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Lera25 [3.4K]
3 years ago
11

Help!!

Physics
2 answers:
scZoUnD [109]3 years ago
7 0

If Fg=mg=ma and, Fg(planetX)=1/5Fg(earth)

then the time would be 5x of the time as gravity is acceleration. So 3.9s*5=19.5s

As the force of gravity is less, then the acceleration of masses is also less, therefore it will take more time for the object to fall by the factor of the force of gravity difference

Neko [114]3 years ago
4 0

D= 1/2 g T^2

T = √(2D/g)

If g changes to g/5, then T increases by the factor √5 .

If it was originally 3.9 sec, then the new T is 8.72 sec.

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The density of gasoline is 730 kg/m3 at 0°C. Its average coefficient of volume expansion is 9.60 10-4(°C)−1. Assume 1.00 gal of
kipiarov [429]

Answer: 0.4911 kg

Explanation:

We have the following data:

\rho_{0\°C}= 730 kg/m^{3} is the density of gasoline at 0\°C

\beta=9.60(10)^{-4} \°C^{-1} is the average coefficient of volume expansion

We need to find the extra kilograms of gasoline.

So, firstly we need to transform the volume of gasoline from gallons to m^{3}:

V=8.50 gal \frac{0.00380 m^{3}}{1 gal}=0.0323 m^{3} (1)

Knowing density is given by: \rho=\frac{m}{V}, we can find the mass m_{1} of 8.50 gallons:

m_{1}=\rho_{0\°C}V

m_{1}=(730 kg/m^{3})(0.0323 m^{3})=23.579 kg (2)

Now, we have to calculate the factor f by which the volume of gasoline is increased with the temperature, which is given by:

f=(1+\beta(T_{f}-T_{o})) (3)

Where T_{o}=0\°C is the initial temperature and T_{f}=21.7\°C is the final temperature.

f=(1+9.60(10)^{-4} \°C^{-1}(21.7\°C-0\°C)) (4)

f=1.020832 (5)

With this, we can calculate the density of gasoline at 21.7\°C:

\rho_{21.7\°C}=730 kg/m^{3} f=(730 kg/m^{3})(1.020832)

\rho_{21.7\°C}=745.207 kg/m^{3} (6)

Now we can calculate the mass of gasoline at this temperature:

m_{2}=\rho_{21.7\°C}V (7)

m_{2}=(745.207 kg/m^{3})(0.0323 m^{3}) (8)

m_{2}=24.070 kg (9)

And finally calculate the mass difference \Delta m:

\Delta m=m_{2}-m_{1}=24.070 kg-23.579 kg (10)

\Delta m=0.4911 kg (11) This is the extra mass of gasoline

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BabaBlast [244]
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A coin released at rest from the top of a tower hits the ground after falling 1.5 s. What is the speed of the coin as it hits th
Inessa05 [86]

initially coin is at rest and then it drop for total time t = 1.5 s

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v_f = v_i + at

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Answer:

Answer:

D) by using military force.

Explanation:

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You need to know the specific heat capacity of air.
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