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krek1111 [17]
4 years ago
12

2. What makes improvisation'a foundation for interpretation

Physics
1 answer:
GuDViN [60]4 years ago
5 0
Two types of theatrical improvisation: spontaneous and structural. I hope this is right
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Calculate the speed of a satellite moving in a stable circular orbit about the Earth at a height of 4930 km .
svet-max [94.6K]

The speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

<h3>Speed of the satellite</h3>

v = √GM/r

where;

  • M is mass of Earth
  • G is universal gravitation constant
  • r is distance from center of Earth = Radius of earth + 4930 km

v = √[(6.626 x 10⁻¹¹ x 5.97 x 10²⁴) / ((6371 + 4930) x 10³)]

v = 5,916.36 m/s

Thus, the speed of the satellite moving in a stable circular orbit about the Earth is 5,916.36 m/s.

Learn more about speed here: brainly.com/question/6504879

#SPJ1

5 0
2 years ago
If a reaction starts with 30 grams how many should it end with?
GrogVix [38]
30 grams because of conservation
6 0
3 years ago
Read 2 more answers
A thundercloud has an electric charge of 48.8 C near the top of the cloud and –41.7 C near the bottom of the cloud. The magnitud
IceJOKER [234]

Answer: 1.51 km

Explanation:

<u>Coulomb's Law:</u> The electrostatic force between two charge particles Q: and Q2 is directly proportional to product of magnitude of charges and inversely proportional to square of separation distance between them.

Or,   \vec{F}=k \frac{Q_{1} Q_{2}}{r^{2}}

Where Q1 and Q2 are magnitude of two charges and r is distance between them:

<u>Given:</u>

Q1 = Charge near top of cloud = 48.8 C

Q2 = Charge near the bottom of cloud = -41.7 C

Force between charge at top and bottom of cloud (i.e. between Q: and Q2) (F) = 7.98 x 10^6N

k = 8.99 x 109Nm^2/C^2

<u>So,</u>

\begin{aligned}&7.98 \times 10^{6}=\left(8.99 \times 10^{9} \mathrm{Nm}^{2} / \mathrm{C}^{2}\right) \frac{48.8 \mathrm{C} \times 41.7 \mathrm{C}}{\mathrm{r}^{2}} \\&r=\sqrt{\frac{1.8294 \times 10^{13}}{7.98 \times 10^{6}}}=1.514  \times 10^{3} \mathrm{~m}=1.51 \mathrm{~km}\end{aligned}

Therefore, the separation between the two charges (r) = 1.51 km

3 0
2 years ago
A disk of mass M and radius R rotates at angular velocity ω0. Another disk of mass M and radius r is dropped on top of the rotat
AleksandrR [38]

Answer:

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

Explanation:

As we know that there is no external torque on the system of two disc

then the angular momentum of the system will remains conserved

So we will have

L_i = L_f

now we have

L_i = (\frac{1}{2}MR^2)\omega_o

also we have

L_f = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now from above equation we have

(\frac{1}{2}MR^2)\omega_o  = (\frac{1}{2}MR^2 + \frac{1}{2}Mr^2)\omega

now we have

\omega = \frac{MR^2\omega_o}{(MR^2 + Mr^2)}

\omega = \frac{(R^2\omega_o}{(R^2 + r^2)}

6 0
3 years ago
a gold smith claims that an ornament is made up of pure gold of density 19.32g c/m2 . A chain made from gold weighs 25.25g . the
Lina20 [59]

Answer:

lllllllllloooooollll

3 0
3 years ago
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