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2 years ago

Does the orbital period of a planet depend on the mass of the planet or on the mass of the star that it orbits?

Physics

1 answer:

jasenka [17]2 years ago

8 0

Answer:

The orbital period of a planet depends on the mass of the planet.

Explanation:

A less massive planet will take longer to complete one period than a more massive planet.

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What kind of research in teaching of physical science can be done???

11Alexandr11 [23.1K]

Answer:

Physical sciences are those academic disciplines that aim to uncover the underlying laws of nature - often written in the language of mathematics. It is a collective term for areas of study including astronomy, chemistry, materials science and physics.

Explanation:

3 0

3 years ago

Car A starts out traveling at 35.0 km/h and accelerates at 25.0 km/h2 for 15.0 min. Car B starts out traveling at 45.0 km/h and

lawyer [7]

1 kilometre=1000 metre

1 hour = 3600 second

$1\ km/hr=\frac{1000}{3600} m/s$

$1\ km/hr=\frac{5}{18} m/s$

The initial velocity of car A is 35.0 km/hr i.e

$35.0\ km/hr=35*\frac{5}{18} m/s$

= 9.72 m/s

The initial velocity of car B is 45 km/hr =12.5 m/s

The initial velocity of car C is 32 km/hr = 8.89 m/s

The initial velocity of car D is 110 km/hr=30.56 m/s

The acceleration of car A is given as $25\ km/hr^2$

$=\ 25*\frac{1000}{3600*3600} m/s^2$

$=0.00192901234 m/s^2$

The time taken by car A = 15 min.

From equation of kinematics we know that-

v= u+at [Here v is the final velocity and a is the acceleration and t is the time]

Final velocity of A, v = 9.72 m/s +[0.00192901234×15×60]m/s

=11.456111106 m/s

The acceleration of B is given as $15\ km/hr^2$

$=0.00115740740740 m/s^2$

The time taken by car B =20 min

The final velocity of B is -

v= u+at

= u-at [Here a is negative due to deceleration]

=12.5 m/s +[0.0011574074074×20×60]

=13.8888888.....

=13.9

The acceleration of C is given as $40\ km/hr^2$

$=\ 0.003086419753 m/s^2$

The time taken by car C =30 min

The final velocity of C is-

v = u+at

=8.89 m/s+[0.003086419753×30×60] m/s

=14.4455555555..m/s

=14.45 m/s

The car C is decelerating.The deceleration is given as- $60\ km/hr^2$

$=0.0046296296296m/s^2$

The time taken by car D= 45 min.

The final velocity of the car D is-

v =u+at

=30.56 -[0.00462962962962×45×60]m/s

=18.06 m/s

Hence from above we see that the magnitude of final velocity car C and B is close to 15 m/s. The car C is very close as compared to car B.

3 0

3 years ago

Match the boundary with its characteristic

kondor19780726 [428]

<span>Matching the boundary with its characteristics 1. Convergent - C. Compression 2. Divergent - B. Along ocean ridges 3. Transform - A. Along strike-slip faults The compression that occur in the convergent boundary causes the reverse fault in the earth crust. So in the divergent boundary two crust plates move apart causing a normal fault along the ocean ridges. The faults in the transform boundary happens at the place where plates slide laterally.</span>

7 0

3 years ago

Read 2 more answers

How far does a car travel as it accelerates from 12 m/s to 18 m/s at 3 m/s??

yulyashka [42]

Answer:

Explanation:

I ASSUME you mean acceleration is 3 m/s²

v² = u² + 2as

s = (v² - u²) / 2a

s = (18² - 12²) / (2(3))

s = 30 m

to verify we can see that the acceleration time is

t = (18 - 12) / 3 = 2 s

s = 0 + 12(2) + ½(3)2² = 30 m

4 0

2 years ago

A capacitor consists of a set of two parallel plates of area A separated by a distance d . This capacitor is connected to a batt

Elis [28]

Answer:

if we double the distance the energy stored will be doubled also

Explanation:

The energy stored in a capacitor is given as

Energy stored =1/2(cv²)

= 1/2(Qv)

Where c = capacitance

Q= charge

But the electric field is expressed as

E= v/d

where v= voltage

d= distance

v=Ed

Substituting into any equation above say

Energy stored =1/2(Qv)

Substituting v=Ev

Energy stored =1/2(QEd)

From the equation above it shows that if we double the distance The energy stored will be doubled also

5 0

3 years ago