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adell [148]
4 years ago
8

The δe of a system that releases 12.4 j of heat and does 4.2 j of work on the surroundings is __________ j.

Physics
1 answer:
Vedmedyk [2.9K]4 years ago
5 0

The answer for this problem is clarified through this, the system is absorbing (+). And now see that it uses that the SURROUNDINGS are doing 84 KJ of work. Any time a system is overshadowing work done on it by the surroundings the sign will be +. So it's just 12.4 KJ + 4.2 = 16.6 KJ.

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A block of mass m1 = 3.5 kg moves with velocity v1 = 6.3 m/s on a frictionless surface. it collides with block of mass m2 = 1.7
maxonik [38]
First, let's find the speed v_i of the two blocks m1 and m2 sticked together after the collision.
We can use the conservation of momentum to solve this part. Initially, block 2 is stationary, so only block 1 has momentum different from zero, and it is:
p_i = m_1 v_1
After the collision, the two blocks stick together and so now they have mass m_1 +m_2 and they are moving with speed v_i:
p_f = (m_1 + m_2)v_i
For conservation of momentum
p_i=p_f
So we can write
m_1 v_1 = (m_1 +m_2)v_i
From which we find
v_i =  \frac{m_1 v_1}{m_1+m_2}= \frac{(3.5 kg)(6.3 m/s)}{3.5 kg+1.7 kg}=4.2 m/s

The two blocks enter the rough path with this velocity, then they are decelerated because of the frictional force \mu (m_1+m_2)g. The work done by the frictional force to stop the two blocks is
\mu (m_1+m_2)g  d
where d is the distance covered by the two blocks before stopping.
The initial kinetic energy of the two blocks together, just before entering the rough path, is
\frac{1}{2} (m_1+m_2)v_i^2
When the two blocks stop, all this kinetic energy is lost, because their velocity becomes zero; for the work-energy theorem, the loss in kinetic energy must be equal to the work done by the frictional force:
\frac{1}{2} (m_1+m_2)v_i^2 =\mu (m_1+m_2)g  d
From which we can find the value of the coefficient of kinetic friction:
\mu =  \frac{v_i^2}{2gd}= \frac{(4.2 m/s)^2}{2(9.81 m/s^2)(1.85 m)}=0.49
3 0
3 years ago
PLEASE HELP ASAP...TIMED TEST. PLEASE ANSWER WITH AT LEAST A PARAGRAPH!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
charle [14.2K]

Answer: I am pretty sure that you should pick radio waves.

Explanation: The scientist should use radio waves. I think this because you can use the radio waves to analyze the signals from outer space. This will work much better than anything there, to analyze it the best possible.

The best I could do.

8 0
3 years ago
two clear, colorless liquids are poured together. A bright yellow solid forms. What physical properties have changed?
jek_recluse [69]
Change in state(from liquid to solid) and change in colour I believe.
3 0
3 years ago
What is the relative size and composition of the universe, a galaxy, and a solar system? Is the universe endless?
vichka [17]
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7 0
3 years ago
Read 2 more answers
A river flows with a speed of 0.600 m/s. A student first swims upriver 0.500 km, then turns around and returns to his starting p
DerKrebs [107]

Answer:

a) 1111.0 seconds

b) 833.3 s

c) Because of proportions

Explanation:

a) Total time of round trip is the sum of time upriver and time downriver

t_{total}=t_{up}+t_{down}

Time upriver is calculated with the net speed of student and 0.500 km:

t_{up}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=v_{relative to river}+v_{river}=-1.2+0.6=-0.6 m/s\\t_{up}=\frac{500 m}{0.6 m/s}=833.3 s

(Becareful with units 0.5 km= 500m) Similarly of downriver:

t_{down}=\frac{d_{istance}}{|v_{swimmer}|} ;\\v_{swimmer}=1.2+0.6=1.8 m/s\\t_{down}=\frac{500 m}{1.8 m/s}=277.7 s

So the sum is:

t_{total}=1111.0s

b) Still water does not affect student speed, so total time would be simply:

t_{total}=\frac{1000 m}{1.2 m/s}=833.3 s

c) For the upriver trip, student moved half the distance in half speed of the calculation in b), so it kept the same ratio and therefore, same time. So the aditional time is actually the downriver.  

6 0
3 years ago
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