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-Dominant- [34]

2 years ago

6

Explain in detail what causes tides and give one reason why it is important for humans to monitor tides.

2 answers:

Softa [21]2 years ago

7 0

Wishy washy is correct she helps :)

Klio2033 [76]2 years ago

3 0

The gravitational attraction to the moon causes the Earth's oceans to bulge out in the direction of the moon. Another bulge occurs on the opposite side, since the Earth itself is pulled towards the moon (and away from the water on the far side). Since the Earth rotates, this happens twice a day. These are tides.

As for the importance of monitoring tides, one reason is that commercial and recreational fishermen use their knowledge of the tides and tidal currents to help them improve their catches. Monitoring the tides is a very important part of their livelihood.

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Explain whether a particle moving in a straight line with constant speed does or does not have an acceleration. b) Explain wheth

Lelechka [254]

Answer:

A: No because it is nor changing speed or direction

B: Yes because it changes direction even though the speed is constant

Please Give Brainliest

5 0

2 years ago

Two insulated copper wires of similar overall diameter have very different interiors. One wire possesses a solid core of copper,

Marrrta [24]

Answer:

a

Solid Wire $I = 0.01237 \ A$

Stranded Wire $I_2 = 0.00978 \ A$

b

Solid Wire $R = 0.0149 \ \Omega$

Stranded Wire $R_1 = 0.0189 \ \Omega$

Explanation:

Considering the first question

From the question we are told that

The radius of the first wire is $r_1 = 1.53 mm = 0.0015 \ m$

The radius of each strand is $r_0 = 0.306 \ mm = 0.000306 \ m$

The current density in both wires is $J = 1750 \ A/m^2$

Considering the first wire

The cross-sectional area of the first wire is

$A = \pi r^2$

= > $A = 3.142 * (0.0015)^2$

= > $A = 7.0695 *10^{-6} \ m^2$

Generally the current in the first wire is

$I = J*A$

=> $I = 1750*7.0695 *10^{-6}$

=> $I = 0.01237 \ A$

Considering the second wire wire

The cross-sectional area of the second wire is

$A_1 = 19 * \pi r^2$

=> $A_1 = 19 *3.142 * (0.000306)^2$

=> $A_1 = 5.5899 *10^{-6} \ m^2$

Generally the current is

$I_2 = J * A_1$

=> $I_2 = 1750 * 5.5899 *10^{-6}$

=> $I_2 = 0.00978 \ A$

Considering question two

From the question we are told that

Resistivity is $\rho = 1.69* 10^{-8} \Omega \cdot m$

The length of each wire is $l = 6.25 \ m$

Generally the resistance of the first wire is mathematically represented as

$R = \frac{\rho * l }{A}$

=> $R = \frac{ 1.69* 10^{-8} * 6.25 }{ 7.0695 *10^{-6} }$

=> $R = 0.0149 \ \Omega$

Generally the resistance of the first wire is mathematically represented as

$R_1 = \frac{\rho * l }{A_1}$

=> $R_1 = \frac{ 1.69* 10^{-8} * 6.25 }{5.5899 *10^{-6} }$

=> $R_1 = 0.0189 \ \Omega$

3 0

2 years ago

At 900.0 K, the equilibrium constant (Kp) for the following reaction is 0.345. 2SO2(g)+O2(g)→2SO3(g) At equilibrium, the partial

elena55 [62]

Answer : The partial pressure of $SO_3$ is, 67.009 atm

Solution : Given,

Partial pressure of $SO_2$ at equilibrium = 30.6 atm

Partial pressure of $O_2$ at equilibrium = 13.9 atm

Equilibrium constant = $K_p=0.345$

The given balanced equilibrium reaction is,

$2SO_2(g)+O_2(g)\rightleftharpoons 2SO_3(g)$

The expression of $K_p$ will be,

$K_p=\frac{(p_{SO_3})^2}{(p_{SO_2})^2\times (p_{O_2})}$

Now put all the values of partial pressure, we get

$0.345=\frac{(p_{SO_3})^2}{(30.6)^2\times (13.9)}$

$p_{SO_3}=67.009atm$

Therefore, the partial pressure of $SO_3$ is, 67.009 atm

6 0

2 years ago

A car travels 5 km due east, then 10 km due north, and then 5

Rufina [12.5K]

Answer:

idk what the answer plz help!!!

Explanation:

8 0

3 years ago

A beaker contain 200mL of water<br> What is its volume in cm3 and m3

Bogdan [553]

The volumes are 200cm3 and 0.0002m3

7 0

3 years ago

Read 2 more answers