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kakasveta [241]
2 years ago
10

The planet Neptune orbits the Sun. Its orbital radius is 30.130.130, point, 1 astronomical units (\text{AU})(AU)left parenthesis

, start text, A, U, end text, right parenthesis.
Assuming Neptune's orbit is circular, what is the distance it travels in a single orbit around the Sun?
Physics
1 answer:
lord [1]2 years ago
5 0

Answer:

The distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>

Explanation:

As it is given that the Neptune's orbit is circular, the formula that we have to use is the circumference of a circle in order to find the distance it travels in a single orbit around the Sun. In other words, you can say that the circumference of the circle is <em>equivalent</em> to the distance it travels around the Sun in a single orbit.

<em>The circumference of the circle = Distance Travelled (in a single orbit) = 2*π*R ---- (A)</em>

Where,

<em>R = Orbital radius (in this case) = 30.1 AU</em>

<em />

Plug the value of R in the equation (A):

<em>(A) => The circumference of the circle = 2*π*(30.1)</em>

<em> The circumference of the circle = </em><em>60.2π</em>

Therefore, the distance the planet Neptune travels in a single orbit around the Sun is <em>60.2π </em><em>AU.</em>

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A violin with string length 32 cm and string density 1.5 g/cm resonates in its fundamental with the first overtone of a 2.0-m or
love history [14]

Answer:

T=1022.42 N

Explanation:

Given that

l = 32 cm ,μ = 1.5 g/cm

L =2 m  ,V= 344 m/s

The pipe is closed so n= 3 ,for first over tone

f=\dfrac{nV}{4L}

f=\dfrac{3\times 344}{4\times 2}

f= 129 Hz

The tension in the string given as

T = f²(4l²) μ

Now by putting the values

T = f²(4l²) μ

T = 129² x (4 x 0.32²)  x 1.5 x  10⁻³ x 100

T=1022.42 N

6 0
2 years ago
How did the team determine that the body was placed in a wood chipper?
jenyasd209 [6]
What team are you talking about
3 0
2 years ago
Read 2 more answers
Keisha looks out the window from a tall building at her friend Monique standing on the ground, 8.3 m away from the side of the b
Salsk061 [2.6K]

Answer:

Explanation:

GIVEN DATA:

Distance between keisha and her friend 8.3 m

angle made by keisha toside building 30 degree

height of her friend monique is 1.5 m

from the figure

\Delta ACB

tan 30 = \frac{8.3}{h}

h= \frac{8.3}{tan 30} = 14.376 m

therefore

height of keisha is = h  + 1.5 m

                               = 14.376 + 1.5

= 15.876 \simeq 16 m

therefore option c is correct

5 0
2 years ago
Hoosier Manufacturing operates a production shop that is designed to have the lowest unit production cost at an output rate of 1
abruzzese [7]

Answer:

90.77%

its capacity utilization rate for the month is 90.77%

Explanation:

The capacity utilisation rate can be expressed mathematically as;

Capacity utilisation rate = capacity used/Best operating level × 100%

Given;

Total Number of production time = 205hours

Production output/capacity used = 21400 units

Best operation rate = 115units/hour

Best operation output for the month of July( at best operation level )

=115units/hour × 205 hours = 23575 units

Capacity utilisation rate = 21400/23575 × 100%

= 90.77%

3 0
2 years ago
(Another tomato/skyscraper problem.) You are looking out your window in a skyscraper, and again your window is at a height of 45
Ivan

Answer:

1027.2 m

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity

s = Displacement

a = Acceleration due to gravity = 32.2 ft/s

s=ut+\frac{1}{2}at^2\\\Rightarrow u=\frac{s-\frac{1}{2}at^2}{t}\\\Rightarrow u=\frac{450-\frac{1}{2}\times 32.2\times 2^2}{2}\\\Rightarrow u=192.8\ ft/s

v^2-u^2=2as\\\Rightarrow s=\frac{v^2-u^2}{2a}\\\Rightarrow s=\frac{192.8^2-0^2}{2\times 32.2}\\\Rightarrow s=577.20\ m

The height the tomato would fall is 450+577.2 = 1027.2 m

6 0
3 years ago
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