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Alexus [3.1K]
2 years ago
7

A starship is cruising the interstellar space with a velocity of 2174.799173 miles per second. How far would the starship have t

raveled in 2 years, 9 days and 4.50 hours? Consider 365 days in one year.
Physics
1 answer:
aalyn [17]2 years ago
8 0

Answer:

1.3889 x 10¹¹ miles.

Explanation:

2 Years = 2 x 365 days = 730 days

2 years and 9 days = 739 days .

739 days = 739 x 24 = 17736 hours

2 years 9 days and 4.5 hours = 17736 + 4.5 = 17740.5 hours

17740.5 hours = 17740.5 x 60 x 60 = 63.865 x 10⁶ s

Distance travelled = velocity  x time

2174.799173 x 63.865 x 10⁶ = 1.3889 x 10¹¹ miles.

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A pot on the stove contains 200 g of water at 20°C. An unknown mass of ice that is originally at −10°C is placed in an identical
Mumz [18]

Answer:

a) The mass of the ice is smaller than the mass of the water

b) The ice reaches first 80°C ,

Explanation:

Since the heat Q that should be provided to ice

Q = sensible heat to equilibrium temperature (as ice) + latent heat + sensible heat until final temperature ( as water)

m ice * c ice * ( T equil -T initial  ) + m ice* L + m ice* c water * ( T final - T equil)

and the heat Q that should be provided to water is

Q= m water * c water * ( T final - T equil )

since the rate of heat addition q = constant and the time t taken to reach the final temperature is the same , then the heat absorbed Q=q*t is the same for both, therefore

m water * c water *  ( T final - T equil ) = m ice* [c ice *( T equil -T initial  ) + L + c water * ( T final - T equil)]

m water/ m ice =  [c ice * ( T equil -T initial  )  + L + c water * ( T final - T equil)]/ [ c water * ( T final - T equil)]

m water/ m ice = [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] + 1

since  [c ice * ( T equil -T initial  )  + L ]/[c water * ( T final - T equil) ] >0 , then

m water/ m ice > 1

m water > m ice

so the mass of ice is smaller that the mass of water

b) Since the heat Q that should be provided to the ice, starting from 55°C mass would be

Q ice= m ice * c water * ( T final2 - T final1 )

and for the water mass

Q water = m water * c water * ( T final2 - T final1 )

dividing both equations

Q water / Q ice = m water / m ice >1

thus

Q water > Q ice

since the heat addition rate is constant

Q water = q* t water and Q ice=q* t ice

therefore

q* t water > q* t ice

t water >  t ice

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5 0
3 years ago
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

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<span>The equation of motion for a rocket in vertical flight can be obtained from newton’s second law of motion and is constant-mass system. The equation of motion for a body mass varies with time and mass. When force acts on rocket, the rocket will accelerate in the direction of force. Therefore, force is equal to the change in momentum per change in time. For constant mass, force equals mass times acceleration.</span>

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Anuta_ua [19.1K]
The answer to this question is true.
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emmasim [6.3K]

Answer: 12

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