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LUCKY_DIMON [66]
4 years ago
14

What can happen to the atomic particles when u rub two objects together ​

Physics
1 answer:
MAVERICK [17]4 years ago
3 0
Look it up on google it has the answer
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Evgen [1.6K]

<span>I think you're looking for "spectrum".</span>



7 0
4 years ago
As the car falls off the cliff, what is happening to the kinetic energy of the falling car?
Airida [17]

Answer:

The kinetic energy is increasing as potential energy is converted.

Explanation:

7 0
3 years ago
If a rod attached to the approaching charge if the rod consists of "stiff" spring-like bonds for which atoms undergo small oscil
Hoochie [10]

Answer: hello options related to your question is missing attached below is the missing part of your question

answer: No charge of the length of the bonds expected because the rod did not touch the charge source ( option A )

Explanation:

When the Charge is first, Furthest away and second  and closest to the source charge. <em>The spring like bonds can be said to have No charge of the length  of the bonds expected because the rod did not touch the charge source </em><em>when Furthest away the bond with charge will be less effective </em>

5 0
3 years ago
(8 points) Air is used as the working fluid in a simple ideal Brayton cycle that has a pressure ratio of 12, a compressor inlet
trapecia [35]

Answer:

a ) \dot m = 351.49 kg/s

b)  \dot m_{actual} = 1046.15 kg/s

Explanation:

given data:

pressure ration rp = 12

inlet temperature = 300 K

TURBINE inlet temperature  = 1000 K

AT the end of isentropic process (compression) temperature is

\frac{T_2'}{T_1} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{T_2'}{300} = 12^{\frac{1.4 -1}{1.4}}

T_2' = 610.181 K

AT the end of isentropic process (expansion) temperature is

\frac{T_3}{T_4'} = rp ^{\frac{\gamma -1}{\gamma}}

\frac{1000'}{T_4'} = 12^{\frac{1.4 -1}{1.4}}

T_4' = 491.66 K

isentropic work is given as

w(compressor) = CP (T_2' -T_1)

w = 1.005(610.18 - 300)

w = 311.73 kJ/kg

w(turbine) = 1.005( 1000 - 491.66)

w(turbine) = 510.88 kJ/kg

a) mass flow rate for isentropic process is given as

\dot m = \frac{70000}{510.88 - 311.73}

\dot m = 351.49 kg/s

b) actual mass flow rate uis given as

\dot m_{actual} = \frac{70000}{51.088\times 0.85 - \frac{311.73}{0.85}}

\dot m_{actual} = 1046.15 kg/s

6 0
3 years ago
An artist working on a piece of metal in his forging studio plunges the hot metal into oil in order to harden it. The metal piec
lutik1710 [3]

Answer:

The temperature of the metal is  T_m  =  376.8 ^o C

Explanation:

From the question we are told that

     The mass of the metal is  M =  60 \ kg

     The specific heat of the metal is  c_p  =  0.1027 kcal/(kg \cdot ^oC)

       The mass of the oil is M_o  =  810 \ kg

       The temperature of the oil is  T_o  =  35^oC

       The specific heat of oil is  c_o  =  0.7167 kcal/(kg \cdot ^oC )

       The equilibrium temperature is T_e  =  39 ^oC

According to the law of energy conservation

     Heat lost by metal  =  heat gained by the oil

So  

   The quantity  of heat lost by the metal is mathematically represented as

               Q =  - Mc_p \Delta T

=>            Q =  -Mc_p (T_m  -  T_c)

Where T_ m  the temperature of metal before immersion

The negative sign show heat lost

The quantity  of gained t by the metal is mathematically represented as      

           Q =  M_o c_o \Delta T

=>        Q =  M_o c_o (T_c - T_o)

So  

         Mc_p (T_m  -  T_c)   =   M_o c_o (T_c - T_o)

substituting values

          - 60 * 0.1027 (T_m  - 39)   =   810 * 0.7167 *  (39 - 35)

=>       T_m  =  376.8 ^o C

         

6 0
3 years ago
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