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3 years ago

Please help and quick!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

Physics

1 answer:

andrew-mc [135]3 years ago

3 0

Answer:

yes i know

Explanation:

Measuring Tsunamis. ... It consists of an anchored, ocean-bottom pressure recorder and a companion buoy (called DART, for Deep-Ocean Assessment and Reporting of Tsunamis). The recorder, sitting at a depth of up to 5,000 meters, measures changes in pressure due to changes in water level.

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An infinite sheet of charge, oriented perpendicular to the x-axis, passes through x = 0. It has a surface charge density σ1 = -2

pishuonlain [190]

Answer:

$E_{total}=4.82*10^6N/C$

vector with direction equal to the axis X.

Explanation:

We use the Gauss Law and the superposition law in order to solve this problem.

Superposition Law: the Total Electric field is the sum of the electric field of the first infinite sheet and the Electric field of the second infinite sheet:

$E_{total}=E_1+E_2$

Thanks Gauss Law we know that the electric field of a infinite sheet with density of charge σ is:

$E=\sigma/(2\epsilon_o)$

Then:

$E_{total}=(\sigma_1+\sigma_2)/(2\epsilon_o)=(-2.7*10^{-6}+88*10^{-6})/(2*8.85*10^{-12})=4.82*10^6N/C$

This electric field has a direction in the axis perpendicular to the sheets, that means it has the same direction as the axis X.

7 0

3 years ago

Read 2 more answers

A photovoltaic panel of dimension 2 m × 4 m is installed on the roof of a home. The panel is irradiated with a solar flux of GS

Flura [38]

Answer:

(a) the electrical power generated for still summer day is 1013.032 W

(b)the electrical power generated for a breezy winter day is 1270.763 W

Explanation:

Given;

Area of panel = 2 m × 4 m, = 8m²

solar flux GS = 700 W/m²

absorptivity of the panel, αS = 0.83

efficiency of conversion, η = P/αSGSA = 0.553 − 0.001 K⁻¹ Tp

panel emissivity , ε = 0.90

Apply energy balance equation to determine he electrical power generated;

transferred energy + generated energy = 0

(radiation + convection) + generated energy = 0

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - \eta \alpha_s G_s = 0$

$[\alpha_sG_s-\epsilon \alpha(T_p^4-T_s^4)]-h(T_p-T_\infty) - (0.553-0.001T_p)\alpha_s G_s$

(a) the electrical power generated for still summer day

$T_s = T_{\infty} = 35 ^oC = 308 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_1^4-308^4)]-10(T_p_1-308) - (0.553-0.001T_p_1)0.83*700 = 0\\\\3798.94-5.103*10^{-8}T_p_1^4 - 9.419T_p_1 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_1\\\\T_p_1 = 335.05 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_1)\alpha_s G_s A \\\\P = (0.553-0.001 *335.05)0.83*700*8 \\\\P = 1013.032 \ W$

(b)the electrical power generated for a breezy winter day

$T_s = T_{\infty} = -15 ^oC = 258 \ k$

$[0.83*700-0.9*5.67*10^{-8}(T_p_2^4-258^4)]-10(T_p_2-258) - (0.553-0.001T_p_2)0.83*700 = 0\\\\8225.81-5.103*10^{-8}T_p_2^4 - 29.419T_p_2 = 0\\\\Apply \ \ iteration \ method \ to \ solve \ for \ T_p_2\\\\T_p_2 = 279.6 \ k$

$P = \eta \alpha_s G_s A = (0.553-0.001 T_p_2)\alpha_s G_s A \\\\P = (0.553-0.001 *279.6)0.83*700*8 \\\\P = 1270.763 \ W$

3 0

3 years ago

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, it i

lianna [129]

Answer:

A) the moment of inertia of the system decreases and the angular speed increases.

Explanation:

The complete question is

A merry-go-round spins freely when Diego moves quickly to the center along a radius of the merry-go-round. As he does this, It is true to say that

A) the moment of inertia of the system decreases and the angular speed increases.

B) the moment of inertia of the system decreases and the angular speed decreases.

C) the moment of inertia of the system decreases and the angular speed remains the same.

D) the moment of inertia of the system increases and the angular speed increases.

E) the moment of inertia of the system increases and the angular speed decreases

In angular momentum conservation, the initial angular momentum of the system is conserved, and is equal to the final angular momentum of the system. The equation of this angular momentum conservation is given as

$I_{1} w_{1} = I_{2} w_{2}$ ....1

where $I_{1}$ and $I_{2}$ are the initial and final moment of inertia respectively.

and $w_{1}$ and $w_{2}$ are the initial and final angular speed respectively.

Also, we know that the moment of inertia of a rotating body is given as

$I = mr^{2}$ ....2

where $m$ is the mass of the rotating body,

and $r$ is the radius of the rotating body from its center.

We can see from equation 2 that decreasing the radius of rotation of the body will decrease the moment of inertia of the body.

From equation 1, we see that in order for the angular momentum to be conserved, the decrease from $I_{1}$ to $I_{2}$ will cause the angular speed of the system to increase from $w_{1}$ to $w_{2}$ .

From this we can clearly see that reducing the radius of rotation will decrease the moment of inertia, and increase the angular speed.

7 0

4 years ago

What causes the earths plates to move

Norma-Jean [14]

The intense heat in the earth's core that causes molten rock in the mantle layer to move.

8 0

3 years ago

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What charge do electrons have? Negative Positive Neutral

hoa [83]

Answer electrons are negative. Protons are positive, and neutrons are neutral

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3 years ago