How do like charges affect one another

A large pendulum with a 200-lb gold-plated bob 12 inches in diameter is on display in the lobby of the United Nations building.

Tpy6a [65]

Answer:

2.4s

Explanation:

The length of the pendulum = 75ft

Diameter d = 12 inches

The time period of the pendulum is given as

T = 2pi(L/g)^1/2

Then the time it takes to move from displacement to equilibrium is given as:

t = T/4

= (Pi/2)*(L/g)^1/2

= pi/2 x [(75x0.3048)/9.81]^0.5

= 1.57x[22.86/9.81)^0.5

= 2.4s

2.4 seconds is the least amount of time that it would take.

5 0

3 years ago

List three cases in which potential energy becomes

olga55 [171]

1 Electrical Potential Energy, separating two charged plates will store energy as the plates want to return to their original position.

2 Spring or Elastic can be stretched to store energy as it wants to return to rest 

3 Gravitational energy is stored by moving something (ball or pendulum are both examples of this) against a gravity gradient (lifting an object) that wants to fall back down.

5 0

4 years ago

Help me please i need to get my grade up!

Rashid [163]

I think it's 500 m/s^2

6 0

3 years ago

In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between

kap26 [50]

The separation between the slits is d = 8.96

What is fringe width?

Fringe width is the distance between two consecutive bright spots (maximas, where constructive interference take place)
Or two consecutive dark spots (minimas, where destructive interference take place).

Fringe width is given by β = λL/d

In the first case fringe width is β1 = λLA /d = 625 x 10-9 x 0.36 / ( 1.4 x 10-5 ) = 0.016071428 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.016071428 = 6.222

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 6.

In the second case fringe width is β1 = λLAB /d = 625 x 10-9 x 0.25 / ( 1.4 x 10-5 ) = 0.011160714 m

The total width of the screen is 0.2 m . So, on one side of the central maximum, the width is 0.1 m

No. of fringes in this 0.1m = 0.1 / 0.011160714 = 8.96

So, since there is a bright fringe after every fringe width, the number of bright fringes on one side of central maximum is 8. The ninth one will not be seen since the screen is less a little less in width.

Learn more about fringe width

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The complete question is -

In a setup like that in Figure 27.7, a wavelength of 625 nm is used in a Young's double-slit experiment. The separation between the slits is d = 1.4 × 10-3 m. The total width of the screen is 0.20 m. In one version of the setup, the separation between the double slit and the screen is LA = 0.36 m, whereas in another version it is LB = 0.25 m. On one side of the central bright fringe, how many bright fringes lie on the screen in the two versions of the setup? Do not include the central bright fringe in your counting. --Tm = 3 (Bright fringe) ++m = 0 (Bright fringe) -m = 3 (Bright fringe) Figure 27.7

5 0

1 year ago

The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic ene

Nataliya [291]

Question in proper order

The rotational kinetic energy term is often called the kinetic energy in the center of mass, while the translational kinetic energy term is called the kinetic energy of the center of mass.

You found that the total kinetic energy is the sum of the kinetic energy in the center of mass plus the kinetic energy of the center of mass. A similar decomposition exists for angular and linear momentum. There are also related decompositions that work for systems of masses, not just rigid bodies like a dumbbell.

It is important to understand the applicability of the formula

$Ktot=Kr+Kt$

Which of the following conditions are necessary for the formula to be valid?

a. The velocity vector $v$ must be perpendicular to the axis of rotation

b.The velocity vector $v$ must be perpendicular or parallel to the axis of rotation

c. The moment of inertial must be taken about an axis through the center of mass

Answer:

Option c

Explanation:

$K_{total} = K_{rotational}+K_{translational}$

The first two conditions are untrue, this is because, you can have rotation in any direction and translation in any direction of any collection of masses. Rotational and translational velocities of masses do not depend on each other

The last statement is true because by definition, the moment of inertia, which is a measure of reluctance, is usually taken about a reference point which is the center of mass

4 0

4 years ago

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