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Hunter-Best [27]

2 years ago

6

Find earth's approximate mass from the fact that the moon orbits earth in an average time of 27.3 days at an average distance of

384,000 kilometers. (hint: the moon's mass is only about 180 of earth's.)

1 answer:

Aleks [24]2 years ago

4 0

We can solve the problem by using Kepler's third law, which states:

$\frac{4 \pi^2}{T^2}=\frac{GM}{r^3}$

where T is the period of revolution of the Moon around the Earth, G is the gravitational constant, M the Earth's mass and r the average distance between Earth and Moon.

Using the data of the problem:

$T=27.3 d \cdot 24 \cdot 60 \cdot 60 = 2358720 s=2.36 \cdot 10^6 s$

$r=384000 km=3.84 \cdot 10^8 m$

We can re-arrange the equation and find the Earth's mass:

$M=\frac{4 \pi^2 r^3}{GT^2}=\frac{4 \pi^2 (3.84 \cdot 10^8 m)^3}{(6.67 \cdot 10^{-11})(2.36 \cdot 10^6 s)^2}=6.0 \cdot 10^{24} kg$

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An apple falls from an apple tree growing on a 20° slope. The apple hits the ground with an impact velocity of 16.2 m/s straight

EleoNora [17]

Apple hits the surface with speed 16.2 m/s

The angle made by the apple velocity with normal to the incline surface is given as 20 degree

now the component of velocity which is parallel to the surface and perpendicular to the surface is given as

$v_{perpendicular} = v cos20$

$v_{parallel} = v sin20$

so here we have

$v_{parallel} = 16.2 sin20$

$v_{parallel} = 5.5 m/s$

<em>so its velocity along the incline plane will be 5.5 m/s</em>

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2 years ago

Water exits a garden hose at a speed of 1.2 m/s. If the end of the garden hose is 1.5 cm in diameter and you want to make the wa

Katarina [22]

Answer:

A 93%

Explanation:

$P_1=P_2$ = Pressure will be equal at inlet and outlet

$\rho$ = Density of water = 1000 kg/m³

g = Acceleration due to gravity = 9.81 m/s²

$v_1$ = Velocity at inlet = 1.2 m/s

$v_2$ = Velocity at outlet

$r_1$ = Radius of inlet = $\dfrac{1.5}{2}=0.75\ cm$

$r_2$ = Radius of outlet

From Bernoulli's relation

$P_1+\dfrac{1}{2}\rho v_1^2+\rho gh_1=P_2+\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow \dfrac{1}{2}\rho v_1^2+\rho gh_1=\dfrac{1}{2}\rho v_2^2+\rho gh_2\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}v_1^2+gh_1-gh_2)}\\\Rightarrow v_2=\sqrt{2(\dfrac{1}{2}1.2^2+9.81\times 15)}\\\Rightarrow v_2=17.19709\ m/s$

From continuity equation

$A_1v_1=A_2v_2\\\Rightarrow \pi r_1^2v_1=\pi r_2^2v_2\\\Rightarrow r_2=\sqrt{\dfrac{r_1^2v_1}{v_2}}\\\Rightarrow r_2=\sqrt{\dfrac{0.0075^2\times 1.2}{17.19709}}\\\Rightarrow r_2=0.00198\ m$

The fraction would be

$\dfrac{A_1-A_2}{A_1}\times 100=\dfrac{r_1^2-r_2^2}{r_1^2}\times 100\\ =\dfrac{0.0075^2-0.00198^2}{0.0075^2}\times 100\\ =93.0304\ \%$

The fraction is 93.0304%

8 0

1 year ago

The energy from 0.015 moles of octane was used to heat 250 grams of water. The temperature of the water rose from 293.0 K to 371

arsen [322]

Answer : The correct option is, (B) -5448 kJ/mol

Explanation :

First we have to calculate the heat required by water.

$q=m\times c\times (T_2-T_1)$

where,

q = heat required by water = ?

m = mass of water = 250 g

c = specific heat capacity of water = $4.18J/g.K$

$T_1$ = initial temperature of water = 293.0 K

$T_2$ = final temperature of water = 371.2 K

Now put all the given values in the above formula, we get:

$q=250g\times 4.18J/g.K\times (371.2-293.0)K$

$q=81719J$

Now we have to calculate the enthalpy of combustion of octane.

$\Delta H=\frac{q}{n}$

where,

$\Delta H$ = enthalpy of combustion of octane = ?

q = heat released = -81719 J

n = moles of octane = 0.015 moles

Now put all the given values in the above formula, we get:

$\Delta H=\frac{-81719J}{0.015mole}$

$\Delta H=-5447933.333J/mol=-5447.9kJ/mol\approx -5448kJ/mol$

Therefore, the enthalpy of combustion of octane is -5448 kJ/mol.

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2 years ago

A CCD has a greatest possible pixel value of 4095. what is the bit level of this CCD?

Shtirlitz [24]

<span>Bit level for a CCD (Charged coupled device) with a greatest possible pixel value of 4095:The relationship between the bit level and pixel value is given as:pixel value = 2^bit level.Most charged coupled devices (CCDs) have 8-bit, 16-bit, 32-bit levels.Using simple mathematics, we can see that 2^12 = 4096.Since the maximum number of pixels is 4095, the bit level is 12., i.e. the CCD has 12-bit level.</span>

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2 years ago

Suppose Earth's mass increased but Earth's diame-

navik [9.2K]

Answer: It would increase.

Explanation:

The equation for determining the force of the gravitational pull between any two objects is:

$F = G \frac{m1m2}{r^2}$

Where G is the universal gravitational constant, m1 is the mass of one body, m2 is the mass of the other body, and r^2 is the distance between the two objects' centers squared.

Assuming the Earth's mass but not its diameter increased, in the equation above m1 (the term usually indicative of the object of larger mass) would increase, while the r^2 would not.

Thus, it goes without saying that, with some simple reasoning about fractions, an increasing numerator over a constant denominator would result in a larger number to multiply by G, thus also meaning a larger gravitational strength between Earth and whatever other object is of interest.

7 0

2 years ago