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erastovalidia [21]
3 years ago
8

In deep space, sphere A of mass MA is located at the origin of an x axis and sphere B of mass MB is located on the axis at a poi

nt x1. Sphere B is released from rest while sphere A is held at the origin. (a) What is the gravitational potential energy of the two-sphere system just as B is released
Physics
1 answer:
Nookie1986 [14]3 years ago
6 0

Answer:

The gravitational potential energy of the two-sphere system just as B is released is

U = -[(G)(MA)(MB)/x₁]

where G = Gravitational constant

G = (6.7 × 10⁻¹¹) Nm²/kg²

Explanation:

The gravitational potential energy of two masses (m and M), separated by a distance, d, is given as

U = -(GMm/d)

For our question,

Mass of object 1 = MA

Mass of object 2 = MB

Distance between them = x₁

U = -[(G)(MA)(MB)/x₁]

where G = Gravitational constant

G = (6.7 × 10⁻¹¹) Nm²/kg²

Hope this Helps!!!

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Ksivusya [100]

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A spherical shell of radius 3.59 cm and a cylinder of radius 7.22 cm are rolling without slipping along the same floor. The two
vampirchik [111]

Answer:

(ω₁ / ω₂) = 1.9079

Explanation:

Given

R₁ = 3.59 cm

R₂ = 7.22 cm

m₁ = m₂ = m

K₁ = K₂

We know that

K₁ = Kt₁ + Kr₁ = 0.5*m₁*v₁²+0.5*I₁*ω₁²

if

v₁ = ω₁*R₁

and

I₁ = (2/3)*m₁*R₁² = (2/3)*m*R₁²

∴    K₁ = 0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁²   <em>(I)</em>

then

K₂ = Kt₂ + Kr₂ = 0.5*m₂*v₂²+0.5*I₂*ω₂²

if

v₂ = ω₂*R₂

and

I₂ = 0.5*m₂*R₂² = 0.5*m*R₂²

∴    K₂ = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²   <em>(II)</em>

<em>∵   </em>K₁ = K₂    

⇒   0.5*m*ω₁²*R₁²+0.5*(2/3)*m*R₁²*ω₁² = 0.5*m*ω₂²*R₂²+0.5*(0.5*m*R₂²)*ω₂²

⇒  ω₁²*R₁²+(2/3)*R₁²*ω₁² = ω₂²*R₂²+0.5*R₂²*ω₂²

⇒  (5/3)*ω₁²*R₁² = (3/2)*ω₂²*R₂²

⇒  (ω₁ / ω₂)² = (3/2)*R₂² / ((5/3)*R₁²)

⇒  (ω₁ / ω₂)² = (9/10)*(7.22/ 3.59)²

⇒  (ω₁ / ω₂) = (7.22/ 3.59)√(9/10)

⇒  (ω₁ / ω₂) = 1.9079

8 0
3 years ago
A 7750 kg space probe, moving nose-first toward Jupiter at 179 m/s relative to the Sun, fires its rocket engine, ejecting 72.0 k
Reika [66]

Answer:

179.47m/s

Explanation:

Using the law of conservation of momentum

m1u1 + m2u2 = (m1+m2)v

m1 and m2 are the masses

u1 and u2 are the initial velocities

v is the final velocity

Substitute

7750(179)+72(230) = (7750+72)v

1,387,250+16560 = 7822v

1,403,810 = 7822v

v = 1,403,810/7822

v= 179.47m/s

Hence the final velocity of the probe is 179.47m/s

7 0
3 years ago
Monochromatic light passes through a double slit, producing interference, the distance between the slit centres is 1.2 mm and th
Alik [6]

Answer:

The wavelength of the light is 7200\ \AA.

Explanation:

Given that,

Distance between the slit centers d= 1.2 mm

Distance between constructive fringes \beta= 0.3\ cm

Distance between fringe and screen D= 5 m

We need to calculate the wavelength

Using formula of width

\beta=\dfrac{D\lambda}{d}

Put the value into the formula

0.3\times10^{-2}=\dfrac{5\times\lambda}{1.2\times10^{-3}}

\lambda=\dfrac{0.3\times10^{-2}\times1.2\times10^{-3}}{5}

\lambda=7.2\times10^{-7}\ m

\lambda=7200\ \AA

Hence, The wavelength of the light is 7200\ \AA.

8 0
3 years ago
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