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steposvetlana [31]
3 years ago
13

Do all rocks stay the same throughout their life time

Physics
2 answers:
Leviafan [203]3 years ago
7 0
No they can change through weathering and erosion.
IrinaK [193]3 years ago
3 0
It depends if it is to hot or cold if it is too hot it will melt to cold it could freeze up and brake down easily or if it is hit a lot then it will brake down.
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I have to draw a wave on a separate sheet of paper given the following measurements down below. I'm a bit confused, do I just us
Alex73 [517]
You have to draw a mathematical spatial axes .in order to judge is it right or not .. well you have to draw the crest and trough both of 1 cm in length and the total wavelength (same phase on the wave of 2 cm ) something like this

7 0
3 years ago
. How much work in joules is done by a person who uses a force of 25 N to move a desk 3.0 m?
sergejj [24]
We Know, W = F * s
here, F = 25 N
s = 3 m

Substitute it into the expression,
W = 25 * 3
W = 75 Joule

So, your final answer is 75 J

Hope this helps!

7 0
3 years ago
A student constructed a series circuit consisting of a 12.0-volt battery, a 10.0-ohm lamp, and a
Zolol [24]
     Considering the unknown resistence as R and using the Ohm's First Law, we have:

i= \frac{V}{R_{eq}}  \\ 0.5= \frac{12}{R+10}  \\ R+10=24 \\ R=14-Ohm
 
     The equivalent resistence is given by the resistor series with the lamp resistence.

R_{eq}=R+10 \\ R_{eq}=14+10 \\ \boxed {R_{eq}=24-Ohm}

If you notice any mistake in my english, please let me know, because i am not native.

6 0
2 years ago
200-grams of computer chips with a specific heat of 0.3 kJ/kg·K are initially at 25°C. These chips are cooled by placement in 0.
balu736 [363]

Answer:

a. -0.01324 kJ/K,  b.  = 0.03233 kJ/K , c.  = 0.01909, Yes the process is possible

Explanation:

Heat transfer will occur between the chip and the surrounding fluid. Then, finally they will attain a common equilibrium temperature and heat transfer will stop. Now, if we assume that, after heat transfer, chip will attain the temperature of fluid, that is, -34 C,, So , to check whether this is possible

Amount of energy lost by the chip = m . c . (T(i) - T(f))

= 0.2 x 0.3 (25 + 34) = 3.54 KJ

Now, to evaluate the final state of the fluid, after the heat transfer completion,

Energy Gained = m(mew final – mew initial) = m[(μf+ x . μfg) - μf]

Note that heat transfer will change the internal energy of the fluid. Do not consider enthalpy change, as this is not a problem involving fluid flow in and out of the system

M[(μf+ x . μfg) - μf] = m(xμfg)

<u>Energy gained by the fluid will be equal to the energy lost by the chip (No energy loss to the surroundings)</u>

3.54 = 0.1 . X x 203.29

<u>x = 0.1741, which is the dryness fraction of fluid at the final state.</u>

Observe that the total energy lost by the chips is 3.45 kJ and fluid R-134a has got its value of mew fg at -34 C which is = 203.29 kJ/kg

So for 0.1kg of R-134a

0.1 x μfg= <u>20.329 kJ, which is much greater than 3.45 kJ</u>, therefore, it is certain that the state of fluid will be at -34 C only and at the saturation pressure of 69.56 KPa. So the chip will come to attain the temperature of -34 C.  

a. Write the equation for the change of entropy in the chips

ΔSchips = mchips . c . ln(T2/T1), where mc is the mass of chips, c is the specific heat of chips, T2 is the temperature at state 2 and T1 is the temperature at state 1

Substitute mc = 0.2 kg, c = 0.3kJ/kg.K, T1 = 25 + 273, T2 = -34 + 273

delSchips = 0.2 x 0.3 x ln [(-34+273)/ (25+273)]

= -0.01324 kJ/K

There fore the change in entropy of the chips is -0.01324 kJ/K

b. Entropy change of fluid R- 134a

ΔS2 = m[Sfinal – S initial]

= m[Sf + x . Sfg - Sf]

= 0.2 x (0.1741 x 0.92859)

= 0.03233 kJ/K

c. Calculate the total change in the entropy of the entire system

delS = delSchips + delSR -134a

= -0.01324 + 0.03233

= 0.01909

<u>Since the total change in entropy of the entire system is positive that exactly explains that the actual processes are happening in the direction of increase of entropy therefore, the process is possible.</u>

<u />

6 0
3 years ago
The acceleration due to the earth's gravity, in si units, is 9.8 m/s2. in the absence of air friction, a ball is dropped from re
gogolik [260]
We don't know anything about the amount of distance it travels, but that's okay. The only equation we need here is 

velocity(final) = velocity(initial) + acceleration * time
vf = vi + (a * t)

The ball is dropped from rest, so vi = 0 m/s.
We want it so that the ball hits the ground with a final velocity of 60 m/s, so vf = 60 m/s. 
We are given the acceleration due to gravity, a = 9.8 m/s^2.
We are solving for the time, t = ?.

Now we just plug in the values.
vf = vi + (a * t)
60 m/s = 0 m/s + (9.8 m/s^2)*(t)

60 = 9.8t

60 / 9.8 = t

t = 6.122 s

Hopefully this is the right answer.

7 0
3 years ago
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