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Eva8 [605]
3 years ago
10

Two asteroids are found at the same distance from the Sun. However, one asteroid is twice as massive the other. According to New

ton's Universal Law of Gravitation the force of gravity exerted by the Sun and the more massive asteroid is

Physics
2 answers:
astraxan [27]3 years ago
4 0

Answer:

IS TWICE THAT OF THE GRAVITATIONAL FORCE BETWEEN THE SMALLER ASTEROID AND THE SUN

Explanation:

The equation for gravitational force is:

F = G *  \frac{m1 * m2}{r^2}

where G is the gravitational constant.

Given that distance remains constant, and the mass of the bigger asteroid is bigger, we can get the following relation:

G *  \frac{2*m1 * m2}{r^2} = G *  \frac{m1 * m2}{r^2}

Here we can see that multiplying the mass by 2 gives us 2 times the gravitational force for the bigger asteroid.

Thus, the gravitational force for the bigger asteroid and the sun is two times that of the smaller asteroid and the sun.

Sophie [7]3 years ago
4 0

Explanation:

Below is an attachment containing the solution.

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brainly plus

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Which of the following is a chemical property of a mineral?
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Hardness, color , and luster.

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A square coil of wire of side 3.95 cm is placed in a uniform magnetic field of magnitude 2.25 T directed into the page as in the
Len [333]

Answer:

Explanation:

In this case we shall calculate rate of change of flux in the coli to calculate induced emf .

Flux through the coil  = no of turns x area x magnetic field perpendicular to it

=34 x  2.25 x (3.95 )²x 10⁻⁴ Weber

= 1193.4  x 10⁻⁴Weber

Final flux through the coil after turn by 90°

= 1193.4 x 10⁻⁴ cos 90 ° =0

Change of flux

= 1193.4 x 10⁻⁴ weber.

Time taken = 0.335 s .

Average emf= Rate of change of flux

= change in flux / time

=1193.4 x 10⁻⁴ / .335

= 3562.4 x 10⁻⁴

356.24 x 10⁻³

=356.24 mV.

Current induced = emf induced / resistance

= 356.24/.780

= 456.71 mA.

8 0
3 years ago
During a compaction test in the lab a cylindrical mold with a diameter of 4in and a height of 4.58in was filled. The compacted s
Ray Of Light [21]

Answer:

part a : <em>The dry unit weight is 0.0616  </em>lb/in^3<em />

part b : <em>The void ratio is 0.77</em>

part c :  <em>Degree of Saturation is 0.43</em>

part d : <em>Additional water (in lb) needed to achieve 100% saturation in the soil sample is 0.72 lb</em>

Explanation:

Part a

Dry Unit Weight

The dry unit weight is given as

\gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}

Here

  • \gamma_d is the dry unit weight which is to be calculated
  • γ is the bulk unit weight given as

                                              \gamma =weight/Volume \\\gamma= 4 lb / \pi r^2 h\\\gamma= 4 lb / \pi (4/2)^2 \times 4.58\\\gamma= 4 lb / 57.55\\\gamma= 0.069 lb/in^3

  • w is the moisture content in percentage, given as 12%

Substituting values

                                              \gamma_{d}=\frac{\gamma}{1+\frac{w}{100}}\\\gamma_{d}=\frac{0.069}{1+\frac{12}{100}} \\\gamma_{d}=\frac{0.069}{1.12}\\\gamma_{d}=0.0616 lb/in^3

<em>The dry unit weight is 0.0616  </em>lb/in^3<em />

Part b

Void Ratio

The void ratio is given as

                                                e=\frac{G_s \gamma_w}{\gamma_d} -1

Here

  • e is the void ratio which is to be calculated
  • \gamma_d is the dry unit weight which is calculated in part a
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72

Substituting values

                                              e=\frac{G_s \gamma_w}{\gamma_d} -1\\e=\frac{2.72 \times 0.04}{0.0616} -1\\e=1.766 -1\\e=0.766

<em>The void ratio is 0.77</em>

Part c

Degree of Saturation

Degree of Saturation is given as

S=\frac{G w}{e}

Here

  • e is the void ratio which is calculated in part b
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

Substituting values

                                      S=\frac{G w}{e}\\S=\frac{2.72 \times .12}{0.766}\\S=0.4261

<em>Degree of Saturation is 0.43</em>

Part d

Additional Water needed

For this firstly the zero air unit weight with 100% Saturation is calculated and the value is further manipulated accordingly. Zero air unit weight is given as

\gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}

Here

  • \gamma_{zav} is  the zero air unit weight which is to be calculated
  • \gamma_w is the water unit weight which is 62.4 lb/ft^3 or 0.04 lb/in^3
  • G is the specific gravity which is given as 2.72
  • w is the moisture content in percentage, given as 12% or 0.12 in fraction

                                      \gamma_{zav}=\frac{\gamma_w}{w+\frac{1}{G}}\\\gamma_{zav}=\frac{0.04}{0.12+\frac{1}{2.72}}\\\gamma_{zav}=\frac{0.04}{0.4876}\\\gamma_{zav}=0.08202 lb/in^3\\

Now as the volume is known, the the overall weight is given as

weight=\gamma_{zav} \times V\\weight=0.08202 \times 57.55\\weight=4.72 lb

As weight of initial bulk is already given as 4 lb so additional water required is 0.72 lb.

4 0
3 years ago
The atomic mass number of copper is A=64. Assume that atoms in solid copper form a cubic crystal lattice. To envision this, imag
Stells [14]

Answer:

<em>0.228 nm</em>

<em></em>

Explanation:

Atomic mass number of copper = 64

but an atomic mass unit = 1.66 x 10^-27 kg

therefore, the mass of the copper atom m = 64 x 1.66 x 10^-27 kg = 1.06 x 10^-25 kg

The number of atoms in this mass n = ρ/m

where ρ is the density of copper = 8.96 x 10^3 kg/m^3

==> n = (8.96 x 10^3)/(1.06 x 10^-25) = 8.45 x 10^28 atoms/m^3

We know that the volume occupied by this amount of atoms n = a^{3}

where a is the lattice constant

equating, we have

8.45 x 10^28 = a^{3}

a = 4.389 x 10^9

we also know that

d =  1/a

where d is the smallest distance between the two copper atom.

d = 1/(4.389 x 10^9) = 2.28 x 10^-10 m

==> <em>0.228 nm</em>

5 0
3 years ago
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