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koban [17]
3 years ago
15

A iron block with a mass of 4.8 kg initially slides over a rough horizontal surface with a speed of 1.2 m/s. Friction slows the

block to rest. While slowing to rest, 85.0% of the kinetic energy of the block is absorbed by the block itself as internal energy. What is the temperature increase of the block
Physics
1 answer:
Harlamova29_29 [7]3 years ago
4 0

Answer:

Explanation:

Kinetic energy of block will be converted into heat energy by friction .

Heat energy produced = 1/2 m v²

= .5 x 4.8 x 1.2²

= 3.456 J

85% of energy is converted into heat energy , so heat energy produced

= .85 x 3.456 = 2.9376 J .

If Q heat is given to m mass of object having s as specific heat and Δt is increase in temperature

Q = msΔt

specific heat of iron s = 462 J / kg C

Putting the values ,

2.9376 = 4.8 x 462 x Δt

Δt = 13.24 x 10⁻⁴ ⁰C.

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25,000 Feet = 7620m

PE = mgh where m is mass, g is gravity accel: 9.8 n h is height

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3 years ago
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A submersible pump is put under the water at the bottom of a well and is used to push water up through a pipe. What minimum outp
Maslowich

Answer:

695800 N/m^2 or Pa

Explanation:

Height of the water from the ground H  =  71 m

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Density is a physical property that relates the mass of a substance to its volume. A. Calculate the density, in g/mL , of a liqu
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Answer:

A) 0.660 g/ml

B) 1.297 ml

C) 0.272 g

Explanation:

Every substance, body or material has mass and volume, however the mass of different substances occupy different volumes.  This is where density D appears as a  physical characteristic property of matter that establishes a relationship between the mass m of a body or substance and the volume V it occupies:

D=\frac{m}{V} (1)

Knowing this, let's begin with the answers:

<h2 /><h2>Answer A:</h2>

Here the mass is m=0.155g and th volume V=0.000235L=0.235mL

Solving (1) with these values:

D=\frac{0.155g}{0.235mL} (2)

D=0.660g/mL (3)

<h2>Answer B:</h2>

In this case the mass of a sample is m=4.71g and its density is D=3.63g/mL.

Isolating V from (1):

V=\frac{m}{D} (4)

V=\frac{4.71g}{3.63g/mL} (5)

V=1.297mL (5)

<h2>Answer C:</h2>

In this case the volume of a sample is V=0.293mL and its density is D=0.930g/mL.

Isolating m from (1):

m=D.V (6)

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