A stuntman of mass 55 kg is to be launched horizontally out of a spring-

Every few hundred years most of the planets line up on the same side of the Sun.(Figure 1)Calculate the total force on the Earth

mylen [45]

Answer: $3.7 \times 10^{-4} N$

Explanation:

The gravitational pull between two object is given by:

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

Where M and m are the masses of the object, r is the distance between the masses and G = 6.67× 10⁻¹¹ m³kg⁻¹ s⁻² is the gravitational constant.

We have to calculate the net force on Earth due to Venus, Jupiter and Saturn when they are in one line. It means when they are the closest distance.

$F_{net] = G\frac{M_eM_v}{r_v^2}+G\frac{M_eM_j}{r_j^2}+G\frac{M_eM_s}{r_s^2}$

Mass of Earth, Me = 5.98 × 10²⁴ kg

Mass of Venus, Mv = 0.815 Me

Mass of Jupiter, Mj = 318 Me

Mass of Saturn, Ms = 95.1 Me

closest distance between Earth and Venus, rv = 38 × 10⁶ km = 0.25 AU

closest distance between Jupiter and Earth, rj = 588 × 10⁶ km = 3.93 AU

closest distance between Earth and Saturn, rs = 1.2 × 10⁹ km = 8.0 AU

where 1 AU = 1.5 × 10¹¹ m

Inserting the values:

$F_{net} = G\frac{M_e\times 0.815 M_e}{(0.25AU)^2}+G\frac{M_e\times 318 M_e}{(3.93AU)^2}+G\frac{M_e\times 95.1 M_e}{(8.0AU)^2}\\ \Rightarrow F_{net} = \frac{(GM_e^2)}{(1AU)^2}(\frac{0.815}{0.25^2}+\frac{318}{3.93^2}+\frac{95.1}{8.0^2})=\frac{6.67\times 10^{-11} \times (5.98\times 10^{24})^2}{(1.5\times 10^{11})^2}(35.1) = 3.7 \times 10^{-4} N$

4 0

3 years ago

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The diffusion rate for a solute is 4.0 x 10^-11 kg/s in a solvent- filled channel that has a cross-sectional area of 0.50 cm^2 a

zlopas [31]

Answer:

$s = 9.6\times 10^{-12}kg/s$

Explanation:

Given:

Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s

Area of cross-section = 0.50 cm²

Length of channel =0.25 cm

Now for the new channel

Area of cross-section = 0.30 cm²

Length of channel =0.10 cm

let the Solute Diffusion rate of new channel = s

now equating the diffusion rate per unit volume for both the channels

$\frac{4\times 10^{-11}}{0.50\times 0.25}=\frac{s}{0.30\times 0.10}$

thus,

$s = 9.6\times 10^{-12}kg/s$

7 0

3 years ago

Is noise energy a kind of energy or not

Volgvan

Yes. sound waves are produced by energy.

4 0

3 years ago

If your face is 62 cm in front of a plane mirror, where is the image of your face located?

timurjin [86]

Answer:

62 cm is in front of the mirror

Explanation:

This is answers

3 0

3 years ago

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13. Suppose that the strength of the electric field about an isolated point charge has a certain value at a distance of 1 m. How

VMariaS [17]

Answer: E/4 ( one - fourth of it electric field)

Explanation:

The electric field of a point charge is given below as

E =kq/r²

E = electric field,

K = electric constant

q = magnitude of electric charge

r = distance between point charge and electric field.

It can be seen that only E and r are the only variable here and also, E is inversely proportional to r²

Which implies that

E = k/r² , k = E × r²

E1 ×(r1)² = E2 × (r2)²

Let E1 = E, r1 =1, r2 = 2 and E2 =?

Let us substitute the parameters

E × 1 = E2 × 2²

E × 1 = E2 × 4

E = E2 × 4

E2 = E/4

Which implies that the electric field at the second distance (r =4) is one fourth of the initial electric field.

5 0

3 years ago