Ask question

Ask question

All categories

emmainna [20.7K]

3 years ago

7

Extreme tides are... a. extremely rare b. a normal occurrence c. caused only by hurricanes d. not really tides I think the answe

r may be b. Please check

1 answer:

Serjik [45]3 years ago

7 0

<span>Extreme tides happen twice a month. They are caused by the earth, Sun, and Moon all being in a straight alignment. Although they are not extremely rare, extreme tides are not normal occurrences. The answer is D.</span>

You might be interested in

Two point charges of -7uC and 4uC are a distance of 20

aivan3 [116]

Answer:

Approximately 0.979 J.

Explanation:

Assume that the two charges are in vacuum. Apply the coulomb's law to find their initial and final electrical potential energy $\mathrm{EPE}$ .

$\displaystyle \mathrm{EPE} = \frac{k \cdot q_1 \cdot q_2}{r}$ ,

where

The coulomb's constant $k = 8.99\times 10^{9}\; \rm N\cdot m^{2} \cdot C^{-2}$ ,
$q_1$ and $q_2$ are the sizes of the two charges, and
$r$ is the separation of (the center of) the two charges.

Note that there's no negative sign before the fraction.

Make sure that all values are in SI units:

$q_1 = -7\rm \;\mu C = -7\times 10^{-6}\; C$ ;
$q_2 = 4\rm \;\mu C = 4\times 10^{-6}\; C$ ;

Initial separation: $\rm 20\; cm = 0.20\; cm$ ;
Final separation: $\rm 90\; cm = 0.90\; cm$ .

Apply Coulomb's law:

Initial potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.20}\\&= \rm -1.2586\; J\end{aligned}$ .

Final potential energy:

$\begin{aligned} \frac{k \cdot q_1 \cdot q_2}{r} &= \frac{8.99\times 10^{9}\times (-7\times 10^{-6})\times 4\times 10^{-6}}{0.90}\\&= \rm -0.279689\; J\end{aligned}$ .

The final potential energy is less negative than the initial one. In other words, the two particles gain energy in this process. The energy difference (final minus initial) will be equal to the work required to move them at a constant speed.

$\begin{aligned}\text{Work required} &= \text{Final EPE} - \text{Initial EPE}\\&= \rm -0.279689\; J - (-1.2586\; J)\\&\approx 0.979\; J\end{aligned}$ .

8 0

3 years ago

Read 2 more answers

An air bubble has a volume of 1.70 cm³ when it is released by a submarine 115 m below the surface of a lake. What is the volume

Natalija [7]

Answer:

V = 20.67 cm³

Explanation:

In this case, let's apply the Boyle's law which is:

P1V1 = P2V2

Where P1 and V1 would be condition in the water, and P2 and V2 would be the condition at the surface.

By logic, at the surface, pressure should be equals to 1 atm or 1.01x10^5 N/m²

We know the volume of the bubble at first which is 1.70 cm³ and we need to calculate V2. We know how much is P2, but we don't know the value of P1, which is the pressure of the bubble below the sea; this can be calculated using Pascal's principle which is the following expression:

P1 = Po + dgh

Where:

Po: innitial pressure, which we can assume is 1 atm

d = density of the substance, in this case, water (1000 kg/m³)

g = gravity (9.8 m/s²)

h = distance of the bubble from the surface (115 m)

Now replacing this data in the boyle's law we have the following:

P1V1 = P2V2

V2 = P1V1/P2

V2 = (Po + dgh) * V1 / P2

Replacing the data we have:

V2 = (1.01x10^5 + 1000*9.8*115) * 1.7 / 1.01x10^5

V2 = 2,087,600 / 1.01x10^5

V2 = 20.67 cm³

5 0

3 years ago

Determine the minimum applied force p required to move wedge a to the right. the spring is compressed a distance of 175 mm. negl

BARSIC [14]

<span>b. The coefficient of static friction for all contacting surfaces is Î¼s=0.35. neglect friction at the rollers.

</span>

6 0

3 years ago

Aspirin is a compound composed of carbon, hydrogen, and oxygen atoms

vekshin1

Answer:

A homogeneous Mixture

Explanation:

The acid that contains the acetylsalicylic acid is a <u>mixture,</u> but it isnt a compound. though aspirin is. (hopefully this helps? qwq)

8 0

2 years ago

A car and a train move together along straight, parallel paths with the same constant cruising speed v(initial). At t=0 the car

kogti [31]

Answer:

$t_1 = \frac{v_i}{a_i}$

$t_2 = \frac{v_i}{a_i}$

Δd = $v_it_1 = v_i^2/a_i$

Explanation:

As $v(t) = v_i + at$ , when the car is making full stop, $v(t_1) = 0$ . $a = -a_i$ . Therefore,

$0 = v_i - a_it_1\\v_i = a_it_1\\t_1 = \frac{v_i}{a_i}$

Apply the same formula above, with $v(t_2) = v_i$ and $a = a_i$ , and the car is starting from 0 speed, we have

$v_i = 0 + a_it_2\\t_2 = \frac{v_i}{a_i}$

As $s(t) = vt + \frac{at^2}{2}$ . After $t = t_1 + t_2$ , the car would have traveled a distance of

$s(t) = s(t_1) + s(t_2)\\s(t_1) = (v_it_1 - \frac{a_it_1^2}{2})\\ s(t_2) = \frac{a_it_2^2}{2}$

Hence $s(t) = (v_it_1 - \frac{a_it_1^2}{2}) + \frac{a_it_2^2}{2}$

As $t_1 = t_2$ we can simplify $s(t) = v_it_1$

After t time, the train would have traveled a distance of $s(t) = v_i(t_1 + t_2) = 2v_it_1$

Therefore, Δd would be $2v_it_1 - v_it_1 = v_it_1 = v_i^2/a_i$

8 0

3 years ago