Hot water is poured into a mug and the mug gets hot. This is an example of which type of energy transfer?

PLEASE HELPPP Count the number of atoms of each element in this molecule: 2H2K(OH2)2

yulyashka [42]

Answer:

H = 6 atoms

K= 2 atoms

O = 2 atoms

7 0

3 years ago

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A 20 ft ladder leans against a wall. The bottom of the ladder is 3 ft from the wall at time t=0 and slides away from the wall at

stellarik [79]

Answer: 0.516 ft/s

Explanation:

Given

Length of ladder L=20 ft

The speed at which the ladder moving away is v=2 ft/s

after 1 sec, the ladder is 5 ft away from the wall

So, the other end of the ladder is at

$\Rightarrow y=\sqrt{20^2-5^2}=19.36\ ft$

Also, at any instant t

$\Rightarrow l^2=x^2+y^2$

differentiate w.r.t.

$\Rightarrow 0=2xv+2yv_y\\\\\Rightarrow v_y=-\dfrac{x}{y}\times v\\\\\Rightarrow v_y=-\dfrac{5}{19.36}\times 2=0.516\ ft/s$

5 0

3 years ago

A sound wave was determined to have a frequency of 0.3 Hz, speed of 150 cm/s, and amplitude of 2 cm. Find its wavelength.

fredd [130]

Answer:

5 m

Explanation:

From the question,

v = λf....................... Equation 1

Where v = speed of the sound wave, λ = wavelength of the sound wave, f = frequency of the sound wave.

make λ the subject of the equation

λ = v/f..................... Equation 2

Given: v = 150 cm/s = 1.5 m/s, f = 0.3 hz.

Substitute these values into equation 2

λ = 1.5/0.3

λ = 5 m.

7 0

3 years ago

A rotating viscometer consists of two concentric cylinders –an inner cylinder of radius Rirotating at angular velocity (rotation

RSB [31]

Answer:

b) the result we got can be termed approximation because we are neglecting the shear stress acting on the two ends of the cylinder. Here we have considered only the share stress acting on the curved surface area only.

Explanation:

check attachment for solution to A

4 0

3 years ago

Carbon dioxide (CO2) gas in a piston-cylinder assembly undergoes three processes in series that begin and end at the same state

topjm [15]

Answer:

a) W =400 kJ

b) W = 0 kJ

c) W =-160.944 KJ

Explanation:

Given

Process 1 ---> 2

The relation of the process P = constant

Pressure of point (1) P1 = 10 bar = P2

Volume of point (1) V1 = 1 m^3

Volume of point (2) V2 =4 m^3

The relation of the process V = constant

Process 2 ---> 3

The relation of the process V = constant

V3 = V2

Pressure of point (3) P3 = 10 bar

Volume of point (3) V3 = 4 m^3

Process 3 ---> 1 

The relation of the process PV = constant

Required

Sketch the processes on the PV coordinates

The work for each process in kJ

Solution

The work is defined by

W= $\int\limits^a_b {x} \, dx$

a=V2

b=V1

x=P

dx=dV

Process 1 ---> 2

P3 = P4 = 5 bar

W= $\int\limits^a_b {x} \, dx$

a=V3

b=V2

x=4

dx=dV

putting the value of a, b, x, dx in above integral

W=400 kJ

Process 2 ---> 3 

V = constant Then there is no change in the volume,hence W = 0 kJ

Process 3 ---> 1

By substituting with point (1) --> 5 x .2 = C ---> C = 1 P = 5V^-1

W= $\int\limits^a_b {x} \, dx$

a=V1

b=V3

x=1V^-1

dx=dV

putting the value of a, b, x, dx in above integral

W=| ln V | limit a and b

= -160.944 KJ

5 0

3 years ago