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prisoha [69]
1 year ago
8

you add equal amounts of heat to two identical cylinders containing equal amounts of the same ideal gas. cylinder a is allowed t

o expand, while cylinder b is not. Part A How do the temperature changes of the two cylinders compare?
Physics
1 answer:
OleMash [197]1 year ago
3 0

Temperature rise will be there in cylinder B more than in cylinder A because of internal energy.

what is internal energy?

The sum of the kinetic and chemical potential energies of all the particles in the system is the internal energy. Particles accelerate and pick up kinetic energy when energy is applied to increase the temperature.

Briefing:

Cylinder A uses the heat it absorbs to both work while expanding and to increase internal energy (or temperature).

While cylinder B solely uses the heat it absorbs to increase its internal energy 

As a result, cylinder B's temperature rise is greater than cylinder A's.

To know more about internal energy visit:

brainly.com/question/11278589

#SPJ4

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yulyashka [42]
It’s burning wax. All of the other options are physical changes .
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3 years ago
A dragster runs the quarter mile in 8.96 s. What is the car's velocity (in ft/s) at the finish line?
lara [203]
The first thing you should know to answer this question is the following conversion:
 1mi = 5280feet
 We have then that the speed is:
 v = ((1/4) * (5280)) / (8.96)
 v = 147.32 feet / s
 Answer:
 the car's velocity (in ft / s) at the finish line is 147.32 feet / s
7 0
3 years ago
Complete the equation to show the radioactive decay of carbon-14 to nitrogen-14
blsea [12.9K]

Answer:

The beta decay takes place.

Explanation:

The reaction of radioactivity of carbon 14 to nitrogen 14 is

There is a beta decay.  

The reaction is

C_{6}^{14}\rightarrow N_{7}^{14}+\beta _{-1}^{0}+ energy

Here some energy is released in form of neutrino.

7 0
3 years ago
How long does it take for 4 coulombs of charge to pass through a cross
Galina-37 [17]

Answer: 2 seconds

Explanation:

Given that,

Time (T) = ?

Charge (Q) = 4 coulombs

current (I) = 2 Amps

Since charge depends on the amount of current flowing through the wire in a given time, hence

Charge = Current x Time

Q = IT

4 coulombs = 2 Amps x Time

Time = 4 coulombs / 2 Amps

Time = 2 seconds

Thus, it takes 2 seconds for the current to flow through the wire

4 0
3 years ago
An 85,000 kg stunt plane performs a loop-the-loop, flying in a 260-m-diameter vertical circle. at the point where the plane is f
konstantin123 [22]
A) When the plane is flying straight down, there are three forces acting on it:
- the centripetal force  F=m \frac{v^2}{r}, directed toward the center of the circle (so, horizontally)
- the weight of the plane: W=mg, downward, so vertically
- a third force, given by the propulsion of the plane, which is accelerating it towards the ground (because the problem says that the plane has an acceleration of a=12 m/s^2 towards the ground)

The radius of the circle is r= \frac{260 m}{2} = 130 m, so the centripetal force acting on the plane is
F_c=m \frac{v^2}{r} = \frac{(85000 kg)(55 m/s)^2}{130 m}=1.98 \cdot 10^6 N
On the vertical axis, we have two forces: the weight
W=mg=(85000 kg)(9.81 m/s^2)=8.34 \cdot 10^5 N
and the other force F given by the propulsion. Since we know that their sum should generate an acceleration equal to a=12 m/s^2, we can find the magnitude of this other force F by using Newton's second law:
F+mg=ma
F=m(a-g)=(85000kg)(12 m/s^2-9.81 m/s^2)=1.86 \cdot 10^5 N

So, the net force acting on the plane will be the resultant of the centripetal force (acting in the horizontal direction) and the two forces W and F (acting in the vertical direction):
R= \sqrt{(F_c^2+(W+F)^2}=
= \sqrt{(1.98\cdot 10^6N)^2+(8.34 \cdot 10^5N+1.86 \cdot 10^5 N)^2}  =2.23 \cdot 10^6 N

(b) The tangent of the angle with respect to the horizontal is the ratio between the sum of the forces in the vertical direction (taken with negative sign, since they are directed downward) and the forces acting in the horizontal direction, so:
\tan \theta =  \frac{-(W+F)}{F_c}= -0.5
And so, the angle is
\theta = \arctan (-0.5)=-26.8 ^{\circ}
 
7 0
3 years ago
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