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juin [17]
3 years ago
8

The latent heat of fusion of alcohol is 25 kcal/kg and its melting point is -114 o C. It has a specific heat of 0.60 in its liqu

id state. Calculate the energy needed to melt 2.0 kg of solid alcohol at -114 o C and raise it to a temperature of -14 o C.
Physics
1 answer:
UNO [17]3 years ago
3 0

Answer:

=170kcal

Explanation:

We first calculate the amount of energy required to melt the alcohol using the formula: MLf, where Lf is the latent heat of fussion

We then calculate amount of heat required to raise the temperature of liquid alcohol to -14° C using MC∅.  We then add the two.

Thus ΔH=MLf+MC∅

ΔH=2kg×25kcal/kg+ 2kg×(0.6kcal/kg.K×(-14-⁻114)

=50kcal+120kcal

=170kcal

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Negatively charged particle that orbits the nucleus
Ede4ka [16]

Answer: Electrons.

Explanation:

Electrons are the negatively charged particles that orbit the nucleus of an atom. Protons, on the other hand, are the positively charged particles that orbit the nucleus of an atom.

in the picture, the protons and neutrons are “in” the nucleus and you’ll also see the the electrons are orbiting all around the nucleus.

If this helped please mark me brainliest.

6 0
3 years ago
consider the free-body diagram. if you want the box to move, the force applied while dragging must be greater than the
NeTakaya

Answer:

Force of static friction between the two surfaces

Explanation:

When two surfaces come into contact, they exert a force that resist the sliding of the two surfaces. This force is called static friction.

This force is given by the relation

                                       F_{s}=\mu_{s}\eta

Where,

                             μ - coefficient of static friction

                             η - normal force acting on the body

When a force acts on a body placed on a rough surface, it doesn't do any work if the applied force was less than the force of static friction.

So, in order to move the body, the applied force should be greater than the force of static friction.

6 0
3 years ago
What is the final concentration of DD at equilibrium if the initial concentrations are [A][A]A_i = 1.00 MM and [B][B]B_i = 2.00
pentagon [3]

Answer:

A) Concentration of A left at equilibrium of we started the reaction with [A] = 2.00 M and [B] = 2.00 M is 0.55 M.

B) Final concentration of D at equilibrium if the initial concentrations are [A] = 1.00 M and [B] = 2.00 M is 0.90 M.

[D] = 0.90 M

Explanation:

With the first assumption that the volume of reacting mixture doesn't change throughout the reaction.

This allows us to use concentration in mol/L interchangeably with number of moles in stoichiometric calculations.

- The first attached image contains the correct question.

- The solution to part A is presented in the second attached image.

- The solution to part B is presented in the third attached image.

8 0
3 years ago
The cable of the 1800kg elevator cab in Fig. 8−56 snaps when the cab is at rest at the first floor, where the cab bottom is a di
drek231 [11]

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

a ) The speed of the cab just before it hits the spring,

Ki + Pi = K final + P final + W

1 / 2 m vo² + m g hi = 1 / 2 m v² + m g h + f d

0 + ( 1800 * 9.8 * 3.7 m ) = ( 1 / 2 * 1800 * v² ) + 0 + ( 4400 * 3.7 )

v = 7.4 m / s

b ) The maximum distance x that the spring is compressed,

Ki + Pi = K final + P final + W + Fs

1 / 2 m vo² + m g x = 1 / 2 m v² + m g h + f d + 1 /2 k x²

( 1/2 * 1800 * 7.4² ) + ( 1800 * 9.8 * x ) =0 + 0+ ( 4400 * x ) + ( 1/2 * 1800 * x² )

75000 x² + 13420 x - 50625 = 0

x = 0.9 m

c ) The distance that the cab will bounce back up the shaft,

Ki + Pi + Fs = K final + P final + W

1 / 2 m vo² + m g x + 1 /2 k x² = 1 / 2 m v² + m g h + f d

0 + 0 + ( 1 / 2 * 0.15 * 0.9² ) = 0 + ( 1800 * 9.8 * h ) + ( 4400 * h )

h = 2.8 m

Therefore,

a ) The speed of the cab just before it hits the spring = 7.4 m / s

b ) The maximum distance x that the spring is compressed = 0.9 m

c ) The distance that the cab will bounce back up the shaft = 2.8 m

To know more about law of conservation of energy

brainly.com/question/12050604

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3 0
11 months ago
Kepler’s third law can be used to derive the relation between the orbital period, P (measured in days), and the semimajor axis,
NikAS [45]
Kepler's 3rd law is given as
P² = kA³
where
P = period, days
A = semimajor axis, AU
k = constant

Given:
P = 687 days
A = 1.52 AU

Therefore
k = P²/A³ = 687²/1.52³ = 1.3439 x 10⁵ days²/AU³

Answer:  1.3439 x 10⁵ (days²/AU³)

8 0
3 years ago
Read 2 more answers
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