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MatroZZZ [7]
2 years ago
8

Suppose Earth orbited a star whose mass was double the mass of the sun. If the radius of Earth’s orbit remained the same as it i

s now, then compared with the gravitational force between Earth and the sun, the gravitational force between Earth and the star would be—
half as much
the same
two times as much
four times as much
Physics
1 answer:
Vaselesa [24]2 years ago
7 0

Answer:

Two times as much

Explanation:

The equation for gravitational force is: Fg = GMm/r^2 with G being the universal gravitational constant.

So to make things easier we'll set r equal to 1 since it's a constant as well as G.

Then we're left with Fg=Mm with M being the mass of the sun and m being the mass of the earth.

So if m is constant and supposedly equals 1 then Fg=M so Fg is proportional to M therefore if M doubles then Fg doubles.

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The capacitor is now reconnected to the battery, and the plate separation is restored to d. A dielectric plate is slowly moved i
Paha777 [63]

Answer:

U2 = KAε0V2 / (2d)

Explanation:

The dielectric constant K just replaces the "3″ from Part B.

8 0
3 years ago
The core of a certain reflected reactor consist of a cylinder 10 ft high 10 ft in diameter The measured maximum-to-average flux
Westkost [7]

Answer:

The maximum power density in the reactor is 37.562 KW/L.

Explanation:

Given that,

Height = 10 ft = 3.048 m

Diameter = 10 ft = 3.048 m

Flux = 1.5

Power = 835 MW

We need to calculate the volume of cylinder

Using formula of volume

V =\pi r^2 h

Put the value into the formula

V=\pi\times(1.524)^2\times 3.048

V= 22.23\m^3

V = 22.23\times10^{3}\ Liter

We need to calculate the maximum power density in the reactor

Using formula of power density

P=\dfrac{E}{V}

Where, P = power density

E = energy

V = volume

Put the value into the formula

P=\dfrac{835\times10^{6}}{22.23\times10^{3}}

P=37561.85 = 37.562\times KW/L

Hence, The maximum power density in the reactor is 37.562 KW/L.

6 0
3 years ago
The pressure on a volume of liquid V = 1.0 mº at the surface is approximately equal to the atmospheric pressure Patm = 1.00 x 10
Alexus [3.1K]

Answer:-2.86*10⁻⁴

Explanation: Use the equation change in volume = (change in pressure * original volume) / Bulks Modulus. ΔV = (-Δp*V₀) / B

Plugging in your numbers, you should get ΔV = (-2.29*10⁷*1) / (8*10¹⁰) = -2.86*10⁻⁴

ΔP = P₂-P₁  ----> ΔP = 2.30*10⁷ - 1.00*10⁵ = 2.29*10⁷

3 0
2 years ago
An upright object 2.80 cm tall is placed 16.0 cm away from the vertex of a concave mirror with a center of curvature of 24.0 cm.
horrorfan [7]

Answer:

f = 12 cm

Explanation:

<u>Center of Curvature</u>:

The center of that hollow sphere, whose part is the spherical mirror, is known as the ‘Center of Curvature’ of  mirror.

<u>The Radius of Curvature</u>:

The radius of that hollow sphere, whose part is the spherical mirror, is known as the ‘Radius of Curvature’ of  mirror. It is the distance from pole to the center of curvature.

<u>Focal Length</u>:

The distance between principal focus and pole is called ‘Focal Length’. It is denoted by ‘F’.

The focal length of the spherical (concave) mirror is approximately equal to half of the radius of curvature:

f = \frac{R}{2}

where,

f = focal length = ?

R = Radius of curvature = 24 cm

Therefore,

f = \frac{24\ cm}{2}

<u>f = 12 cm</u>

8 0
2 years ago
A string, stretched between two fixed posts, forms standing-wave resonances at 325 Hz and 390 Hz. What is the largest possible v
Pavel [41]

Answer:

65

Explanation:

The resonant frequencies for a fixed string is given by the formula  nv/(2L).  

Where n is the multiple .

v is speed in m/s .

The difference between any two resonant frequencies is given by v/(2L)= fn+1 – fn

fundamental frequency means n=1

i.e  fn+1 – fn = 390 -325

                      =  65

3 0
3 years ago
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