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3 years ago

Is there a formula for the 2nd Law of Therodynamics, is so name it?

Physics

1 answer:

IrinaK [193]3 years ago

8 0

∆S>_closed integral of dQ/T

There are many equations for different situations of entropy but this is a general one

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2. How can people know that they've learned?

photoshop1234 [79]

Answer: People know that they jap e learned by their mistakes or by someone ya being them something

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2 years ago

Sediment laid down by glacial meltwater is called

Mrac [35]

<span>Your answer isStratified drift</span>

6 0

3 years ago

Read 2 more answers

As compared to the asthenosphere, the lithosphere is ____________. a. hotter and more able to flow b. hotter and less able to fl

Mnenie [13.5K]

Answer:

its a solid but can flow

Explanation:

those answers to choose from are wrong

8 0

2 years ago

Doe anyone get this

8_murik_8 [283]

Answer:

we know a = F/ M

Explanation:

2 m/s²
0.19 m/s²
9.25 m/ s²
0.04 m/s²
100.39 m/s²

4 0

2 years ago

a person throws a ball upward into the air with an initial velocity of 20 m/s. calculate (a) how high it goes, and (b) how long

Phantasy [73]

Answer:

a) about 20.4 meters high

b) about 4.08 seconds

Explanation:

Part a)

To find the maximum height the ball reaches under the action of gravity (g = 9.8 m/s^2) use the equation that connects change in velocity over time with acceleration.

$a=\frac{Vf-Vi}{t}$

$-9.8 \frac{m}{s} =\frac{Vf-Vi}{t}$

In our case, the initial velocity of the ball as it leaves the hands of the person is Vi = 20 m/s, while thw final velocity of the ball as it reaches its maximum height is zero (0) m/s. Therefore we can solve for the time it takes the ball to reach the top:

$-9.8 =\frac{0-20}{t}\\t=\frac{20}{9.8} s = 2.04 s$

Now we use this time in the expression for the distance covered (final position Xf minus initial position Xi) under acceleration:

$Xf-Xi=Vi*t+\frac{1}{2} a t^{2} \\Xf-Xi=20*(2.04)-\frac{1}{2} 9.8*2.04^{2}\\Xf-Xi=20.408 m$

Part b) Now we use the expression for distance covered under acceleration to find the time it takes for the ball to leave the person's hand and come back to it (notice that Xf-Xi in this case will be zero - same final and initial position)

$Xf-Xi=0=20*(t)-\frac{1}{2} 9.8*t^{2$

To solve for "t" in this quadratic equation, we can factor it out as shown:

$0= t(20-\frac{9.8}{2} t)$

Therefore there are two possible solutions when each of the two factors equals zero:

1) t= 0 (which is not representative of our case) , and

2) the expression in parenthesis is zero:

$0= 20-\frac{9.8}{2} t\\t=\frac{20*2}{9.8} = 4.08 s$

7 0

3 years ago