Given what we know, as with all substances, the particles in the bowl of hot soup will slow down as they cool after being poured.
<h3>Why do particles slow when cooled?</h3>
- This has to do with energy.
- The soup particles are excited by heat energy.
- As the soup is poured, its heat energy radiates and it loses energy.
- Decreased energy causes the particles to slow down.
Therefore, the motion of the particles of the soup decrease or slows down as the soup cools due to the loss of energy they are experiencing. The soup radiates heat energy into the room or plate, causing the particles to lose energy and slow down.
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Yea it would be barrier. since the stream has a cover for the water
Answer: A. The total displacement divided by the time and C. The slope of the ant's displacement vs. time graph.
Explanation:
Hi! The question seems incomplete, but I found the options on the internt:
A. The total displacement divided by the time.
B. The slope of the ant's acceleration vs. time graph.
C. The slope of the ant's displacement vs. time graph.
D. The average acceleration divided by the time.
Now, since we know the ant is travelling at a constant speed, its average velocity
will be expressed by the following equation:

Where:
is the ant's total displacement
is the time it took to the ant to travel to the kitchen
Hence one of the correct options is: A. The total displacement divided by the time
On the other hand, this can be expressed by a displacement vs. time graph graph, where the slope of that line leads to the equation written above. So, the other correct option is:
C. The slope of the ant's displacement vs. time graph.
Answer:
kinetic energy = 0.1168 J
Explanation:
From Hooke's law, we know that ;
F = kx
k = F/x
We are given ;
Mass; m = 1.95 kg
Spring stretch; d = x = 0.0865
So, Force = mg = 1.95 × 9.81
k = 1.95 × 9.81/0.0865 = 221.15 N/m
Now, initial energy is;
E1 = mgL + ½k(x - L)²
Also, final energy; E2 = ½kx² + ½mv²
From conservation of energy, E1 = E2
Thus;
mgL + ½k(x - L)² = ½kx² + ½mv²
Making the kinetic energy ½mv² the subject, we have;
½mv² = mgL + ½k(x - L)² - ½kx²
We are given L=0.0325 m
Plugging other relevant values, we have ;
½mv² = (1.95 × 9.81 × 0.0325) + (½ × 221.15(0.0865 - 0.0325)² - ½(221.15 × 0.0865²)
½mv² = 0.62170875 + 0.3224367 - 0.82734979375
½mv² = 0.1168 J
Answer:
yes driver exceed the car speed
Explanation:
given data
speed of car = 38 m/s
speed limit = 75.0 mi/h
to find out
Is the driver exceeding the speed limit
solution
we know car speed is 38 m/s and limit is 75 mi/hr
so for compare the speed limit we convert limit and make them same
as we know
1 m/s = 2.236 mi/hr
so
car speed 38 m/s = 38 × 2.236 = 85.003 mi/hr
as this car speed is exceed the speed limit that is 75 mi/hr
yes driver exceed the car speed