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3 years ago

How are mountains and plateaus alike?

Physics

2 answers:

mr Goodwill [35]3 years ago

7 0

I think its " both have level surfaces"

jonny [76]3 years ago

3 0

Both have level surfaces lol hope this helps

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7.22 Ignoring reflection at the air–water boundary, if the amplitude of a 1 GHz incident wave in air is 20 V/m at the water surf

Serga [27]

Answer:

z = 0.8 (approx)

Explanation:

given,

Amplitude of 1 GHz incident wave in air = 20 V/m

Water has,

μr = 1

at 1 GHz, r = 80 and σ = 1 S/m.

depth of water when amplitude is down to 1 μV/m

Intrinsic impedance of air = 120 π Ω

Intrinsic impedance of water = $\dfrac{120\pi}{\epsilon_r}$

Using equation to solve the problem

$E(z) = E_0 e^{-\alpha\ z}$

E(z) is the amplitude under water at z depth

E_o is the amplitude of wave on the surface of water

z is the depth under water

$\alpha = \dfrac{\sigma}{2}\sqrt{\dfrac{(120\pi)^2}{\Epsilon_r}}$

$\alpha = \dfrac{1}{2}\sqrt{\dfrac{(120\pi)^2}{80}}$

$\alpha =21.07\ Np/m$

now ,

$1 \times 10^{-6} = 20 e^{-21.07\times z}$

$e^{21.07\times z}= 20\times 10^{6}$

taking ln both side

21.07 x z = 16.81

z = 0.797

z = 0.8 (approx)

5 0

3 years ago

Pls helppp. Is this right?

andreyandreev [35.5K]

Answer:

yes you are totally right

7 0

3 years ago

Read 2 more answers

Calculate 8 ∙ 10^-4 divided by 2 ∙ 10^2.

Digiron [165]

8 ∙ 10^-4 / 2 ∙ 10^2 = (8/2) ∙ ((10^-4)*(10^-2)) = <span>4 ∙ 10^-6</span>

7 0

3 years ago

A force F with arrow applied to an object of mass m1 produces an acceleration of 3.10 m/s2. The same force applied to a second o

Bas_tet [7]

Answer:

(a) The value of the ratio m₁/m₂ is 0.581

(b) the acceleration of the combined masses is 1.139 m/s²

Explanation:

Given;

The acceleration of force applied to M₁, a₁ = 3.10 m/s²

The same force applied to M₂ has acceleration, a₂ = 1.80 m/s²

Let this force = F

According Newton's second law of motion;

F = ma

(a) the value of the ratio m₁/m₂

since the applied force is same in both cases, M₁a₁ = M₂a₂

$\frac{m_1}{m_2} = \frac{a_2}{a_1} \\\\\frac{m_1}{m_2} = \frac{1.8}{3.1} \\\\\frac{m_1}{m_2} = 0.581$

(b) the acceleration of m₁ and m₂ combined as one object under the action force F

F = ma

$a = \frac{F}{M} \\\\a = \frac{F}{m_1 + m_2} \\\\a = \frac{F}{0.581m_2 + m_2}\\\\a = \frac{F}{1.581m_2}$

$But, m_2 = \frac{F}{a_2} \\\\a = \frac{F}{1.581m_2} = \frac{F*a_2}{1.581F} \\\\a = \frac{a_2}{1.581} \\\\a = \frac{1.8}{1.581} = 1.139 \ m/s^2$

Therefore, the acceleration of the combined masses is 1.139 m/s²

5 0

3 years ago

If an object has a mass of 14 kg, what is it’s weight

Alex_Xolod [135]

Answer:

Weight = 137.2 kg.m/ $s^{2}$

Explanation:

Given

mass of an object, m = 14 kg

Weight (W) is given by the formula:

W = mg , where m is mass of the object and g is acceleration due to gravity and the value is 9.8 m/ $s^{2}$

Hence W = 14 kg x 9.8 m/ $s^{2}$

= 137.2 kg.m/ $s^{2}$

3 0

3 years ago