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Anna71 [15]
2 years ago
8

In an experiment in which molten naphthalene is allowed to cool, the cooling curve shown below was obtained, the temperature 80∘

C is known as
A. cooling temperature

B. boiling point

C. melting point

D. vaporization point

Physics
1 answer:
notka56 [123]2 years ago
7 0

Answer:

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Explanation:

In an experiment in which molten naphthalene is allowed to cool, the cooling cur...

Question

In an experiment in which molten naphthalene is allowed to cool, the cooling curve in fig 5 was obtained, the temperature

80

∘

<h2>C.</h2>

is known as

Options

A) cooling temperature

B) boiling point

C) melting point √√√√√√√√√√※※※ ↑↑ANSWER

D) vaporization point

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Proper design of automobile braking systems must account for heat buildup under heavy braking. Part A Calculate the thermal ener
AVprozaik [17]

Answer:

1838216 J

Explanation:

95 km/h = 26.39 m/s

40 km/h = 11.11 m/s

Initial kinetic energy

= .5 x 1600 x(26.39)²

= 557145.67 J

Final kinetic energy

= .5 x 1600 x ( 11.11)²

= 98745.68 J

Loss of kinetic energy

= 458400 J

Loss of potential energy

= mg x loss of height

= 1600 x 9.8 x 340 sin 15

= 1379816 J

Sum of Loss of potential energy and Loss of kinetic energy

=  1379816 + 458400

= 1838216 J

This is the work done by the friction . So this is heat generated.

8 0
3 years ago
A net force of +15 N changes the momentum of an object by +100 kg-m/s. What is the time over which the force is applied? (please
sattari [20]
As momentum / time = force
so; time = 100÷15

so your answer is 6.7 !!
3 0
3 years ago
. An object whose mass is 375 lb falls freely under the influence of gravity from an initial elevation of 253 ft above the surfa
lozanna [386]

Answer:

(a) Vf = 128 ft/s

(b) K.E = 122.8 Btu

Explanation:

(a)

In order to find the velocity of the object just before striking the surface of earth or the final velocity, we use 3rd equation of motion:

2gh = Vf² - Vi²

where,

g = 32.2 ft/s²

h = height = 253 ft

Vf = Final Velocity = ?

Vi = Initial Velocity = 10 ft/s

Therefore,

(2)(32.2 ft/s²)(253 ft) = Vf² - (10 ft/s)²

16293.2 ft²/s² + 100 ft²/s² = Vf²

Vf = √(16393.2 ft²/s²)

<u>Vf = 128 ft/s</u>

<u></u>

(b)

The kinetic energy of the object before it hits the surface of earth is given by:

K.E = (0.5)(m)(Vf)²

where,

m = mass of object = 375 lb

K.E = Kinetic energy of object before it strikes the surface of earth = ?

Therefore,

K.E = (0.5)(375 lb)(128 ft/s)²

K.E = 3073725 lb.ft²/s²

Now, converting this to Btu:

K.E = (3073725 lb.ft²/s²)(1 Btu/25037 lb.ft²/s²)

<u>K.E = 122.8 Btu</u>

3 0
3 years ago
needhelpp101 Why is the gravitational potential energy of an object 1 meter above the moon’s surface less than its potential ene
Vlad1618 [11]

If you remember the formula for potential energy,
then this question is a piece-o-cake.

  <em>Potential energy = (mass) x (<u>acceleration of gravity</u>) x (height) .</em>

-- The object's mass is the same everywhere.
-- You said that the height is the same both times.
-- How about the acceleration of gravity ? 

Compared to gravity on Earth, it's only  16.5 percent as much on the Moon. 
So naturally, from the formula, you'd expect the Potential Energy to be less
on the Moon.

4 0
3 years ago
Find the voltage across the 15 Ω resistor. .<br> 35 Ω<br> 20 Ω<br> 15 Ω<br> 10 V
almond37 [142]

Answer:

35 Ω

Explanation:

3 0
2 years ago
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