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2 years ago

7

Of orbital radius, orbital velocity, orbital circumference, or radial area swept through, which of these stays constant accordin

g to one of Kepler’s Laws?
a. orbital distance
b. radial area swept through
c. none of these
d. orbital velocity

1 answer:

kow [346]2 years ago

5 0

Answer:

dyow

Explanation:

sorrrey

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The speeds of 22 particles are as follows: two at 5.30 cm/s, four at 1.40 cm/s, six at 7.14 cm/s, eight at 1.52 cm/s, two at 7.6

Leona [35]

Answer:

Average velocity is 3.93 cm/s

Root mean square velocity is 4.79 cm/s

Velocity peak to peak is 6.28 cm/s

Explanation:

Speed of 2 particles = 5.3 cm/s

Speed of 4 particles = 1.4 cm/s

Speed of 6 particles = 7.14 cm/s

Speed of 8 particles = 1.52 cm/s

Speed of 2 particles = 7.68 cm/s

$v_{avg}=\frac{2\times 5.3+4\times 1.4+6\times 7.14+8\times 1.52+2\times 7.68}{22}\\\Rightarrow v_{avg}=\frac{86.56}{22}\\\Rightarrow v_{avg}=3.93\ cm/s$

Average velocity is 3.93 cm/s

$v_{rms}=\sqrt{\frac{2\times 5.3^2+4\times 1.4^2+6\times 7.14^2+8\times 1.52^2+2\times 7.68^2}{22}}\\\Rightarrow v_{rms}=\sqrt{\frac{507.34}{22}}\\\Rightarrow v_{rms}=4.79\ cm/s$

Root mean square velocity is 4.79 cm/s

$v_p=7.68-1.4=6.28\ cm/s$

Velocity peak to peak is 6.28 cm/s

8 0

2 years ago

The components of a 15 meters per second velocity at an angle of 60 degrees above the horizontal are?

mafiozo [28]

Answer:

x-component of velocity: 7.5 m/s

y-component of velocity: 13 m/s

Explanation:

This problem is pure trigonometry. Assuming you know trig, there are only a couple of steps to solving this problem. First, split the velocity into components; recall that any vector not directed along an axis has x and y components. Then, remember that sinΘ = opposite/hypotenuse. Applying this to your scenario, you get sin60° = vy/15. Multiplying this out gives you vy=15sin60. Put this into a calculator (make sure it's set to degree mode because the angle in this problem is in degrees) and you should get 12.99, which you can round up to 13 m/s. This is the velocity in the y-direction.

The procedure to find the x-velocity is very similar, but instead of using sine, we will use the cosine of theta. Recall that cosΘ=adjacent/hypotenuse. Once again plugging this scenario's numbers into that, you end up with cos60 = vₓ/15. Multiplying this out gives you vₓ = 15cos60. Once again, plug this into your calculator. 7.5 m/s should be your answer. This is the velocity in the x-direction.

By the way, a quick way to find the components of a vector, whether it's velocity, force, or whatever else, is to use these functions. Generally, if the vector points somewhere that's not along an axis, you can use this rule. The x-component of the vector is equal to hypotenuse*cosΘ and the y-component of the vector is equal to hypotenuse*sinΘ.

8 0

1 year ago

Which is a major threat to biodiversity? A. Red-billed oxpeckers eat ticks off of impalas' coats in the grasslands of East Afric

SIZIF [17.4K]

B. Mining in the Guinean Forests of West Africa to provide diamond and gold jewelry for humans.

4 0

3 years ago

Please help me guys never mind the calculations

vlada-n [284]

The shape is connected in parallel so;

5.1) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{2} + \frac{1}{3} \\ \frac{1}{R} = \frac{3 + 2}{6} = \frac{5}{6} \\ R = \frac{6}{5} = 1.2 \: \: ohm$

5.2) Ans;

$\frac{1}{R} = \frac{1}{R1} + \frac{1}{R2} \\ \frac{1}{R} = \frac{1}{8} + \frac{1}{10} \\ \frac{1}{R} = \frac{5 + 4}{40} = \frac{9}{40} \\ R = \frac{40}{9} = 4.4 \: \: ohm$

I hope I helped you^_^

7 0

2 years ago

A 0.50 kg toy is attached to the end of a 1.0 m very light string. The toy is whirled in a horizontal circular path on a frictio

xenn [34]

Answer:

The maximum speed will be 26.475 m/sec

Explanation:

We have given mass of the toy m = 0.50 kg

radius of the light string r = 1 m

Tension on the string T = 350 N

We have to find the maximum speed without breaking the string

For without breaking the string tension must be equal to the centripetal force

So $T=\frac{mv^2}{r}$

So $350=\frac{0.5\times v^2}{1}$

$v^2=700$

v = 26.475 m /sec

So the maximum speed will be 26.475 m/sec

6 0

2 years ago