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s344n2d4d5 [400]

3 years ago

14

A lump of steel of mass 10kg at 627 degree Celsius is dropped in 100kg oil at 30 degree Celsius . the specific heat of steel And

oil are 0.5kj/kg.k and 3.5kj/kg.k calculate the entropy change in steel,oil and in the universe.

1 answer:

Naily [24]3 years ago

8 0

Answer:

700J

Explanation:

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if you could see the electric field coming off a proton you would notice that electric field lines are

lbvjy [14]

<span>you would notice that they're going out from the proton.</span>

8 0

3 years ago

¿Cuál es el volumen de un cubo de acero si se sabe que su masa es 50 hg, si la densidad del acero es 7850 kg/m3? Exprese el resu

Goshia [24]

Answer:

Conceptos y Magnitudes en

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Luciano LarozeNicols Porras Gonzalo Fuster

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7 0

3 years ago

By what distance do two objects carrying 1.0 C of charge each have to be separated before the electric force exerted on each obj

NNADVOKAT [17]

Answer:

Distance between both the object will be 404.51 m

Explanation:

We have given charge on two objects $q_1=q_2=1C$

Coulomb force between the two objects is given F = 5.5 N

We have to fond the distance between both the object so that force between them is 5.5 N

According to coulomb law force between two charge particle is $F=\frac{1}{4\pi \epsilon _0}\frac{q_1q_2}{r^2}$

So $5.5=\frac{9\times 10^9\times 1\times 1}{r^2}$

$r^2=16.36\times 10^8$

r = 404.51 m

So distance between both object will be 404.51 m

6 0

3 years ago

A 62 kg boy and a 37 kg girl use an elastic rope while engaged in a tug-of-war on a friction-less icy surface. If the accelerati

Paraphin [41]

Answer:

The magnitude of the acceleration of the boy toward the girl is $1.31\ m/s^2$

Explanation:

It is given that,

Mass of the boy, $m_1=62\ kg$

Mass of the girl, $m_2=37\ kg$

The acceleration of the girl toward the boy is, $a_2=2.2\ m/s^2$

To find,

The acceleration of the boy toward the girl.

Solution,

Let $a_1$ is the magnitude of the acceleration of the boy toward the girl. We know that force acting on one object to other are equal in magnitude but opposite in direction. So,

$F_1=-F_2$

$m_1a_1=-m_2a_2$

$a_1=-\dfrac{m_2a_2}{m_1}$

$a_1=-\dfrac{37\times 2.2}{62}$

$a_1=-1.31\ m/s^2$

$|a_1|=1.31\ m/s^2$

So, the magnitude of the acceleration of the boy toward the girl is $1.31\ m/s^2$

7 0

3 years ago

Help me with this question please

SpyIntel [72]

Answer:

the answer is A, 9.5 units

4 0

3 years ago