In addition to hot surface temperatures, which characteristic makes mercy different from earth

Both positive and negative charges will spread the leaves of an uncharged electroscope.

hoa [83]

The answer is A)TRUE.

7 0

3 years ago

The earth has a net electric charge that causes a field at points near its surface equal to 150 N/C and directed in toward the c

maria [59]

To solve this problem we will start using the concepts related to the electric field, from there we will find the load exerted on the body. Through this load it will be possible to make a sum of forces in balance to find the load that a human supports. Finally with these values it will be possible to find the repulsive force. We will proceed as follows,

The electric field is

$E= \frac{kQ}{R^2}$

Here,

k = Coulomb's Constant

Q = Charge

R = Distance (At this case from the center of mass of the earth to the surface)

Rearranging to find the charge,

$Q = \frac{ER^2}{k}$

Replacing,

$Q = frac{(150)(6.38*10^6)}{8.99*10^9}$

$Q = 6.79*10^5 C$

Since the electric field is directed towards the center of earth, the charge is negative.

PART A) Once the load is found we can proceed to apply the balance of Forces, for which the electrostatic force must be equivalent to the weight, this in order to satisfy the balance, therefore

$F_w = F_e$

$mg = \frac{kQq}{R^2}$

Replacing,

$(62)(9.8) = \frac{(8.99*10^9)(q)(-6.79*10^5)}{(6.38*10^6)^2}$

Solving for q,

$q = -4.056C$

PART B) Finally using the given distance and the values of the found load we can find the repulsive Force, which is

$F =\frac{kq^2}{d^2}$

$F = \frac{(8.99*10^9)(-4.056)^2}{110^2}$

$F = 1.22*10^7N$

PART C) The answer is no. According to the information found, we can conclude that traveling through an electric field is not viable because there is a repulsive force of great magnitude acting on the body.

3 0

3 years ago

Two trucks have the same velocity but different masses. Truck 1 has a mass of 200kg and truck 2 has a mass of 500kg . Which truc

Readme [11.4K]

Your answer would be Truck 2.

8 0

3 years ago

A section of a parallel-plate air waveguide with a plate separation of 7.11 mm is constructed to be used at 15 GHz as an evanesc

adell [148]

Answer:

the required minimum length of the attenuator is 3.71 cm

Explanation:

Given the data in the question;

we know that;

$f_{c_1$ = c / 2a

where f is frequency, c is the speed of light in air and a is the plate separation distance.

we know that speed of light c = 3 × 10⁸ m/s = 3 × 10¹⁰ cm/s

plate separation distance a = 7.11 mm = 0.0711 cm

so we substitute

$f_{c_1$ = 3 × 10¹⁰ / 2( 0.0711 )

$f_{c_1$ = 3 × 10¹⁰ cm/s / 0.1422 cm

$f_{c_1$ = 21.1 GHz which is larger than 15 GHz { TEM mode is only propagated along the wavelength }

Now, we determine the minimum wavelength required

Each non propagating mode is attenuated by at least 100 dB at 15 GHz

so

Attenuation constant TE₁ and TM₁ expression is;

∝₁ = 2πf/c × √( ( $f_{c_1$ / f)² - 1 )

so we substitute

∝₁ = ((2π × 15)/3 × 10⁸ m/s) × √( (21.1 / 15)² - 1 )

∝₁ = 3.1079 × 10⁻⁷

∝₁ = 310.79 np/m

Now, To find the minimum wavelength, lets consider the design constraint;

20log₁₀ $e^{-\alpha _1l_{min$ = -100dB

we substitute

20log₁₀ $e^{-(310.7np/m)l_{min$ = -100dB

$l_{min$ = 3.71 cm

Therefore, the required minimum length of the attenuator is 3.71 cm

6 0

3 years ago

Interference is evidence of the particle nature of light.<br><br> True<br> False

Virty [35]

It's true answer that interference is the evidence of the particle nature of light..

3 0

3 years ago

Read 2 more answers