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ICE Princess25 [194]
4 years ago
11

The internal energy of a system is always increased by __________. a adding heat to the system and having the system do work on

the surroundings b having the system do work on the surroundings c withdrawing heat from the system d a volume compression e adding heat to the system
Physics
1 answer:
Nitella [24]4 years ago
8 0
There are two correct options:
<span>d) a volume compression
e) adding heat to the system

In fact, we can have a look at the first law of thermodynamics:
</span>\Delta U = Q-W
<span>where
</span>\Delta U is the variation of internal energy
<span>Q is the heat absorbed by the system
W is the work done by the system on the surrounding

It is important to keep in mind the right sign convention for Q and W: Q is positive when the heat is absorbed by the system, while it is negative when it is released by the system; W is positive when it is done by the system on the surrounding, and negative when it is done by the surrounding on the system.

Using this convention, we can see that:
d) when there is a volume compression, the work is done by the surrounding on the system, so the work is negative, and the variation of internal energy is positive due to the negative sign -W in the first law of thermodynamics
e) when heat is added to the system, Q is positive, so the variation of internal energy is positive as well</span>
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Compared to the amount of heat jupiter receives from the sun, jupiter radiates _____ heat.
uysha [10]
C 
http://education.seattlepi.com/planet-gives-off-2-times-much-heat-receives-sun-5420.html
3 0
3 years ago
A 0.58 kg mass is moving horizontally with a speed of 6.0 m/s when it strikes a vertical wall. The mass rebounds with a speed of
Sunny_sXe [5.5K]

Answer:

5.8\; {\rm kg\cdot m \cdot s^{-1}}.

Explanation:

If the mass of an object is m and the velocity of that object is v, the linear momentum of that object would be m\, v.

Assume that the initial velocity of the mass is positive (6.0\; {\rm m\cdot s^{-1}}.) However, the direction of the velocity is reversed after the impact. Thus, the sign of the new velocity of the object would be negative- the opposite of that of the initial velocity. The new velocity would be (-4.0\; {\rm m\cdot s^{-1}}).

Thus, the change in the velocity of the mass would be:

\begin{aligned}& (\text{Change in Velocity}) \\ =\; & (\text{Final Velocity}) - (\text{Initial Velocity}) \\ =\; & (-4.0\; {\rm m\cdot s^{-1}}) - (6.0\; {\rm m\cdot s^{-1}}) \\ =\; & (-10\; {\rm m\cdot s^{-1})\end{aligned}.

The change in the linear momentum of the mass would be:

\begin{aligned} & \text{change in momentum} \\ =\; & (\text{mass}) \times (\text{change in velocity}) \\ =\; & 0.58\; {\rm kg} \times (-10\; {\rm m\cdot s^{-1}}) \\  =\; & (-5.8\; {\rm kg \cdot m \cdot s^{-1}})\end{aligned}.

Thus, the magnitude of the change of the linear momentum would be 5.8\; {\rm kg \cdot m \cdot s^{-1}}.

7 0
3 years ago
A cannon of mass 5.71 x 103 kg is rigidly bolted to the earth so it can recoil only by a negligible amount. The cannon fires a 7
zheka24 [161]

Answer:

541.14 m/s

Explanation:

We are given that

Mass of cannon=m_1=5.71\times 10^3 kg

Mass of shell,m_2=73.5 kg

Initial velocity of shell,v=547 m/s

We have to find the velocity of shell fired from this loose cannon.

According to law of conservation of momentum

m_1v_1+m_2v_2=m_1u_1+m_2u_2

Initial momentum of system=0

m_1v_1=-m_2v_2

v_1=-\frac{m_2v_2}{m_1}

When the cannon is bolted to the ground then only shell moves and kinetic energy of system equals to kinetic energy of shell

Kinetic energy of shell,K.E=\frac{1}{2}mv^2=\frac{1}{2}\times 73.5(547)^2}=1.09\times 10^7 J

K.E of shell=\frac{1}{2}m_1v^2_1+\frac{1}{2}m_2v^2_2

K.E of shell=\frac{1}{2}m_1(-\frac{m_2v_2}{m_1})^2+\frac{1}{2}m_2v^2_2

K.E of shell=\frac{1}{2}(\frac{m^2_2v^2_2}{m_1}+\frac{1}{2}m_2v^2_2)

2K.E of shell=m_2v^2_2(\frac{m_2}{m_1}+1)

Velocity of shell fired from this loose cannon,v_2=\sqrt{\frac{2k.E}{m_2(\frac{m_2}{m_1}+1)}

v_2=\sqrt{\frac{2\times 1.09\times 10^7}{73.5(\frac{73.5}{5.71\times 10^3}+1)}

v_2=541.14m/s

Hence, the velocity of shell fired from this loose cannon would be 541.14 m/s

5 0
3 years ago
At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particu
Ne4ueva [31]

Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = E_{h}

Let the vertical electric field component = E_{v}

The formula for light intensity is given by:

I = \frac{E_{m} ^{2} }{2c \mu}..............................(1)

E_{m} is the resolution of the vertical and horizontal components, E_{h} and    E_{v}

E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}..................(2)

Light intensity before the glasses were put on:

I_{1}  = \frac{E_{m} ^{2} }{2c \mu_{1} }.............................(3)

Put equation (2) into equation (3)

I_{1}  = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }.............................(4)

After the glasses were put on the horizontal component vanishes, i.e. E_{h} = 0

I_{2}  = \frac{ E_{v} ^{2}}{2c \mu_{2} }...................................(5)

Divide equation (5) by equation (4)

\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}...............................(6)

But E_{h} = 2.3E_{v}......................(7)

Insert equation (7) into (6)

\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2}  }{(2.3E_{v})^{2}   + E_{v} ^{2}   } \\\frac{I_{2} }{I_{1} } =  \frac{E_{v}^{2}  }{5.29E_{v}^{2}   + E_{v} ^{2}   }\\\frac{I_{2} }{I_{1} } =  \frac{E_{v}^{2}  }{6.29E_{v}^{2}  }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\

\frac{I_{2} }{I_{1} }= 0.159

b) When the sunbather lies on his side, the vertical component vanishes, i.e E_{v} = 0

\frac{I_{2} }{I_{1} } = \frac{E_{h} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}

\frac{I_{2} }{I_{1} } = \frac{(2.3E_{v} )^{2}  }{E_{v} ^{2} +(2.3E_{v} )^{2}}

\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2}  }{E_{v} ^{2} +5.29E_{v}^{2} }

\frac{I_{2} }{I_{1} } = \frac{5.29E_{v}^{2}  }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } = \frac{5.29}{6.29} \\\frac{I_{2} }{I_{1} } = 0.84

8 0
3 years ago
Newtons third law is called law of action and reaction?
TiliK225 [7]

Answer:

His third law states that for every action (force) in nature there is an equal and opposite reaction. In other words, if object A exerts a force on object B, then object B also exerts an equal and opposite force on object A. ... In reaction, a thrusting force is produced in the opposite direction.

Explanation:

7 0
3 years ago
Read 2 more answers
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