Why does jupiter have several distinct cloud layers?

How was the water level by rock affected by wave?<br> -Tsunami-

krok68 [10]

The answer is:

The water level by rock affected by wave as:

if the direction of the wave is going to the rock naturally by the effect of the tide which comes in, the water level will rise and increase if the wave is escaping or receding the water level will decrease and will be lower.

when the tsunami is a tidal wave which is a series of waves in a water body caused by the displacement of a large volume of water.

7 0

3 years ago

At what time was the person at a position of 0m?

Anni [7]

Answer: The person was not at a position of "0" at any time. The person started at 10 metres from the starting line. The explanation below shows how to use the standard formula for position when the initial position is not "0". It is noteworthy that the standard expression of the formula for distance travelled does not include a variable (e.g. "d") for distance at the start (when t(time) = 0)

Explanation: At time = 0, the start, the person was at 10m distance from the starting line. Therefore, to use the standard equation, "s + ut + 1/2att (t squared, that is), distance from starting line = 10 + s, that is, total distance from starting line equals initial position, 10 metres, plus "s" (distance travelled from t = 0 to t = 1) in metres.

for the section of the graph from "0" seconds (t = 0) to 1 second (t = 1):

s = ut + 1/2att

the initial position is 10 metres.

s = 10

the distance is constant from t = 0 to t = 1, therefore the velocity for the whole of that section of graph must be 0.

u = 0

there is no change in the velocity from t = 0 to t= 1, therefore the acceleration for the first section of the graph must be 0.

a = 0

s = ut + 1/2att

= (0 x 1) + 1/2 (0 x 1 x 1)

= (0) + 1/2 (0)

= 0

total distance from starting line (position) equals initial position plus change in position (distance travelled).

at t = 1,

position = 10 + 0

= 10 metres

The whole of the graph can be analysed using this process for each straight section of the graph separately, adding "s" for each section to the previous total of distance from starting line.

using "d" for initial distance from starting line ( position ), d1 for distance from starting line at t = 1, d2 for distance from starting line at t = 2, etcetera:

section 1, t = 0 to t = 1:

d1 (t=0 to t=1) = 10 + s (t=0 to t=1).

section 2, t= 1 to t = 2:

d2 (t=0 to t=2) = 10 + s (t=0 to t=1) + s (t=1 to t=2).

etcetera.

8 0

2 years ago

The starships of the Solar Federation are marked with the symbol of the Federation, a circle, whereas starships of the Denebian

Llana [10]

Complete question

The complete question is shown on the first uploaded image

Answer:

The velocity is $v = c* \sqrt{1 - \frac{1}{n^2} }$

Explanation:

From the question we are told that

a = nb

The length of the minor axis of the symbol of the Federation, a circle, seen by the observer at velocity v must be equal to the minor axis(b) of the Empire's symbol, (an ellipse)

Now this length seen by the observer can be mathematically represented as

$h = t \sqrt{1 - \frac{v^2}{c^2} }$

Here t is the actual length of the major axis of of the Empire's symbol, (an ellipse)

So t = a = nb

and b is the length of the minor axis of the symbol of the Federation, (a circle) when seen by an observer at velocity v which from the question must be the length of the minor axis of the of the Empire's symbol, (an ellipse)

i.e h = b

So

$b = nb [\sqrt{1 - \frac{v^2}{c^2} } ]$

$[\frac{1}{n} ]^2 = 1 - \frac{v^2}{c^2}$

$v^2 =c^2 [1- \frac{1}{n^2} ]$

$v^2 =c^2 [\frac{n^2 -1}{n^2} ]$

$v = c* \sqrt{1 - \frac{1}{n^2} }$

6 0

3 years ago

You're carrying a 3.0-m-long, 24 kg pole to a construction site when you decide to stop for a rest. You place one end of the pol

kozerog [31]

Answer:

$F=133N$

Explanation:

From the question we are told that:

Length $l=3.0m$

Mass $m=24kg$

Distance from Tip $d=35cm$

Generally, the equation for Torque Balance is mathematically given by

$mg(l/2)=F(l-d)$

$2*9.81(3/2)=F(3-35*10^-2)$

Therefore

$F=133N$

8 0

2 years ago

When melting of a metamorphic rock occurs, it changes into what?

Thepotemich [5.8K]

Answer:

The upper limit of metamorphism occurs at the pressure and temperature of wet partial melting of the rock in question. Once melting begins, the process changes to an igneous process rather than a metamorphic process. During metamorphism the protolith undergoes changes in texture of the rock and the mineral make up of the rock.

Explanation:

i hope this helps

8 0

2 years ago