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Nadya [2.5K]
3 years ago
15

A refrigerator removes 55.0 kcal of heat from the freezer and releases 73.5 kcal through the condenser on the back.How much work

was done by the compressor
Physics
1 answer:
sammy [17]3 years ago
5 0

Here refrigerator removes 55 kcal heat from freezer

Refrigerator releases 73.5 kcal heat to surrounding

So here we can use energy conservation principle by II Law of thermodynamics

the law says that

Q_1 = Q_2 + W

here we know that

Q_1 = heat released to the surrounding

Q_2 = heat absorbed from freezer

W = work done by the compressor

now using above equation we can write

73.5 = 55 + W

W = 73.5 - 55

W = 18.5 kcal

So here compressor has to do 18.5 k cal work on it

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43 kg bear slides, from rest, 15 m down a lodgepole pine tree, moving with a speed of 5.5 m/s just before hitting the ground. (a
olga2289 [7]

Answer:

-6327.45 Joules

650.375 Joules

378.47166 N

Explanation:

h = Height the bear slides from = 15 m

m = Mass of bear = 43 kg

g = Acceleration due to gravity = 9.81 m/s²

v = Velocity of bear = 5.5 m/s

f = Frictional force

Potential energy is given by

P=mgh\\\Rightarrow P=43\times -9.81\times 15\\\Rightarrow P=-6327.45\ J

Change that occurs in the gravitational potential energy of the bear-Earth system during the slide is -6327.45 Joules

Kinetic energy is given by

K=\frac{1}{2}mv^2\\\Rightarrow K=\frac{1}{2}\times 43\times 5.5^2\\\Rightarrow K=650.375\ J

Kinetic energy of the bear just before hitting the ground is 650.375 Joules

Change in total energy is given by

\Delta E=fh=-(\Delta K+\Delta P)\\\Rightarrow fh=-(650.375-6327.45)\\\Rightarrow fh=5677.075\\\Rightarrow f=\frac{5677.075}{h}\\\Rightarrow f=\frac{5677.075}{15}\\\Rightarrow f=378.47166\ N

The frictional force that acts on the sliding bear is 378.47166 N

5 0
3 years ago
Read 2 more answers
n the figure, a uniform ladder 12 meters long rests against a vertical frictionless wall. The ladder weighs 400 N and makes an a
Leto [7]

Since the ladder is standing, we know that the coefficient of friction is at least something. This [gotta be at least this] friction coefficient can be calculated. As the man begins to climb the ladder, the friction can even be less than the free-standing friction coefficient. However, as the man climbs the ladder, more and more friction is required. Since he eventually slips, we know that friction is less than what's required at the top of the ladder.

 

The only "answer" to this problem is putting lower and upper bounds on the coefficient. For the lower one, find how much friction the ladder needs to stand by itself. For the most that friction could be, find what friction is when the man reaches the top of the ladder.

Ff = uN1

Fx = 0 = Ff + N2

Fy = 0 = N1 – 400 – 864

N1 = 1264 N

Torque balance

T = 0 = N2(12)sin(60) – 400(6)cos(60) – 864(7.8)cos(60)

N2 = 439 N

Ff = 439= u N1

U = 440 / 1264 = 0.3481

3 0
3 years ago
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Calculate the acceleration of a helium nucleus (charge +2e, where e-electronic charge), when it enters an electric field of stre
konstantin123 [22]

Answer:

3.8 x 10^10 m/s^2

Explanation:

Charge, q = 2 e =  2 x 1.6 x 10^-19 C = 3.2 x 10^-19 C

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the force due to electric filed on a charge particle is given by

F = q x E

Where, q be the charge on the charged particle and E be the strength of electric field.

By substituting the values

F = 3.2 x 10^-19 x 790

F = 2528 x 10^-19 N

According to the Newton's second law

F = m x a

Where, me be the mass and a be the acceleration

By substituting the values

2528\times 10^{-19}=6.645\times 10^{-27}\times a

a = 3.8 x 10^10 m/s^2

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Answer:

25N

Explanation:

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What word can be used to describe the compression of a longitudal wave
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Answer:

Mechanical longitudinal waves are also called compressional or compression waves, because they produce compression and rarefaction when traveling through a medium, and pressure waves, because they produce increases and decreases in pressure.

Explanation:

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