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vova2212 [387]
2 years ago
10

There are devices to put in a light socket that control the current through a lightbulb, thereby increasing its lifetime. Which

of the following strategies would increase the lifetime of a bulb without making it dimmer? A. Reducing the average current through the bulb B. Limiting the maximum current through the bulb C. Increasing the average current through the bulb D. Limiting the minimum current through the bulb
Physics
1 answer:
Dmitrij [34]2 years ago
4 0

Answer: B

Explanation:

Limiting the maximum current through the bulb. This will help in preserving or improving the bulb's lifetime and also this won't have an effect on the brightness of the bulb as brightness is affected by the average value. Although brightness is a factor of current, reducing the maximum current won't have any bearing on the average current the bulb is getting.

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The motion of a particle connected to a spring is described by x = 10 sin (pi*t). At
Gekata [30.6K]

To solve this problem we will first apply the principle of energy conservation, for which the elastic potential energy must be the same as the elastic kinetic energy of the simple harmonic movement of the system.

From this conservation it will be possible to find in total terms the total displacement of the system and thus replace it in the given function to find the time.

The potential energy stored would be

PE= \frac{1}{2} kx^2

The kinetic energy would be

KE = \frac{1}{2}m\omega (A^2-x^2)

Here,

A = 10m

\omega = \pi rad/s

Now assuming the planned conservation of energy

PE = KE

\frac{1}{2}kx^2 =\frac{1}{2} m\omega^2 (A^2-x^2)

\frac{1}{2}kx^2 = \frac{1}{2} m(x\frac{k}{m})(A^2-x^2)

x^2 = A^2 -x^2

2x^2 = A^2

x = \frac{A}{\sqrt{2}}

x = \frac{10}{\sqrt{2}}

So now using the equation previously given to describe the motion of the particle we have to

\frac{10}{\sqrt{2}} = 10 sin(xt)

\frac{1}{\sqrt{2}} = sin (xt)

sin^{-1} (\frac{1}{\sqrt{2}} ) = \pi t

\frac{\pi}{4} = \pi t

t = \frac{1}{4}s= 0.25s

Therefore the correct option is B.

6 0
1 year ago
A quantum system has three energy levels, so three wavelengths appear in its emission spectrum. the shortest observed wavelength
daser333 [38]
The wavelength emitted is indirectly proportional to the difference in the change in the energy level. For the wavelength 278 nm the change in energy level is significantly high. Further change in energy level is indicated by 454nm light but the difference in energy level for this wavelength to be emitted  is not greater than the previous one. There is a possibility that these subsystems have now very low energy which should result in wavelengths ranging from 700 to 900 nm. There is another possibility that there is some metastable subsystems in the system which may cause LASER emission.
3 0
2 years ago
Assume that when you stretch your torso vertically as much as you can, your center of mass is 1.0 m above the floor. The maximum
Elenna [48]

1) 0.77 m

2) 0.23 m

Explanation:

1)

Here we want to find the time elapsed for crouching in order to jump and reach a height of 2.0 m above the floor, starting from 1.0 m above the floor.

First of all, we start by calculating the speed required to jump up to a height of 2.0 m. Since the total energy is conserved, the initial kinetic energy is converted into gravitational potential energy, so:

\frac{1}{2}mv^2 = mgh

where

m is the mass of the man

v is the speed after jumping

g=9.8 m/s^2 is the acceleration due to gravity

h = 2.0 - 1.0 = 1.0 m is the change in height

Solving for v,

v=\sqrt{2gh}=\sqrt{2(9.8)(1.0)}=4.43 m/s

In the acceleration phase, we know that the initial velocity is

u=0

And the force exerted on the floor is 2.3 times the gravitational force, so

F=2.3 mg

This means the net force on you is

F_{net} = F-mg=2.3mg-mg=1.3 mg

because we have to consider the force of gravity acting downward.

So the acceleration of the man is

a=\frac{F_{net}}{m}=\frac{1.3mg}{m}=1.3g

Now we can use the  following suvat equation to find the displacement in the acceleration phase, which is how low the man has to crouch in order to jump:

v^2-u^2=2as

where s is the quantity we want to find. Solving for s,

s=\frac{v^2-u^2}{2a}=\frac{4.43^2-0}{2(1.3g)}=0.77 m

2)

At the beginning, we are told that the height of the center of mass above the floor is

h = 1.0 m

During the acceleration phase and the crouch, the height of the center of mass of the body decreases by

\Delta h = -0.77 m

This means that the lowest point reached by the center of mass above the floor during the crouch is

h'=h+\Delta h = 1.0 - 0.77 = 0.23 m

This value seems unpractical, since it is not really easy to crouch until having the center of mass 0.23 m above the ground.

3 0
2 years ago
Can a small child play with fat child on the seesaw?Explain how?
RoseWind [281]
Yes a small child can play with fat child in the seesaw because if the the load distance is decreased the effort will increase. That's means if the distance between the fat boy and the fulcrum is decreased the small child needs less effort.so,he can play
8 0
2 years ago
Read 2 more answers
A rigid, uniform bar with mass mmm and length bbb rotates about the axis passing through the midpoint of the bar perpendicular t
Pie

Answer:

I = \frac{mvb}{6}

Explanation:

we know angular velocity in terms of moment of inertia and angular speed

       L = Iω ....                        (1)

moment of inertia of rod rotating about its center of length b

 

      I = \frac{ mb^2}{12}  ........               .(2)  

using         v = ωr  

where w is angular velocity

and r is radius of  rod which is equal to b

        so we get  2v =  ωb  

                            ω  = 2v/b  .................            (3)    

here velocity is two time because two opposite ends  are moving opposite with a velocity v so net velocity will be 2v

put second and third equation in ist equation

                 L   =   \frac{mb^2}{12}×\frac{2v}{b}

              so final answer will be      L  =   \frac{mvb}{6}

7 0
2 years ago
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