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2 years ago

using mass and distance, identify and compare the sun's and moon's contribution to the formation of tides on earth

Physics

1 answer:

vekshin12 years ago

7 0

Answer:

Based on its mass, the sun's gravitational attraction to the Earth is more than 177 times greater than that of the moon to the Earth.

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Gravity on the surface of the moon is only 1/6 as strong as gravity on the Earth. What is the weight of a 19 kg object on the Ea

Dahasolnce [82]

The weight of anything in any place is

(mass of the thing) x (acceleration of gravity in that place).

-- On Earth, the acceleration of gravity is about 9.807 m/s²

Weight of 19 kg of mass is (19 kg) x (9.807 m/s²) = 186.3 newtons

-- On the Moon, the acceleration of gravity is about 1.623 m/s²

Weight of the same 19 kg of mass is (19 kg) x (1.623 m/s²) = 30.8 newtons

7 0

3 years ago

Sound travels faster through rigid material like steel rather than through spongy materials like rubber. Why?

drek231 [11]

since sound travels using mechanical waves and needs a material medium to propagate and since mechanical waves spread through vibrations ...and since hard materials have their atoms packed closely....they need to vibrate with a smaller amplitude to pass on the wave....thus sound travels faster in a denser medium than a less dense one.

3 0

3 years ago

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

6 0

3 years ago

As a laudably skeptical physics student, you want to test Coulomb's law. For this purpose, you set up a measurement in which a p

Oksana_A [137]

Answer:

a.Attractive

$2.31531\times 10^{-16}\ N$

Explanation:

When it comes to charges, the charges which are alike repel each other and the charges which are different will attract each other.

Here, there is a proton and electron which are different particles hence, they will attract each other.

$q_1=q_2$ = Charge of electron and proton = $1.6\times 10^{-19}\ C$

r = Distance between them = 997 nm

k = Coulomb constant = $8.99\times 10^{9}\ Nm^2/C^2$

Force is given by

$F=\dfrac{kq_1q_2}{r^2}\\\Rightarrow F=\dfrac{8.99\times 10^9\times 1.6\times 10^{-19}\times 1.6\times 10^{-19}}{(997\times 10^{-9})^2}\\\Rightarrow F=2.31531\times 10^{-16}\ N$

The force of attraction between the particles will be $2.31531\times 10^{-16}\ N$

8 0

2 years ago

A bond between a positively-charged particle and a negatively-charged particle is a(n) _______ bond.

weeeeeb [17]

The answer is an ionic bond.

6 0

3 years ago