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Savatey [412]
3 years ago
13

How much heat is absorbed by 3 kg honey baked ham as energy from the oven causes its temperature to change from 10°C to 60°C? (S

pecific heat capacity of ham 3287 cal/kgºC)?
Physics
1 answer:
sergiy2304 [10]3 years ago
3 0

Answer:

The answer is 493.05kJ

Explanation:

given

mass m= 3kg

t1=  10°C

t2= 60°C

Specific heat capacity of ham =3287 cal/kgºC

Required,

The quantity of heat, Q

Step two:

the formula is

Q= mcΔT

substitute

Q= 3*3287(60-10)

Q=9861*50

Q=493050 J

Q= 493.05kJ

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Answer:

Answer:

New speed of the 22-kg block is 1.57 m/s

Explanation:

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g In 1920, Stern and Gerlach performed an experiment that first demonstrated Group of answer choices energy quantization. space
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3 years ago
A 1.5 kg ball is dropped from a height of 2.Gm. Assuming energy is
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6 0
2 years ago
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A dockworker loading crates on a ship finds that a 21-kg crate, initially at rest on a horizontal surface, requires a 73-N horiz
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1) Static friction coefficient: 0.355

The crate is initially at rest. The crate remains at rest until the horizontal pushing force is less than the maximum static frictional force.

The maximum static frictional force is given by

F_s = \mu_s mg

where

\mu_s is the static coefficient of friction

m = 21 kg is the mass of the crate

g = 9.8 m/s^2 is the acceleration due to gravity

The horizontal force required to set the crate in motion is 73 N: this means that this is the value of the maximum static frictional force. So we have

F_s=73 N

Using this information into the previous equation, we can find the coefficient of static friction:

\mu_s = \frac{F}{mg}=\frac{73 N}{(21 kg)(9.8 m/s^2)}=0.355

2) Kinetic friction coefficient: 0.267

Now the crate is in motion: this means that the kinetic friction is acting on the crate, and its magnitude is

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So we can write Newton's second law as

F-F_k = ma = 0

And by substituting (1), we can find the value of the coefficient of kinetic friction:

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5 0
3 years ago
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