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Hoochie [10]
3 years ago
6

The acronym laser stands for light amplification by ____ emission of radiation

Physics
1 answer:
AleksAgata [21]3 years ago
6 0
<span>light amplification by stimulated emission of radiation </span>
You might be interested in
If a nearsighted person has a far point df that is 3.50 m from the eye, what is the focal length f1 of the contact lenses that t
Anastaziya [24]

Answer:

f1 = -3.50 m

Explanation:

For a nearsighted person an object at infinity must be made to  appear  to be at his far point which is 3.50 m away. The image of an object at infinity must be formed on the same side of the lens as the object.

∴ v = -3.5 m

Using mirror formula,

i/f1 = 1/v + 1/u

Where f1 = focal length of the contact lens, v = image distance = -3.5 m, u =         object distance = at infinity(∞) = 1/0

∴ 1/f1 = (1/-3.5) + 1/infinity

  Note that, 1/infinity = 1/(1/0) = 0/1 =0.

∴ 1/f1 = 1/(-3.5) + 0

  1/f1 = 1/(-3.5)

Solving the equation by finding the inverse of both side of the equation.

∴ f1 = -3.50 m

 Therefore a converging lens of focal length  f1 = -3.50 m

would be needed by the person to see an object at infinity clearly

8 0
3 years ago
A ball is attached to a string of length 3 m to make a pendulum. The pendulum is placed at a location that is away from the Eart
Musya8 [376]

1) 0.61 m/s^2

2) 13.9 s

Explanation:

1)

The acceleration due to gravity is the acceleration that an object in free fall (acted upon the force of gravity only) would have.

It can be calculated using the equation:

g=\frac{GM}{r^2} (1)

where

G is the gravitational constant

M=5.98\cdot 10^{24} kg is the Earth's mass

r is the distance of the object from the Earth's center

The pendulum in the problem is at an altitude of 3 times the radius of the Earth (R), so its distance from the Earth's center is

r=4R

where

R=6.37\cdot 10^6 m is the Earth's radius

Therefore, we can calculate the acceleration due to gravity at that height using eq.(1):

g=\frac{GM}{(4R)^2}=\frac{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})0.}{(4\cdot 6.37\cdot 10^6)^2}=0.61 m/s^2

2)

The period of a simple pendulum is the time the pendulum takes to complete one oscillation. It is given by the formula

T=2\pi \sqrt{\frac{L}{g}}

where

L is the length of the pendulum

g is the acceleration due to gravity at the location of the pendulum

Note that the period of a pendulum does not depend on its mass.

For the pendulum in this problem, we have:

L = 3 m is its length

g=0.61 m/s^2 is the acceleration due to gravity (calculated in part 1)

Therefore, the period of the pendulum is:

T=2\pi \sqrt{\frac{3}{0.61}}=13.9 s

4 0
3 years ago
Which country’s data center industry taps latest in energy-efficient technology?.
SCORPION-xisa [38]

Answer:

N i g e r i a

Explanation:

Hope that helps!!!

Have A good day/night!!

3 0
2 years ago
6. In a competition, Mike lifts a 100 kg weight from the floor in 2 sec. Brian also
zaharov [31]

Answer:

Explanation:

Analogy:

Who does more work:

Work done is defined as the product of force and distance. Work is done when force moves a body in a particular direction.

In this case, work done is the same thing as the potential energy of the weight.

           Work done = potential energy = m x g x h

 Mass of weight = 100kg for both Mike and Brian

g is the acceleration due to gravity

h is the height

Since they lifted the weight through the same height;

        Work done by Mike = 100gh

        Work done by Brian = 100gh

Both Mike and Brian does equal work

Who generates more power:

Power is the rate at which work is done;

          Power  = \frac{Work done}{time}

Power generated by Mike = \frac{100gh}{2}  = 50gh

Power generated by Brian = \frac{100}{3} = 33.33gh

We can see that Mike generated more power

6 0
3 years ago
A piston having 7.23 g of steam at 110°C increases its temperature by 35°C. At the same time it expands from a volume of 2.00 L
My name is Ann [436]

Answer : The value of q,w,\Delta U\text{ and }\Delta U is 505 J, -599 J, -94 J and -693 J respectively.

Explanation : Given,

Mass of steam = 7.23 g

Initial temperature = 110^oC

Final temperature = (110+35)^oC=145^oC

Initial volume = 2 L

Final volume = 8 L

External pressure = 0.985 bar

Heat capacity of steam = 1.996 J/g.K

First law of thermodynamic : It states that the energy can not be created or destroyed, it can only change or transfer from one state to another state.

As per first law of thermodynamic,

\Delta U=q+w

First we have to calculate the heat absorbed by the system.

Formula used :

Q=m\times c\times \Delta T

or,

Q=m\times c\times (T_2-T_1)

where,

Q = heat absorbed by the system = ?

m = mass of steam = 7.23 g

C_p = heat capacity of steam = 1.966J/g.K

T_1 = initial temperature  = 110^oC=273+110=383K

T_2 = final temperature  = 145^oC=273+145=418K

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

Q=7.23g\times 1.966J/g.K\times (418-383)K

Q=505J

Now we have to calculate the work done.

Formula used :

w=-p_{ext}dV\\\\w=-p_{ext}(V_2-V_1)

where,

w = work done  = ?

p_{ext} = external pressure = 0.985 bar = 0.985 atm   (1 bar = 1 atm)

V_1 = initial volume of gas = 2.00 L

V_2 = final volume of gas = 8.00 L

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

w=-p_{ext}(V_2-V_1)

w=-(0.985atm)\times (8.00-2.00)L

w=-5.91L.atm=-5.91\times 101.3J=-599J

conversion used : (1 L.atm = 101.3 J)

Now we have to calculate the change in internal energy of the system.

\Delta U=q+w

\Delta U=505J+(-599J)

\Delta U=-94J

Now we have to calculate the change in enthalpy of the system.

Formula used :

\Delta H=\Delta U+P\Delta V

\Delta H=\Delta U+w

\Delta H=(-94J)+(-599J)

\Delta H=-693J

Therefore, the value of q,w,\Delta U\text{ and }\Delta U is 505 J, -599 J, -94 J and -693 J respectively.

4 0
3 years ago
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