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sasho [114]
3 years ago
13

3. What would the seasons be like if Earth's orbit had the eccentricity value of Mars?

Physics
2 answers:
Dmitriy789 [7]3 years ago
4 0
According to Kepler's laws, the mass of a planet has almost no effect on its orbit; the mass of the sun is what controls things. Even though Earth is 10 times heavier than Mars, it would still trundle along Mars's old path. Both Mars and Earth are perpetually falling toward the sun, and all falling bodies fall at the same rate.
densk [106]3 years ago
3 0

Answer:

123456789

Explanation:

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The displacement vector A and B when added together , give the resultant vector R so that R= A+B use the data in the drawing and
Salsk061 [2.6K]
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.

Let:

b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°

We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.

Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.

In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
 
5 0
3 years ago
A uniform magnetic field passes through a horizontal circular wire loop at an angle 19.5 ∘ from the vertical. The magnitude of t
nlexa [21]

To solve this problem, we will apply the concepts related to Faraday's law that describes the behavior of the emf induced in the loop. Remember that this can be expressed as the product between the number of loops and the variation of the magnetic flux per unit of time. At the same time the magnetic flux through a loop of cross sectional area is,

\Phi = BA Cos \theta

Here,

\theta = Angle between areal vector and magnetic field direction.

According to Faraday's law, induced emf in the loop is,

\epsilon= -N \frac{d\Phi }{dt}

\epsilon = -N \frac{(BAcos\theta)}{dt}

\epsilon = -NAcos\theta \frac{dB}{dt}

\epsilon = -N\pi r^2 cos\theta \frac{d}{dt} ( ( 3.75 T ) + ( 3.05T/s ) t + ( -6.95 T/s^2 ) t^2)

\epsilon = -N\pi r^2 cos\theta( (3.05T/s)-(13.9T/s)t )

At time t = 5.71s,  Induced emf is,

\epsilon = -(1) \pi (0.220m)^2 cos(19.5\°)(  (3.05T/s)-(13.9T/s)(5.71s))

\epsilon = 10.9V

Therefore the magnitude of the induced emf is 10.9V

4 0
3 years ago
Read 2 more answers
What kind of friction occurs as a fish swims through water?
Paul [167]
It must be sliding friction, because the fish is already in motion.
7 0
3 years ago
Approximate the Sun as a uniform sphere of radius 6.96 X 108 m, rotating about its central axis with a period of 25.4 days. Supp
In-s [12.5K]

Answer:

T = 184 seconds

Explanation:

First in order to solve this, we need to know which is the expression to calculate the period. This is an exercise of angular velocity, so:

T = 2π/w

Where w: angular speed (in rad/s)

So, let's calculate first the innitial angular speed:

w = 2π/T

Converting days to seconds:

25.4 days * 24 h/day * 3600 s/h = 2,194,560 s

Then the angular speed:

w = 2π / 2,194,560 = 2.863x10^-6 rad/s

Now, the innitial angular momentum is:

I = (2/5)Mr² replacing data:

I = 2/5* (6.96x10^8)² * M = 1.94x10^17m² * M

so the initial angular momentum would be:

L = Iω = 2.863x10^-6 * 1.94x10^17 M

L = 5.55x10^11 m²/s * M = final angular momentum

Now the  final I = 2/5Mr²

Final I = 2/5 * (6.37x10^6)² * M  = 1.62x10^13m² * M

Then 5.55x10^11m²/s * M = 1.62x10^13m² * M * ω → M cancels

ω = 3.42x10^-2 rad/s

Then the new period

T = 2π/ω = 2*3.14 / 3.42x10^-2

T = 184 seconds

8 0
3 years ago
If a 5 N force pushes a 20 kg mass on a frictionless surface, how fast is the mass going in 5 seconds.
stira [4]

Explanation:

Let me know if you have questions

3 0
3 years ago
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