What is a similarity for heat and temperature

Physics

2 answers:

Alex787 [66]3 years ago

7 0

It is the subtance betwen the person or the object

MatroZZZ [7]3 years ago

3 0

Heat is a measure of the internal energy or kinetic energy of the particles in a body. While the temperature is a measure of how hot a body is.
Note also that the higher the internal energy the higher the measure of the heat and also the higher the temperature of the body.

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A 12.0-cm long cylindrical rod has a uniform cross-sectional area A = 5.00 cm2. However, its density increases linearly from 2.6

andriy [413]

Answer:

(a) The constants required describing the rod's density are B=2.6 and C=1.325.

(b) The mass of the road can be found using $A\int_0^{12}\left(B+Cx)dx$

Explanation:

(a) Since the density variation is linear and the coordinate x begins at the low-density end of the rod, we have a density given by

$2.6\frac{g}{cm^3}+\frac{18.5\frac{g}{cm^3}-2.6\frac{g}{cm^3}}{12 cm}x = 2.6\frac{g}{cm^3}+1.325x\frac{g}{cm^2}$

recalling that the coordinate x is measured in centimeters.

(b) The mass of the rod can be found by having into account the density, which is x-dependent, and the volume differential for the rod:

$m=\int\rho dv=\int\left(B+Cx\right)Adx=5\int_0^{12}\left(2.6+1.325x\right)dx=126.6$ ,

hence, the mass of the rod is 126.6 g.

7 0

3 years ago

Question 2

Tju [1.3M]

B. True give me branliest

8 0

3 years ago

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slader A transcontinental flight of 4890 km is scheduled to take 40 min longer westward than eastward. The airspeed of the airpl

11111nata11111 [884]

Answer:

$v_{s}=65.2km/h$

Explanation:

Given data

Flight distance S=4890 km

Time difference Δt=t₂-t₁=40 min

Air speed of plane=980 km/h

To find

Speed of jet stream

Solution

When moving in the same direction as the jet stream time taken as t₁=d/(v+vs),v is velocity of plane and vs is velocity of plane

While moving in opposite direction t₂=d/(v+vs)

$t_{2}-t_{1}=\frac{d}{(v-v_{s}) } - \frac{d}{(v+v_{s}) }\\t_{2}-t_{1}=\frac{d(v+v_{s})-d(v-v_{s})}{(v-v_{s})(v+v_{s})} \\t_{2}-t_{1}=\frac{2dv_{s}}{(v)^{2} -(v_{s})^{2} }\\0.666667h=\frac{2(4890km)v_{s}}{(980km/h)^{2} -(v_{s})^{2} }\\0.666667((980km/h)^{2} -(v_{s})^{2})=9780v_{s}\\640267-0.666667(v_{s})^{2}-9780v_{s}=0\\0.666667(v_{s})^{2}+9780v_{s}-640267=0$