 42.924 m/s

Answer:
(a) The bolt takes 0.6122 s to pass through the last 26% of its fall.
(b) When the bolt begins the last 26% of its fall, its speed is 36.92 m/s (nearest hundredth).
(c) Just before it strikes the ground, the speed of the bolt is 42.94 m/s (nearest hundredth).

8 0
3 years ago
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