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Butoxors [25]
3 years ago
15

What is thermodynamic?​

Physics
2 answers:
Dominik [7]3 years ago
8 0

Answer:

thermodynamics is the branch of physics which deals with the study of heat and other forms energy and their mutual relationship(relation ship between them)

Explanation:

i hope this will help you :)

Cerrena [4.2K]3 years ago
5 0

Answer:

Thermodynamics is a branch of physics which deals with the energy and work of a system. It was born in the 19th century as scientists were first discovering how to build and operate steam engines. Thermodynamics deals only with the large scale response of a system which we can observe and measure in experiment.

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4) (5 points) Given are the magnitudes and orientations (with respect to x-axis) of 3
Kazeer [188]

Expand each vector into their component forms:

\vec A=(4.5\,\mathrm N)(\cos\theta_A\,\vec\imath+\sin\theta_A\,\vec\jmath)=(2.58\,\vec\imath+3.69\,\vec\jmath)\,\mathrm N

Similarly,

\vec B=(-1.23\,\vec\imath+0.860\,\vec\jmath)\,\mathrm N

\vec C=(-3.44\,\vec\imath-4.91\,\vec\jmath)\,\mathrm N

Then assuming the resultant vector \vec R is the sum of these three vectors, we have

\vec R=\vec A+\vec B+\vec C

\vec R=(-2.09\,\vec\imath-0.368\,\vec\jmath)\,\mathrm N

and so \vec R has magnitude

\|\vec R\|=\sqrt{(-2.09)^2+(-0.368)^2}\,\mathrm N\approx2.12\,\mathrm N

and direction \theta_R such that

\tan\theta_R=\dfrac{-0.368}{-2.09}\implies\theta_R=-170^\circ=190^\circ

5 0
3 years ago
A positive charge +q1 is located to the left of a negative charge -q2. On a line passing through the two charges, there are two
AfilCa [17]

Answer:

please the answer below

Explanation:

(a) If we assume that our origin of coordinates is at the position of charge q1, we have that the potential in both points is

V_1=k\frac{q_1}{r-1.0}-k\frac{q_2}{1.0}=0\\\\V_2=k\frac{q_1}{r+5.2}-k\frac{q_2}{5.2}=0\\\\

k=8.89*10^9

For both cases we have

k\frac{q_1}{r-1.0}=k\frac{q_2}{1.0}\\\\q_1(1.0)=q_2(r-1.0)\\\\r=\frac{q_1+q_2}{q_2}\\\\k\frac{q_1}{r+5.2}=k\frac{q_2}{5.2}\\\\q_1(5.2)=q_2(r+5.2)\\\\r=\frac{5.2q_1-5.2q_2}{q_2}

(b) by replacing this values of r in the expression for V we obtain

k\frac{q_1}{\frac{5.2(q_1-q_2)}{q_2}+5.2}=k\frac{q_2}{5.2}\\\\\frac{q_1}{q_2}=\frac{(q_1-q_2)}{q_2}-1.0=\frac{q_1-q_2-q_2}{q_2}=\frac{q_1-2q_2}{q_2}

hope this helps!!

3 0
3 years ago
Read 2 more answers
block with of mass m is at rest on horizontal frictionless surface at time t=0. A force given by F=Bt+C is applied horizontally
Alexeev081 [22]

Answer:

v_{2} =\frac{1}{2}

Explanation:

From the second law of Newton movement laws, we have:

F=m*a, and we know that a is the acceleration, which definition is:

a=\frac{dv}{dt}, so:

F=m*\frac{dv}{dt}\\\frac{dv}{dt}=\frac{F}{m}=\frac{\frac{1}{2}(t+1)}{4}=\frac{t+1}{8}

The next step is separate variables and integrate (the limits are at this way because at t=0 the block was at rest (v=0):

dv=\frac{1}{8}(t+1)dt\\\int\limits^{v_{2}}_0 \, dv=\int\limits^{2}_{0} {\frac{1}{8}(t+1)} \, dt

v_{2}=\frac{1}{8}*(\frac{t^{2}}{2}+t) (This is the indefinite integral), the definite one is:

v_{2}=\frac{1}{8}*(2+2)=\frac{1}{2}

3 0
3 years ago
Short Answer
STatiana [176]

15 degrees because a glass of water won't do anything to a bath tub of 15 degree water

4 0
3 years ago
HEY CAN ANYONE PLS PLS PLS HELP ME OUT IN DIS I AM STRUGGLING TOO MUCH
meriva
<h3>Answer: 104.5 cubic cm</h3>

=======================================================

Work Shown:

r = radius = 1.045 cm

h = height = 30.48 cm

pi = 3.141 approximately

V = volume of cylinder

V = pi*r^2*h

V = 3.141*(1.045)^2*30.48

V = 104.547940002

V = 104.5 cubic cm

6 0
3 years ago
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