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lisabon 2012 [21]
3 years ago
7

I just need to reassurance, would they both be negative?

Physics
2 answers:
Serggg [28]3 years ago
6 0
Answer:

Yes


Explanation:
kirill [66]3 years ago
5 0

Answer:

yep

Explanation:

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Since the aluminum bar is not an isolated system, the second law of thermodynamics cannot be applied to the bar alone. Rather, i
max2010maxim [7]

Answer:

ΔS total ≥ 0 (ΔS total = 0 if the process is carried out reversibly in the surroundings)

Explanation:

Assuming that the entropy change in the aluminium bar is due to heat exchange with the surroundings ( the lake) , then the entropy change of the aluminium bar is, according to the second law of thermodynamics, :

ΔS al ≥ ∫dQ/T

if the heat transfer is carried out reversibly

ΔS al =∫dQ/T  

in the surroundings

ΔS surr ≥ -∫dQ/T = -ΔS al → ΔS surr ≥ -ΔS al = - (-1238 J/K) = 1238 J/K

the total entropy change will be

ΔS total = ΔS al + ΔS surr

ΔS total ≥ ΔS al + (-ΔS al) =

ΔS total ≥ 0

the total entropy change will be ΔS total = 0 if the process is carried out reversibly in the surroundings

4 0
4 years ago
Which type of bond (ionic, covalent, or metallic) does the phrase "opposite attract" apply to best? Explain.
snow_tiger [21]
Ionic is where the bonding of elements attract opposites
6 0
3 years ago
Read 2 more answers
A sinusoidal wave has the following wave function: y(x,t) = (2.5 m) sin((3.0 m ) x - (24 s-')t + Tt/2] What is the frequency of
vladimir1956 [14]

Answer:

Frequency of the wave is 3.8 hertz.

Explanation:

It is given that, the sinusoidal wave has following wave equation as :

y(x,t)=(2.5\ m)sin[(3\ m^{-1})x-(24\ s^{-1})t+\pi/2]

The general equation for the sinusoidal wave is :

y=A\ sin(kx-\omega t+\phi)

Where

A is the amplitude

k is the constant

\omega is the angular frequency

\phi is the phase difference

Since, \omega=24\ s^{-1}

2\pi f=24

f=\dfrac{24}{2\pi}=3.81\ Hz

or

f = 3.8 Hz

So, the frequency of the wave is 3.8 hertz. Hence, this is the required solution.

8 0
3 years ago
A girl throws a ball of mass 0.80 kg against a wall. The ball strikes the wall horizontally with a speed of 25 m/s, and it bounc
erma4kov [3.2K]

Answer:

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

Explanation:

Applying the impulse-momentum equation;

Impulse = change in momentum

Ft = m∆v

F = (m∆v)/t

Where;

F = force

t = time

m = mass

∆v = v2 - v1 = change in velocity

Given;

m = 0.80 kg

t = 0.050 s

The ball strikes the wall horizontally with a speed of 25 m/s, and it bounces back with this same speed.

v2 = 25 m/s

v1 = -25 m/s

∆v = v2 - v1 = 25 - (-25) m/s = 25 +25 = 50 m/s

Substituting the values;

F = (m∆v)/t

F = (0.80×50)/0.05

F = 800N

the magnitude of the average force exerted on the wall by the ball is 800N

4 0
3 years ago
Two identical conducting spheres are placed 80.0 cm apart. One is given a charge of 5.8 C and the other is given a charge of 6.4
Zepler [3.9K]

Answer: 5.214(10)^{11} N

Explanation:

According to <u>Coulomb's Law:</u>  

<em>"The electrostatic force F_{E} between two point charges q_{1} and q_{2} is proportional to the product of the charges and inversely proportional to the square of the distance d that separates them, and has the direction of the line that joins them".</em>

<em />

Mathematically this law is written as:

F_{E}=K\frac{q_{1}.q_{2}}{d^{2}}  

Where:

F_{E}  is the electrostatic force

K=8.99(10)^{9} Nm^{2}/C^{2} is the Coulomb's constant  

q_{1}=5.8 C and q_{2}=6.4 C are the electric charges

d=80 cm \frac{1 m}{100 cm}=0.8 m is the separation distance between the charges

Solving:

F_{E}=8.99(10)^{9} Nm^{2}/C^{2}\frac{(5.8 C)(6.4 C)}{(0.8 m)^{2}}  

F_{E}=5.214(10)^{11} N    

7 0
4 years ago
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