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Shtirlitz [24]
3 years ago
15

Was the dirty job a chimney maker? TrueOr false.

Physics
2 answers:
svetlana [45]3 years ago
8 0
Yes it is true hope this helped
Cloud [144]3 years ago
6 0

Answer:

True or False

Explanation:

True I think

You might be interested in
A metal wire is in thermal contact with two heat reservoirs at both of its ends. Reservoir 1 is at a temperature of 781 K, and r
andreev551 [17]

Answer:2.517 J/K

Explanation:

Given

Reservoir 1 Temperature T_1=781 K

Reservoir 2 Temperature T_2=335 K

Let Q is the amount of heat Flows i.e. Q=1477 J

thus change in Entropy is given by \frac{\sum Q}{T}

\Delta S=\frac{\sum Q}{T}=-\frac{Q}{T_1}+\frac{Q}{T_2}

\Delta S=\frac{\sum Q}{T}=-\frac{1477}{781}+\frac{1477}{335}

\Delta S=\frac{\sum Q}{T}=-1.891+4.4089

\Delta S=\frac{\sum Q}{T}=2.517 J/K                              

6 0
2 years ago
What is the displacement of a car in 50 s if it is travelling with a velocity of 25 m/s?
Schach [20]

Answer:

v =25 m/s

time= 50 s

Velocity =Displacement/Time

Displacement = Velocity × Time

S = 25×50

s=1250m

Explanation:

v =25 m/s

time= 50 s

Velocity =Displacement/Time

Displacement = Velocity × Time

7 0
2 years ago
a man counts 6 waves on a pond in 10 seconds. the distance between them is 40 cm. what is their speed?​
gavmur [86]

Answer:

so a man counts 6 waves on a pound in 10 second

Explanation:

6×10 = 60

60/40

so the answer is3

7 0
2 years ago
Review. For a certain type of steel, stress is always proportional to strain with Young's modulus 20 × 10¹⁰ N/m² . The steel has
timurjin [86]

1.58\times 10^{-4}\ \mathrm{s}$ is the time interval elapses before the back end of the rod receives the message that it should stop.

Given:

Length of the rod, L = 80 cm = 0.800 m

Young's modulus, Y = 20 \times 10^{10}\;N/m^{2}

steel density, \rho = 7.86 \times 10^{3}\;kg/m^{3}

The speed of the wave in the rod is,

$v = \sqrt{\frac{Y}{\rho}} = \sqrt{\frac{20\times 10^{10}\ \mathrm{N/m^2}}{7.86\times 10^3\ \mathrm{kg/m^3}}} = 5044\ \mathrm{m/s}$

Consequently, the length of the rod's end is traveled by the wave in at

$t=\frac{L}{v} = \frac{0.800\ \mathrm{m}}{5044\ \mathrm{m/s}} = 1.58\times 10^{-4}\ \mathrm{s}$

Hence, 1.58\times 10^{-4}\ \mathrm{s}$ is the time interval elapses before the back end of the rod receives the message that it should stop.

<h3>What are Newtons Laws?</h3>

The three fundamental laws of classical mechanics known as Newton's laws of motion describe how an object's motion and the forces acting on it interact. The following paraphrase of these statutes is available

Unless a force acts upon a body, it remains at rest or in continual straight-line motion.

When a force acts on a body, the force is equal to the time rate at which the body's momentum changes.

When two bodies exert force on one another, the direction and amount of the force are opposed.

Isaac Newton first identified the three laws of motion in his 1687 book Philosophize Naturalis Principia Mathematica (Mathematical Principles of Natural Philosophy).

They served as the cornerstone for classical mechanics as Newton used them to examine and explain the motion of numerous physical objects and systems. The conceptual foundations of classical physics have been reconstructed in several ways since Newton, utilizing various mathematical techniques that have revealed insights that were hidden in the original, Newtonian formulation.

To know more about Newtons Laws, visit:

brainly.com/question/27573481

#SPJ4

7 0
1 year ago
shows two parallel nonconducting rings with their central axes along a common line. Ring 1 has uniform charge q1 and radius R; r
Ainat [17]

Answer:

\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}

Explanation:

Given that

Charge on ring 1 is q1 and radius is R.

Charge on ring 2 is q2 and radius is R.

Distance ,d= 3 R

So the total electric field at point P is given as follows

Given that distance from ring 1 is R

\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(d-R)}{(R^2+(d-R)^2)^{3/2}}=0

\dfrac{1}{4\pi\epsilon _o}\dfrac{q_1R}{(R^2+R^2)^{3/2}}-\dfrac{1}{4\pi\epsilon _o}\dfrac{q_2(3R-R)}{(R^2+4R^2)^{3/2}}=0

\dfrac{q_1R}{(2R^2)^{3/2}}-\dfrac{q_2(2R)}{(5R^2)^{3/2}}=0

\dfrac{q_1}{q_2}=2\left(\dfrac{2}{5}\right )^{3/2}

8 0
3 years ago
Read 2 more answers
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