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Radda [10]
3 years ago
9

Given: Q1 = +10 uc = 1.0 x 10-5C

Physics
1 answer:
ser-zykov [4K]3 years ago
5 0

Answer:

-0.038 N

Explanation:

F=K Q1 Q2/r^2 by COULOMB'S LAW

F= 9×10^9×1×10^-5×-1.5×10^-5/(6)^2

F= -0.038 N

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How long does it take an object to travel 375 m at a rate of 25 m/s?
Leni [432]

Answer:

15s

Explanation:

Given parameters:

Distance traveled  = 375m

Speed = 25m/s

Unknown:

Time taken = ?

Solution:

To solve this problem, we make time the subject of the speed equation.

    Speed  = \frac{distance}{time}  

  Time  = \frac{distance}{speed}  

 Now insert the parameters and solve;

  Time  = \frac{375}{25}   = 15s

3 0
3 years ago
Derivation 1.2 showed how to calculate the work of reversible, isothermal expansion of a perfect gas. Suppose that the expansion
stellarik [79]

Answer:

(a) The work done by the gas on the surroundings is, 17537.016 J

(b) The entropy change of the gas is, 73.0709 J/K

(c) The entropy change of the gas is equal to zero.

Explanation:

(a) The expression used for work done in reversible isothermal expansion will be,

where,

w = work done = ?

n = number of moles of gas  = 4 mole

R = gas constant = 8.314 J/mole K

T = temperature of gas  = 240 K

= initial volume of gas  =  

= final volume of gas  =  

Now put all the given values in the above formula, we get:

The work done by the gas on the surroundings is, 17537.016 J

(b) Now we have to calculate the entropy change of the gas.

As per first law of thermodynamic,

where,

= internal energy

q = heat

w = work done

As we know that, the term internal energy is the depend on the temperature and the process is isothermal that means at constant temperature.

So, at constant temperature the internal energy is equal to zero.

Thus, w = q = 17537.016 J

Formula used for entropy change:

The entropy change of the gas is, 73.0709 J/K

(c) Now we have to calculate the entropy change of the gas when the expansion is reversible and adiabatic instead of isothermal.

As we know that, in adiabatic process there is no heat exchange between the system and surroundings. That means, q = constant = 0

So, from this we conclude that the entropy change of the gas must also be equal to zero.

Explanation:

6 0
2 years ago
Compare the force of air resistance and the force of gravity on an object falling at its terminal velocity.
Sholpan [36]
The viscous force on an object moving through air is proportional to its velocity.
The only forces acting on an object when falling are air resistance and its weight itself. The weight acts vertically downwards whereas air resistance acts vertically upward.
Let F be the viscous force due to air molecules, B be buoyant force due to air and W be the weight of falling object. Initially, the velocity of falling object and hence the viscous force F is zero and the object is accelerated due to force
(W-B). Because of the acceleration the velocity increases and accordingly the viscous force also increases. At a certain instant, the viscous force becomes equal to W-B. The net force then becomes zero and the object falls with constant velocity. This constant velocity is called terminal velocity.
Thus at terminal velocity, air resistance and force of gravity becomes equal.
7 0
3 years ago
How fast is the sixth cosmic velocity?
denpristay [2]
<span>25,000 miles per hour

hope that this helps</span>
3 0
3 years ago
Radiation present in the environment but not produced by humans is called ______.
mafiozo [28]

Answer:

background

Explanation:

5 0
2 years ago
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