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larisa86 [58]
3 years ago
11

Suppose that a planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average o

rbit radius of Mercury. What would be the orbital period of such a planet? (Use the mass of the sun of 1.99 \times 10^{30} kg1.99×10 30 kg and the radius of Mercury's orbit is 5.79 \times 10^{10} m5.79×10 10 m.) Express your answer in units of "days".
Physics
1 answer:
Musya8 [376]3 years ago
6 0

Explanation:

It is given that,

A planet were discovered between the sun and Mercury, with a circular orbit of radius equal to 2/3 of the average orbit radius of Mercury.

Mass of the Sun, M=1.99\times 10^{30}\ kg

Radius of Mercury's orbit, r=5.79\times 10^{10}\ m

Radius of discovered planet, R=\dfrac{2}{3}r

R=\dfrac{2}{3}\times 5.79\times 10^{10}\ m=3.86\times 10^{10}\ m

Let T is the orbital period of such a planet. Using Kepler's third law of planetary motion as :

T^2\propto R^3

T^2=\dfrac{4\pi^2R^3}{GM}

T^2=\dfrac{4\pi^2\times (3.86\times 10^{10})^3}{6.67\times 10^{-11}\times 1.99\times 10^{30}}

T=\sqrt{1.71\times 10^{13}}

T = 4135214.625 s

or

T = 47.86 days

So, the orbital period of such a planet is 47.86 days. Hence, this is the required solution.

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A 150 kg car hits a wall with 45 N of force, what was its acceleration?​
alisha [4.7K]

Answer:

a = 0.3 m/s²

Explanation:

Given: 45 N, 150 kg

To find: a

Formula: a = \frac{F}{m}

Solution: To find a, divide the force by the weight  

A = F ÷ m

=  45 ÷  150

= 0.3 m/s²

Newtons are derived units, equal to 1 kg-m/s². In other words, a single Newton is equal to the force needed to accelerate one kilogram one meter per second squared.

4 0
3 years ago
A train at a constant 79.0 km/h moves east for 27.0 min, then in a direction 50.0° east of due north for 29.0 min, and then west
ivolga24 [154]

Answer:

Magnitude of avg velocity, |v_{avg}| = 18.9 km/h

\theta' = 56.85^{\circ}

Given:

Constant speed of train, v = 79 km/h

Time taken in East direction, t = 27 min = \frac{27}{60} h

Angle, \theta = 50^{\circ}

Time taken in 50^{\circ}east of due North direction, t' = 29 min =  \frac{29}{60} h

Time taken in west direction, t'' = 37 min =  \frac{27}{60} h

Solution:

Now, the displacement, 's' in east direction is given by:

\vec{s} = vt = 79\times \frac{27}{60} = 35.5\hat{i} km

Displacement in  50^{\circ} east of due North for 29.0 min is given by:

\vec{s'} = vt'sin50^{\circ}\hat{i} + vt'cos50^{\circ}\hat{j}

\vec{s'} = 79(\frac{29}{60})sin50^{\circ}\hat{i} + 79(\frac{29}{60})cos50^{\circ}\hat{j}

\vec{s'} = 29.25\hat{i} + 24.54\hat{j} km

Now, displacement in the west direction for 37 min:

\vec{s''} = - vt''hat{i} = - 79\frac{37}{60} = - 48.72\hat{i} km

Now, the overall displacement,

\vec{s_{net}} = \vec{s} + \vec{s'} + \vec{s''}

\vec{s_{net}} = 35.5\hat{i} + 29.25\hat{i} + 24.54\hat{j} - 48.72\hat{i}

\vec{s_{net}} =  16.03\hat{i} + 24.54\hat{j} km

(a) Now, average velocity, v_{avg} is given:

v_{avg} = \frac{total displacement, \vec{s_{net}}}{total time, t}

v_{avg} = \frac{16.03\hat{i} + 24.54\hat{j}}{\frac{27 + 29 + 37}{60}}

v_{avg} = 10.34\hat{i} + 15.83\hat{j}) km/h

Magnitude of avg velocity is given by:

|v_{avg}| = \sqrt{(10.34)^{2} + (15.83)^{2}} = 18.9 km/h

(b) angle can be calculated as:

tan\theta' = \frac{15.83}{10.34}

\theta' = tan^{- 1}\frac{15.83}{10.34} = 56.85^{\circ}

6 0
3 years ago
A train leaves the train station at noon and travels at a constant speed of vt = 50 mi/hr on a straight track. 2 hr later, a car
asambeis [7]

Answer:

350 miles

Explanation:

When the car starts 2 hours later, the train would have a head start of

50 * 2 = 100 miles

The speed of the car relative to the train is

70 - 50 = 20 mi/hr

For the car to catch up with the train, it must cover the 100 miles difference at the rate of 20mi/hr. So the time it would need to cover this difference is

100 / 20 = 5 hours

After 5 hours, the car would have traveled a distance of

5 * 70 = 350 miles which is also the distance from the station to where the car catches up

6 0
3 years ago
The current in a coil with a self-inductance of 1 mH is 2.8 A at t = 0, when the coil is shorted through a resistor. The total r
pogonyaev

Given:

L = 1 mH = 1\times 10^{-3} H

total Resistance, R = 11 \Omega

current at t = 0 s,

I_{o} = 2.8 A

Formula used:

I = I_{o}\times e^-{\frac{R}{L}t}

Solution:

Using the given formula:

current after t = 0.5 ms = 0.5\times 10^{-3} s

for the inductive circuit:

I = 2.8\times e^-{\frac{11}{1\times 10^{-3}}\times 0.5\times 10^{-3}}

I =   2.8\times e^-5.5

I =0.011 A

5 0
3 years ago
An airplane weighing 11,000 N climbs to a
Gennadij [26K]

The power in horsepower is 40.1 hp

Explanation:

We start by calculating the work done by the airplane during the climb, which is equal to its change in gravitational potential energy:

W=(mg)\Delta h

where

mg = 11,000 N is the weight of the airplane

\Delta h = 1.6 km = 1600 m is the change in height

Substituting,

W=(11,000)(1600)=17.6\cdot 10^6 J

Now we can calculate the power delivered, which is given by

P=\frac{W}{t}

where

W=17.6\cdot 10^6 J is the work done

t=9.8 min \cdot 60 = 588 s is the time taken

Substituting,

P=\frac{17.6\cdot 10^6 J}{588}=2.99\cdot 10^4 W

Finally, we can convert the power into horsepower (hp), keeping in mind that

1 hp = 746 W

Therefore,

P=\frac{2.99\cdot 10^4}{746}=40.1 hp

Learn more about power:

brainly.com/question/7956557

#LearnwithBrainly

8 0
3 years ago
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