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Lelu [443]
3 years ago
7

What is a dwarf planet???

Physics
2 answers:
Oliga [24]3 years ago
6 0
Its a very small planet
lara31 [8.8K]3 years ago
3 0
A dwarf planet is a celestical body that orbits the sun, has enough mass to be almost round, has not cleared the neighborhood around it's orbit, and is not a moon. The main difference is that dwarf planets orbit through an uncleared path, such as asteroid belts.
Hope this helps!
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A circular loop of wire with a radius of 15.0 cm and oriented in the horizontal xy-plane is located in a region of uniform magne
bekas [8.4K]

Answer:

Explanation:

Given that,

Radius r = 15cm = 0.15m

Area of the circular loop can be determined using the formula for area of a circle

A = π r²

A = π × 0.15²

A = 0.0708 m²

Magnetic field B = 1.2T in positive z direction

B = 1.2 •k T.

If loop is remove from the field in the time interval

∆t = 2.3ms = 2.3×10^-3s

We want to find the average EMF and it is given as

ε = —∆Φ/∆t

The final flux is zero

Φf = 0

Where magnetic flux is given as

Φi = BA Cosθ

Where θ=0 since the area and the magnetic field point in the same direction.

Φi = BA Cos0

Φi = BA

Φi = 1.2 × 0.0708

Φi = 0.0848 Vs

Then, ε = —∆Φ/∆t

ε = —(Φf — Φi) / ∆t

ε = —(0-0.0848) / (2.3×10^-3)

ε = 0.0848 / (2.3×10^-3)

ε = 36.88 V

The EMF is 36.88 Volts

6 0
3 years ago
Levi and Clara are trying to move a very heavy box. Levi is pushing the box with a force of 30 N, and Clara is pulling the box w
mihalych1998 [28]
<span>There is a net force of 15 N allowing Levi and Clara to move the box would be your answer which is B</span>
5 0
3 years ago
Read 2 more answers
g The international space station has an orbital period of 93 minutes at an altitude (above Earth's surface) of 410 km. A geosyn
krok68 [10]

Answer:

r = 4.21 10⁷ m

Explanation:

Kepler's third law It is an application of Newton's second law where the forces of the gravitational force, obtaining

            T² = (\frac{4\pi }{G M_s} ) r³             (1)

           

in this case the period of the season is

            T₁ = 93 min (60 s / 1 min) = 5580 s

            r₁ = 410 + 6370 = 6780 km

            r₁ = 6.780 10⁶ m

for the satellite

           T₂ = 24 h (3600 s / 1h) = 86 400 s

if we substitute in equation 1

            T² = K r³

            K = T₁²/r₁³

            K = \frac{ 5580^2}{ (6.780 10^6)^2}

            K = 9.99 10⁻¹⁴ s² / m³

we can replace the satellite values

            r³ = T² / K

            r³ = 86400² / 9.99 10⁻¹⁴

            r = ∛(7.4724 10²²)

            r = 4.21 10⁷ m

this distance is from the center of the earth

7 0
3 years ago
The moon is smaller and more dense than the Earth, and has less extreme temperature changes.
Lerok [7]
<span>The moon is smaller and more dense than the Earth, and has less extreme temperature changes. The statement presented is True. In terms of temperature, since there is no atmosphere on the moon, then it has less extreme temperature changes. The moon can reach 253 Fahrenheit in the day and -387 Fahrenheit at night.</span>
8 0
3 years ago
Read 2 more answers
Air at 400 kPa, 980 K enters a turbine operating at steady state and exits at 100 kPa, 670 K. Heat transfer from the turbine occ
Angelina_Jolie [31]

Answer:

a). \frac{\dot{W}}{m}= 311 kJ/kg

b). \frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

Explanation:

a). The energy rate balance equation in the control volume is given by

\dot{Q} - \dot{W}+m(h_{1}-h_{2})=0

\frac{\dot{Q}}{m} = \frac{\dot{W}}{m}+m(h_{1}-h_{2})

\frac{\dot{W}}{m}= \frac{\dot{Q}}{m}+c_{p}(T_{1}-T_{2})

\frac{\dot{W}}{m}= -30+1.1(980-670)

\frac{\dot{W}}{m}= 311 kJ/kg

b). Entropy produced from the entropy balance equation in a control volume is given by

\frac{\dot{Q}}{T_{boundary}}+\dot{m}(s_{1}-s_{2})+\dot{\sigma _{gen}}=0

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+(s_{2}-s_{1})

\frac{\dot{\sigma _{gen}}}{m}=\frac{-\frac{\dot{Q}}{m}}{T_{boundary}}+c_{p}ln\frac{T_{2}}{T_{1}}-R.ln\frac{p_{2}}{p_{1}}

\frac{\dot{\sigma _{gen}}}{m}=\frac{-30}{315}+1.1ln\frac{670}{980}-0.287.ln\frac{100}{400}

\frac{\dot{\sigma _{gen}}}{m}=0.0952+0.4183+0.3978

\frac{\dot{\sigma _{gen}}}{m}=0.9113 kJ/kg-K

5 0
3 years ago
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