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Dmitry_Shevchenko [17]
3 years ago
8

Joan stirred a pot of hot soup with a metal spoon. after a minute or two, the handle of the spoon became too hot to hold. what k

ind of heat transfer does this represent?
Physics
1 answer:
natima [27]3 years ago
4 0

When the metal spoon comes in contact with the hot soup, heat transfer from hot soup to the spoon because of difference in temperature of the two. The kind of heat transfer represented here is conduction. In conduction, heat is transferred from a region of higher temperature to a region of lower temperature when there is a physical contact between the two.

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The greater the pull of gravity on an object, the greater the what of that object
Nikitich [7]

Answer:

weight

Explanation:

" the greater the pull of gravity on an object, the greater the weight of that object." In physics, weight is measured in newtons (N), the common unit for measuring force.

3 0
2 years ago
The speed of light inside a medium is (2.0 x 10^8m/s) What is the index of refraction (n) of the medium?
kondor19780726 [428]

Answer:

Refractive index of a medium = Speed of light in vacuum/Speed of light in a medium

1.5 = 3 x 108 / Speed of light in medium

Speed of light in the medium = 3 x 108 /1.5

= 2 x 108 m/s.

Explanation:

3 0
2 years ago
a 13-gram bullet, moving at 270 m/s, penetrates a 2 kg block of wood and emerges at a speed of 130 m/s. if teh block sits one a
saveliy_v [14]

Answer:

1.52m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute the given values into the formula

0.013(270)+2(130) = (270+130)v

3.51+260 = 400v

263.51 = 400v

v = 400/263.51

v = 1.52m/s

Hence the velocity after the bullet emerges is 1.52m/s

6 0
3 years ago
calculate the time rate of change in air density during expiration. Assume that the lung has a total volume of 6000mL, the diame
kipiarov [429]

Answer:

The time rate of change in air density during expiration is 0.01003kg/m³-s

Explanation:

Given that,

Lung total capacity V = 6000mL = 6 × 10⁻³m³

Air density p = 1.225kg/m³

diameter of the trachea is 18mm = 0.018m

Velocity v = 20cm/s = 0.20m/s

dv /dt = -100mL/s (volume rate decrease)

= 10⁻⁴m³/s

Area for trachea =

\frac{\pi }{4} d^2\\= 0.785\times 0.018^2\\= 2.5434 \times10^-^4m^2

0 - p × Area for trachea =

\frac{d}{dt} (pv)=v\frac{ds}{dt} + p\frac{dv}{dt}

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

-1.225\times2.5434\times10^-^4\times0.20=6\times10^-^3\frac{ds}{dt} +1.225(-1\times10^-^4)

⇒-0.623133\times10^-^4+1.225\times10^-^4=6\times10^-^3\frac{ds}{dt}

           \frac{ds}{dt} = \frac{0.6018\times10^-^4}{6\times10^-^3} \\\\= 0.01003kg/m^3-s

ds/dt = 0.01003kg/m³-s

Thus, the time rate of change in air density during expiration is 0.01003kg/m³-s

3 0
3 years ago
Read 2 more answers
A cricket ball is dropped from a height of 20 m. Calculate: a) the speed of the ball. b) the time it takes to fall through this
AysviL [449]

Answer:

Initial velocity is 0 .. ( since it is just dropped)

now using V²= u² +2gh

=> (Vfinal)² = 0+2*10*20

=> v² = 20*20 = 400

=> v = √400 = 20m/s

for time taken

use V = u+gt

=> t = V/g = 20/10 = 2sec

7 0
2 years ago
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