What happens to the Total Energy as the spring bounce?

An AM radio station broadcasts isotropically (equally in all directions) with an average power of 3.40 kW. A receiving antenna 6

lara [203]

To solve the problem we will apply the concepts related to the Intensity as a function of the power and the area, as well as the electric field as a function of the current, the speed of light and the permeability in free space, as shown below.

The intensity of the wave at the receiver is

$I = \frac{P_{avg}}{A}$

$I = \frac{P_{avg}}{4\pi r^2}$

$I = \frac{3.4*10^3}{4\pi(4*1609.34)^2} \rightarrow 1mile = 1609.3m$

$I = 6.529*10^{-6}W/m^2$

The amplitude of electric field at the receiver is

$I = \frac{E_{max}^2}{2\mu_0 c}$

$E_{max}= \sqrt{2I\mu_0 c}$

The amplitude of induced emf by this signal between the ends of the receiving antenna is

$\epsilon_{max} = E_{max} d$

$\epsilon_{max} = \sqrt{2I \mu_0 cd}$

Here,

I = Current

$\mu_0$ = Permeability at free space

c = Light speed

d = Distance

Replacing,

$\epsilon_{max} = \sqrt{2(6.529*10^{-6})(4\pi*10^{-7})(3*10^{8})(60.0*10^{-2})}$

$\epsilon_{max} = 0.05434V$

Thus, the amplitude of induced emf by this signal between the ends of the receiving antenna is 0.0543V

6 0

3 years ago

In terms of newton first law of motion why is it important to wear a seat belt

lilavasa [31]

Answer:

Seatbelts stop you

Explanation:

Any passengers in the car will also be decelerated to rest if they are strapped to the car by seat belts.

6 0

3 years ago

People in the future may well live inside a rotating space structure that is more than 2 km in diameter. Inside the structure, p

Sergeeva-Olga [200]

Answer:

option B

Explanation:

given,

diameter of the rotating space = 2 Km

Force exerted at the edge of the space = 1 g

force experienced at the half way = ?

As the object is rotating in the circular part

Force is equal to centripetal acceleration.

at the edge

g = ω² r

ω is the angular velocity of the particle

r is the radius.

now, acceleration at the half way

g' = ω² r'

$g' = \omega^2 (\dfrac{r}{2})$

$g' =\dfrac{1}{2}(\omega^2 r)$

$g'=\dfrac{g}{2}$

People at the halfway experience g/2

hence, the correct answer is option B

7 0

3 years ago

Which letter represents the position of maximum potential energy of the pendulum

worty [1.4K]

If you mean the SI Unit of GPE, the answer is J for Joules.
if that's not the question being asked, i would need a little more elaboration please :)

4 0

3 years ago

A man whose mass is 69 kg and a woman whose mass is 52 kg sit at opposite ends of a canoe 5 m long, whose mass is 20 kg. Suppose

dusya [7]

Answer:

the canoe moved 1.2234 m in the water

Explanation:

Given that;

A man whose mass = 69 kg

A woman whose mass = 52 kg

at opposite ends of a canoe 5 m long, whose mass is 20 kg

now let;

x1 = position of the man

x2 = position of canoe

x3 = position of the woman

Now,

Centre of mass = [m1x1 + m2x2 + m3x3] / m1 + m2 + m3

= ( 69×0 ) + ( 52×5) + ( 20× 5/2) / 69 + 52 + 20

= (0 + 260 + 50 ) / ( 141 )

= 310 / 141

= 2.19858 m

Centre of mass is 2.19858 m

Now, New center of mass will be;

52 × 2.5 / ( 69 + 52 + 20 )

= 130 / 141

= 0.9219858 m { away from the man }

To get how far, the canoe moved;

⇒ 2.5 + 0.9219858 - 2.19858

= 1.2234 m

Therefore, the canoe moved 1.2234 m in the water

5 0

2 years ago