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padilas [110]
3 years ago
14

The asteroid, Ida, has a small moon, Dactyl, that orbits at a speed of 5.66 m/s in an orbit of radius 90, 000m . What is Ida's m

ass?

Physics
1 answer:
cricket20 [7]3 years ago
4 0

Answer:

4.3 x 10^16 kg

Explanation:

M = rv^2/G =[90,000 x 5.66^2] / [6.67 x 10^-11]

M = 43,226,446,776,611,694 = 4.3 x 10^16 kg - Ida's mass.

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Two coaxial conducting cylindrical shells have equal and opposite charges. The inner shell has charge +q and an outer radius a,
Leviafan [203]

Answer:

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

Explanation:

As we know that the charge per unit length of the long cylinder is given as

\lambda = \frac{q}{L}

here we know that the electric field between two cylinders is given by

E = \frac{2k\lambda}{r}

now we know that electric potential and electric field is related to each other as

\Delta V = - \int E.dr

\Delta V = -\int_a^b (\frac{2k\lambda}{r})dr

\Delta V = -2k \lambda ln(\frac{b}{a})

\Delta V = \frac{\lambda ln(\frac{b}{a})}{2\pi \epsilon_0}

\Delta V = \frac{q ln(\frac{b}{a})}{2\pi \epsilon_0 L}

7 0
3 years ago
Explain why theories may be changed or replaced overnight time.
vlabodo [156]
They may be changed because they may find evidence of some thing that will change their perspective on things.
5 0
3 years ago
Consider a metal single crystal oriented such that the normal to the slip plane and the slip direction are at angles of 60 and 3
Scilla [17]

Explanation:

Given that,

Angle by the normal to the slip α= 60°

Angle by the slip direction with the tensile axis β= 35°

Shear stress = 6.2 MPa

Applied stress = 12 MPa

We need to calculate the shear stress applied at the slip plane

Using formula of shear stress

\tau=\sigma\cos\alpha\cos\beta

Put the value into the formula

\tau=12\cos60\times\cos35

\tau=4.91\ MPa

Since, the shear stress applied at the slip plane is less than the critical resolved shear stress

So, The crystal will not yield.

Now, We need to calculate the applied stress necessary for the crystal to yield

Using formula of stress

\sigma=\dfrac{\tau_{c}}{\cos\alpha\cos\beta}

Put the value into the formula

\sigma=\dfrac{6.2}{\cos60\cos35}

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3 0
3 years ago
Whiat is quantum numbers describes the size and energy of an orbital?
Levart [38]

Answer:

The answer is the principal Quantum number (n)

Explanation:

The principal quantum number is one of the four quantum numbers associated with an atom.

It is denoted by a number n=1,2,3,4 etc

It tells both size (directly) and energy (indirectly) of an orbital.

When n=1 means it is the closest to the nucleus and is the smallest orbital and with increase in principal quantum number, it depicts that size of the orbital is increasing.

It tells the energy of the orbital as well as smaller number means less distance from nucleus and having less energy. Since electrons requires to absorb energy to jump into higher orbitals making n=2,3,4 etc. Thus electrons in the orbitals with higher n number indicates higher energy orbitals.

7 0
3 years ago
Please help I literally don’t understand
Veseljchak [2.6K]

Answer:

A= 2

B=3

C=4

D=5

E=7

F=8

H=12

7 0
3 years ago
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