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3 years ago

Define the word logrolling in your own words?

Physics

2 answers:

Elan Coil [88]3 years ago

3 0

A crazy sport thats kinda dangerous

Doss [256]3 years ago

3 0

Answer:

Logrolling is the practice of legislators exchanging favors, or quid pro quo, such as vote trading, in order to enact legislation that is of interest to each of them.

<u>OAmalOHopeO</u>

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An instructor’s laser pointer produces a beam of light with a circular cross section of diameter 0.900 mm and a total power outp

il63 [147K]

Answer:

$E_0 = 2180.53N/C.$

$B_0 = 7.27*10^{-6}T.$

$U_{avg} = 4.2*10^{-5}J/m^3.$

Explanation:

The Intensity $I$ of the beam is

$I = P /A$

The diameter of the beam is 0.900mm; therefore, the area is

$A = \pi( \dfrac{0.900*10^{-3}}{2} )^2$

$A = 6.36*10^{-7}m^2$

and since $P = 4.00*10^{-3}W$ , the intensity of the beam is

$I = \dfrac{4.00*10^{-3}W}{6.36*10^{-7}m^2}$

$\boxed{I = 6289.3W/m^2.}$

Now, the intensity $I$ is related to $E_0$ by the relation

$I = \dfrac{E_0^2}{2\mu_0 c}$

solving for $E_0$ we get

$E_0 = \sqrt{2\mu_0 c I}$

putting in the numbers we get:

$E_0 = \sqrt{2(1.26*10^{-6}) (3*10^8) (6289.3)}$

$\boxed{E_0 = 2180.53N/C.}$

The amplitude of magnetic field $B_0$ is related to $E_0$ by

$B_0 = \dfrac{E_0}{c}$

putting in numerical values we get:

$B_0 = \dfrac{2180.53}{3*10^8}$

$\boxed{B_0 = 7.27*10^{-6}T. }$

The average energy density of the laser light is

$U_{avg} = \epsilon_0 E_0^2$

$U_{avg} = (8.85*10^{-12}) (2180.53)^2$

$\boxed{U_{avg} = 4.2*10^{-5}J/m^3.}$

8 0

4 years ago

Which property must be the same in order to stop the transfer of energy between two objects?

juin [17]

Its is temperture
hope this helps :)

7 0

3 years ago

Homer Agin leads the Varsity team in home runs. In a recent game, Homer hit a 34.5 m/s sinking curve ball head on, sending it of

Aneli [31]

Answer:

–77867 m/s/s.

Explanation:

From the question given above, the following data were obtained:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

Acceleration is simply defined as the rate of change of velocity with time. Mathematically, it is expressed as:

Acceleration = (final velocity – Initial velocity) /time

a = (v – u) / t

With the above formula, we can obtain acceleration of the ball as follow:

Initial velocity (u) = 34.5 m/s

Final velocity (v) = –23.9 m/s

Time (t) = 0.00075 s

Acceleration (a) =?

a = (v – u) / t

a = (–23.9 – 34.5) / 0.00075

a = –58.4 / 0.00075

a = –77867 m/s/s

Thus, the acceleration of the ball is –77867 m/s/s.

3 0

3 years ago

A block is at rest on the incline shown in the figure. The coefficients of static and ki- netic friction are μs = 0.62 and μk =

GREYUIT [131]

The frictional force is 218.6 N

Explanation:

The block in the problem is at rest along the inclined surface: this means that the net force acting along the direction parallel to the incline must be zero.

There are two forces acting along this direction:

- The component of the weight parallel to the incline, downward along the plane, of magnitude

$mg sin \theta$