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Rus_ich [418]
3 years ago
12

a) A Kuiper Belt Object orbits the Sun with an average orbital distance of 40 AU. What is its orbital period? Round your answer

to the nearest integer.
Physics
1 answer:
denpristay [2]3 years ago
7 0

Answer:

253 years

Explanation:

a = Distance in Astronomical units from the sun = 40 AU

P = Orbital period in years

From Kepler's law we have

P^2=a^3

\Rightarrow P^2=40^3

\Rightarrow P^2=64000

\Rightarrow P=\sqrt{64000}

\Rightarrow P=252.982212813

\Rightarrow P\approx 253\ years

The orbital period of the Kuiper belt object is 253 years.

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Which sizes influence the adhesive and rolling friction?
julsineya [31]

juju on that beat

Explanation:

4 0
3 years ago
A rock band playing an outdoor concert produces sound at 120 dB 5.0 m away from their single working loudspeaker. What is the so
joja [24]

ANSWER:100dB

f<em>rom the  sound intensity level,the sound intensity is calculated as:</em>

<em />\beta<em>₁=</em>\beta<em>=(10dB)㏒₁₀(l₁/l₀)</em>

<em>inserting numbers:</em>

<em>120dB=(10dB)㏒₁₀[l₁/10⁻¹²W/m⁸] or 12=㏒₁₀[l₁/(10⁻¹²)W/m²</em>

<em>Getting the antilog of both sides and obtain 10¹²=l₁(10⁻¹²W/m²)which </em>

<em>can be used to solve for l₁ and get</em>

<em>          l₁=(10⁻¹²W/m²)(10¹²)=1 W/m²</em>

<em>since the sound  intensity is related to the power and that the power does not change,the sound intensity at any other point can be solved.Plugging-in</em>

<em>ᵃ = 4πr²,into P=l₁ₐ₁=l₂ₐ₂ and get:</em>

<em>l₂=l₁(r₁/r₂)² =(1W/m²)(5/35) =2.04×10⁻²W/m²</em>

<em>since we know the sound intensity at the sound point 2r,the sound intensity level at the point can be solved.We have:</em>

<em>     </em>\beta<em>₂=</em>\beta<em>=(10dB)㏒₁₀(l₂/l₀)=(10dB)㏒₁₀(2.04×10⁻²/1×10⁻²)</em>

<em>    </em>\beta<em>₂=(10dB)㏒₁₀(2.04×10¹⁰)</em>

<em />\beta<em> =(10dB)[㏒₁₀(2.04)+㏒₁₀(10¹⁰)]=10dB[0.32+10]=103dB=100dB</em>

<em />

3 0
4 years ago
A block with a weight of 20 N sits on a table. How much force does the table exert
sasho [114]
The answer would be 20 N
6 0
3 years ago
What is the kinetic energy in Joules of a 1,500kg car traveling at 75mph?What is the kinetic energy in Joules of a 1,500kg car t
Sergio039 [100]

Answer: 2812500 joules

Explanation:

Mass of car = 1500kg

Velocity of car = 75mph

Kinetic energy = ?

Recall that kinetic energy is the energy possessed by a moving object, and it depends on its mass M and velocity, V

Thus, Kinetic energy = 1/2 x mv^2

= 1/2 x 1000kg x (75mph)^2

= 0.5 x 1000kg x (75mph)^2

= 500 x 5625

= 2812500 joules

Thus, the car travels with a kinetic energy of 2812500 joules

5 0
3 years ago
A machine has a mechanical advantage of 5. if 300 newtons of input force is used to produce 3,000 newton meters of work what is
Arte-miy333 [17]
F(in) x d(in)= F(out)x d(out)
A machine has a mechanical advantage of 5.
F(out )= 5xF(in)300 newtons

F(in) = 300 N
F(out)= 5xF(in) = 5x300 N
= 1500 N

Work in = Work out = 3000 N x m3000
N x m = F_in x d_ind_in = 3000 Nxm / F_in = 3000 N x m / 300 N = 10 m3000 Nxm = F_out x d_out d_out = 3000 Nx m / F_out = 3000 N x m / 1500 N = 2 m
8 0
4 years ago
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