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3 years ago

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A 2.3 kg block of copper is heated at atmospheric pressure such that its temperature increases from 6 oC to 90 oC. How much heat

(in J) is absorbed by the block of copper? The specific heat of copper is 0.092 (kcal/(kgoC))

1 answer:

Shtirlitz [24]3 years ago

7 0

Answer:

the heat absorbed by the block of copper is 74368.476J

Explanation:

Hello!

To solve this problem use the first law of thermodynamics that states that the heat applied to a system is the difference between the initial and final energy considering that the mass and the specific heat do not change so we can infer the following equation

Q=mCp(T2-T1)

Where

Q=heat

m=mass=2.3kg

Cp=0.092 kcal/(kg C)=384.93J/kgK

T2=Final temperatura= 90C

T1= initial temperature=6 C

solving

$Q=(2.3kg)(384.93\frac{J}{kgC} )(90C-6C)=74368.476J$

the heat absorbed by the block of copper is 74368.476J

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Un vas plin cu lichid cântăreşte 175kg. Ceea ce reprezintă de 5 ori masa vasului gol. Ştiind că volumul interior al vasului este

Mariana [72]

a) Density of the liquid: $823.5kg/m^3$

b) Weight of the liquid: 1372 N

Explanation:

Translation of the text:

<em>"A full tank with liquid weighs 175kg. Which is 5 times the mass of the empty vessel. Knowing that the inside volume of the vessel is 0.17kl, calculate: </em>

<em>a) the density of the liquid; </em>

<em>b) the weight of the liquid."</em>

a)

We know that the full tank with liquid has a total mass of M = 175 kg. We can write the total mass as

$M=m_L + m_V$ (1)

where

$m_L$ is the mass of the liquid

$m_V$ is the mass of the vessel

We also know that the total mass M is 5 times the mass of the empty vessel, so we have:

$M=5m_V\\m_V=\frac{M}{5}=\frac{175}{5}=35 kg$

which is the mass of the empty vessel.

Therefore, we can find the mass of the liquid only using (1):

$m_L=M-m_V=175-35=140 kg$

The density of the liquid is given by

$d=\frac{m}{V}$

where

m = 140 kg (mass of the liquid)

V = 0.170 kL = 170 L = $0.170 m^3$ (volume of the liquid, which is equal to the volume of the vessel)

So we get

$d=\frac{140}{0.170}=823.5kg/m^3$

b)

The weight of a body is given by

$F=mg$

where

m is its mass

g is the acceleration due to gravity

For the liquid in this problem, we have

m = 140 kg (mass)

$g=9.8 m/s^2$ (acceleration due to gravity)

Therefore, its weight is

$F=(140)(9.8)=1372 N$

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3 years ago

A man has 887.5 J of kinetic energy while running with a velocity of 5 m/s. What is his mass?

monitta

Answer:

The mass of the man is 71 kg

Explanation:

Given;

kinetic energy of the man, K.E = 887.5 J

velocity of the man, v = 5 m/s

The mass of the man is calculated as follows;

K.E = ¹/₂mv²

where;

m is the mass of the man

2K.E = mv²

m = 2K.E / v²

m = (2 x 887.5) / (5)²

m = 71 kg

Therefore, the mass of the man is 71 kg

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3 years ago

Give at least one fact about subatomic particles

victus00 [196]

Answer:

- Particles smaller than atoms are called subatomic particles .

- There are three famous subatomic particles, proton, neutron and electron .

- The study of sub atomic particles are called particle physics

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3 years ago

A factory worker pushes a 30.0-kg crate a distance of 4.5 m along a level floor at constant velocity by pushing horizontally on

SIZIF [17.4K]

(a) 73.5 N

The velocity of the crate is constant: this means that the acceleration is zero (a=0), so according to Newton's second law

$\sum F = ma$

the resultant of the forces must be zero: $\sum F = 0$ (1)

The motion is along the horizontal direction, so we are only interested in the forces acting along this direction. There are two of them:

$F$ , the push applied by the worker

$F_f=-\mu mg$ , the force of friction, with $\mu=0.25$ being the coefficient of friction, $m=30.0 kg$ being the mass of the crate, and $g=9.8 m/s^2$ . The negative sign is due to the fact that the friction acts in the opposite direction to the motion. Eq.(1) then becomes

$F-\mu mg=0\\F=\mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$

So, this is the force that the worker must apply.

(b) 330.8 J

The work done by the pushing force of the worker on the crate is given by:

$W=Fd cos \theta$

where

F = 73.5 N is the force

d = 4.5 m is the displacement

$\theta=0^{\circ}$ is the angle between the direction of the force and the displacement (0 degrees, since they are in same direction)

Substituting, we have

$W=(73.5 N)(4.5 m)(cos 0^{\circ})=330.8 J$

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To calculate the work done by friction, we apply the same formula:

$W=F_f d cos \theta$

where

$F_f = \mu mg=(0.25)(30.0 kg)(9.8 m/s^2)=73.5 N$ is the magnitude of the force of friction

d = 4.5 m is the displacement

$\theta=180^{\circ}$ is the angle between the direction of the force of friction and the displacement (it is 180 degrees since the two are into opposite directions)

Substituting, we find

$W=(73.5 N)(4.5 m)(cos 180^{\circ})=-330.8 J$

So, the work done by friction is negative.

(d) 0 J

As before, the work done by any force on the crate is

$W=F_f d cos \theta$

We notice that both gravity and normal force are perpendicular to the displacement: therefore, $\theta=90^{circ}$ , and so

$cos \theta=0$

which means that the work done by both forces is zero.

(e) 0 J

The total work done on the crate is the sum of the work done by the four forces acting on it, so:

$W=W_{push} + W_{friction}+W_{gravity}+W_{normal}=330.8J-330.8J+0+0=0$

And this is in accordance with the work-energy theorem, which states that the variation of kinetic energy of the crate is equal to the work done on it: since the crate is moving at constant velocity, its variation of kinetic energy is zero, as well as the work done on it.

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Sveta_85 [38]

The centre of mass of the system is the point at which the total

mass of system could be concentrated without changing the moment

of the system.

Centre of Mass is the point at which the whole mass of the system

is assumed to be concentrated.

The general formula for the COM is:

xₙ = Σmₐxₐ / Σmₐ where, a = 1,2,3.........n

Here the term Σ mₐ xₐ is called the first moment of the system and the

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Therefore, from this theory we can say that the moment of the system

remain unchanged while calculating the COM.

Hence, The centre of mass of the system is the point at which the total

mass of system could be concentrated without changing the moment

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8 0

2 years ago