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4 years ago

Does orbital direction of a planet affect its synodic period as seen from earth?

Physics

1 answer:

Mice21 [21]4 years ago

4 0

Answer:

Yes

Explanation:

Yes, the orbital direction of a planet affect its Synodic period as seen from earth. The synodic period of mercury is about 116 days on Earth. Whereas the sidereal period of mercury as seen from Earth is 88 days. This difference is caused by the simultaneous motion of Earth in its orbit. Hence we can say that the orbital direction of a planet affect its synodic period as seen from earth.

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At a distance of 41 ft , an ionizing radiation source delivers 5.0 rem of radiation. How close could you get to the source and s

mash [69]

Answer:

at distance 18.33 ft no biological effects

Explanation:

given data

distance = 41 ft

radiation = 5 rem

to find out

how close get so no biological effects

solution

we know here that intensity of radiation R is inversely proportional to (radiation)²

so here equation will be

$\frac{I1}{I2} = (\frac{R2}{R1})^{2}$ .............1

here I is intensity

so here I1 = 25

because for 0 to 25 there is no detectable effects and for 25 to 100 it will . temporary decrease so

we take 25 because dose is less than 25 so we take that highest value

$\frac{I1}{I2} = (\frac{R2}{R1})^{2}$

$\frac{25}{5} = (\frac{41}{R1})^{2}$

R1 = 18.33 ft

so at distance 18.33 ft no biological effects

6 0

3 years ago

Describe the trip from your home to school using the words position, distance, displacement, and

ozzi

Answer:

The position of my house is a little uphill as compared to the position of my school. The distance I have to travel from my house to school is nearly 2 kilometers. The displacement is in the 2000 m towards the left from my house. The speed of the bus which I usually take is 40 km/ hour.

6 0

3 years ago

Suppose you are at the center of a large freely-rotating horizontal turntable in a carnival funhouse. As you crawl toward the ed

I am Lyosha [343]

Answer:

c. remains the same, but the RPMs decrease.

Explanation:

Because there aren't external torques on the system composed by the person and the turntable it follows that total angular momentum (I) is conserved, that means the total angular momentum is a constant:

$\overrightarrow{L}=constant$

The total angular momentum is the sum of the individual angular momenta, in our case we should sum the angular momentum of the turntable and the angular momentum of a point mass respect the center of the turntable (the person)

$\overrightarrow{L_{turnatble}}+\overrightarrow{L_{person}}=constant$ (1)

The angular momentum of the turntable is:

$\overrightarrow{L_{turnatble}}=I\overrightarrow{\omega}$ (2)

with I the moment of inertia and ω the angular velocity.

The angular momentum of the person respects the center of the turntable is:

$\overrightarrow{L_{person}}=\overrightarrow{r}\times m\overrightarrow{v}$ (3)

with r the position of the person respects the center of the turntable, m the mass of the person and v the linear velocity

Using the fact $v=\omega r$ :

$\overrightarrow{L_{person}}=\overrightarrow{r}\times rm\overrightarrow{\omega}$ (3)

By (3) and (2) on (1) and working only the magnitudes (it's all that we need for this problem):

$I\omega+r^{2}m\omega=constant$

$\omega(I+r^{2}m)=constant$

Because the equality should be maintained, if we increase the distance between the person and the center of the turntable (r), the angular velocity should decrease to maintain the same constant value because I and m are constants, so the RPM's (unit of angular velocity) are going to decrease.

4 0

4 years ago

Which of these quantities needs to have a direction associated with it? Select one: a. force only b. force and acceleration c. m

ehidna [41]

B) force & acceleration

6 0

3 years ago

An object starts from rest and falls freely for 40. meters near the surface of planet P. If the time of fall is 4.0 seconds, wha

Sergeu [11.5K]

Let's use the equations for rectilinear motion. One particular equation is:

d = v₀t +(1/2)at²

Since it has been mentioned that the object starts from rest, then v₀ = 0. Substituting the values,

40 m = (0)(4 s) + (1/2)(a)(4 s)²
Solving for a,
a = 5 m/s²

<em>Therefore, the acceleration due to gravity is 5 m/s².</em>

6 0

4 years ago