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2 years ago

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What is the value of work done on an object when a 0.1x102-newton force moves it 30 meters and the angle between the force and t

he
displacement is 25°?

1 answer:

disa [49]2 years ago

3 0

Answer:306joules

Explanation:

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What is the least possible initial kinetic energy in the oxygen atom could have and still excite the cesium atom?

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K=E[(m+M)/M]

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A 1000-kg car is slowly picking up speed as it goes around a horizontal curve whose radius is 100 m. The coefficient of static f

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Answer:

18.5 m/s

Explanation:

On a horizontal curve, the frictional force provides the centripetal force that keeps the car in circular motion:

$\mu mg = m\frac{v^2}{r}$

where

$\mu$ is the coefficient of static friction between the tires and the road

m is the mass of the car

g is the gravitational acceleration

v is the speed of the car

r is the radius of the curve

Re-arranging the equation,

$v=\sqrt{\mu gr}$

And by substituting the data of the problem, we find the speed at which the car begins to skid:

$v=\sqrt{(0.350)(9.8 m/s^2)(100 m)}=18.5 m/s$

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Describe the energy transformations that occur from the time a skydiver jumps out of a plane until landing on the ground.

Kisachek [45]

When the Skydiver jump out a plane, his Potential Energy is being converted or transform into Kinetic energy due to gravity. Hope this helps

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A closed, rigid container holding 0.2 moles of a monatomic ideal gas is placed over a Bunsen burner and heated slowly, starting

Georgia [21]

Answer:

a) 2250 J

b) 0 J

c) 2250 J

Explanation:

a) Since, the process is isochoric

the change in internal energy

$\Delta U = n C_v(T_f-T_i)$

Here, n = 0.2 moles

Cv = 12.5 J/mole.K

We have to find T_f so we can use gas equation as

$\frac{P_1V_1}{P_2V_2} =\frac{T_i}{T_f}\\Since, V_1=V_2 [isochoric/process]\\\Rightarrow \frac{P_{atm}}{4P_{atm}} = \frac{300}{T_f} \\\Rightarrow T_f = 1200 K$

So, $\Delta U= 0.2\times12.5(1200-300)\\=2250 J$

b) Since, the process is isochoric no work shall be done.

c) By first law of thermodynamics we have

$\Delta U = Q-W\\Since, W = 0\\\Delta U = Q\\Therefore, Q = 2250 J$

Since, Q is positive 2250 J of heat will flow into the system.

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2 years ago

A beam of unpolarized light with intensity I0 falls first upon a polarizer with transmission axis θTA,1 then upon a second polar

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Answer:

The intensity I₂ of the light beam emerging from the second polarizer is zero.

Explanation:

Given:

Intensity of first polarizer = Io/2

For the second polarizer, the intensity is equal:

$I_{2} =\frac{I_{o} }{2} (cos\theta )^{2} =\frac{I_{o} }{2} (cos90)^{2} =0$

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