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3 years ago

All of the planets stay near the ecliptic as they move on the Celestial Sphere

Physics

1 answer:

stellarik [79]3 years ago

3 0

Answer:

True

Explanation:

As the Earth goes around the Sun, it will appear that Earth is stationary and Sun is going around it. One can observe the same in real life as well. This apparent path followed by the Sun is called Ecliptic. The plane consisting Ecliptic is called as Ecliptic plane which is same as the orbital plane of Earth.

All the planets of the Solar system are also going around the Sun. Their orbital plane has negligible tilt with respect to Ecliptic plane. Due to this the planets will always appear near to the Ecliptic as they move on the celestial sphere.

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A railroad track and a road cross at right angles. An observer stands on the road and watches an eastbound train traveling at 60

mamaluj [8]

Answer:

After 4 s of passing through the intersection, the train travels with 57.6 m/s

Solution:

As per the question:

Suppose the distance to the south of the crossing watching the east bound train be x = 70 m

Also, the east bound travels as a function of time and can be given as:

y(t) = 60t

Now,

To calculate the speed, z(t) of the train as it passes through the intersection:

Since, the road cross at right angles, thus by Pythagoras theorem:

$z(t) = \sqrt{x^{2} + y(t)^{2}}$

$z(t) = \sqrt{70^{2} + 60t^{2}}$

Now, differentiate the above eqn w.r.t 't':

$\frac{dz(t)}{dt} = \frac{1}{2}.\frac{1}{sqrt{3600t^{2} + 4900}}\times 2t\times 3600$

$\frac{dz(t)}{dt} = \frac{1}{sqrt{3600t^{2} + 4900}}\times 3600t$

For t = 4 s:

$\frac{dz(4)}{dt} = \frac{1}{sqrt{3600\times 4^{2} + 4900}}\times 3600\times 4 = 57.6\ m/s$

4 0

3 years ago

Aligned magnetic domains are only present in which of the following?

butalik [34]

Answer:

i think c

Explanation:

8 0

3 years ago

Solve the problem.

gulaghasi [49]

As the shock waves travel in concentric outward circles from the epicenter, and the diameter is measured 120 miles,
area of a circle =πr*r

d=120
r=120/2r=6060*60=36003600*π=11309.734
11309.734 square miles

5 0

3 years ago

Water flows into a horizontal, cylindrical pipe at 1.4 m/s. the pipe then narrows until its diameter is halved. what is the pres

inna [77]

According to the Bernoulli's equation,the pressure difference between the wide and narrow ends of the pipe is given by

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )$

Here, $v_{1}$ is the velocity of water through wide ends of cylindrical pipe and $v_{2}$ is the velocity of water through narrow ends of cylindrical pipe.

Given, $v_{1} =1.4 m/s$

Now from equation continuity,

$v_{1} A_{1} = v_{2} A_{2}$ .

Here, $A_{1}$ and $A_{2}$ are cross- sectional areas of wide and narrow ends of cylindrical pipe.

As pipe is circular, so

$v_{1} \pi r^2_{1} = v_{2} \pi r^2_{2}$ .

At the second point, the diameter is halved, which means the radius is also halved. Therefore,

$v_{1} r^2_{1} = v_{2}(\frac{1}{2} r_{1})^2 \\\\ v_{2} = 4 v_{1}$

$v_{2} = 4 \times 1.4 = 5.6 m/s$

Substituting these values with the density of water is $1000 \ kg/m^3$ in pressure difference formula we get.

$\Delta P= \frac{1}{2} \rho ( v^2_{2} - v^2_{1} )=\frac{1}{2}\times 1000 kg/m^3(5.6^2-1.4^2)\\\\ \Delta P = 14700\ Pa$

3 0

3 years ago

Read 2 more answers

Question 14 (1 point)

Karo-lina-s [1.5K]

Answer:

car travel

precipitation

O temperature

Explanation:

Jet streams which is the ability of the object to move at a high speed due to its power is common among some given set of objects. Some are powered by the objects fuel while others are entirely different.

The above given options are actually affected by the jet streams.

5 0

3 years ago