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larisa [96]
3 years ago
6

I will Give Brainliest To whoever actually answers A 500 kg satellite experiences a gravitational force of 3000 N, while moving

in a circular orbit around the earth. Determine the speed of the satellite. and include the correct number of sig figs
Physics
1 answer:
snow_lady [41]3 years ago
7 0

Answer:

9.7\times 10^{-4}\ rad/s

Explanation:

Given:

m=500 kg\\F=3000 N

Radius of earth , R=6371 \times 10^3\ m\\Angular speed =\omega\\We\  know\  that\ \\F= m\times \omega^{2} \times R\\\omega^{2}=\frac{F}{m*R} \\\\=\frac{3000}{500*6371 \times 10^3\ m}

=\frac{6}{6371 \times 10^3\ m}

=9.7\times 10^{-4}\ rad/s

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Which of the following statements best explains why a book resting on a table is in equilibrium?
timofeeve [1]
I'm pretty sure the answer is d.The weight of the book and the table's upward force on the book are equal in magnitude but opposite in direction.

3 0
3 years ago
Calculate the acceleration of gravity as a function of depth in the earth (assume it is a sphere). You may use an average densit
Ber [7]

Solution :

Acceleration due to gravity of the earth, g $=\frac{GM}{R^2}$

$g=\frac{G(4/3 \pi R^2 \rho)}{R^2}=G(4/3 \pi R \rho)$

Acceleration due to gravity at 1000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-1000) \times 5.5 \times 10^3\right)$

  $= 822486 \times 10^{-8}$

  $=0.822 \times 10^{-2} \ km/s$

 = 8.23 m/s

Acceleration due to gravity at 2000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-2000) \times 5.5 \times 10^3\right)$

  $= 673552 \times 10^{-8}$

  $=0.673 \times 10^{-2} \ km/s$

 = 6.73 m/s

Acceleration due to gravity at 3000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-3000) \times 5.5 \times 10^3\right)$

  $= 3371 \times 153.86 \times 10^{-8}$

  = 5.18 m/s

Acceleration due to gravity at 4000 km depths is :

$g=G\left(\frac{4}{3}\pi (R-d) \rho\right)$

$g=6.67 \times 10^{-11}\left(\frac{4}{3}\times 3.14 \times (6371-4000) \times 5.5 \times 10^3\right)$

  $= 153.84 \times 2371 \times 10^{-8}$

  $=0.364 \times 10^{-2} \ km/s$

 = 3.64 m/s

       

3 0
3 years ago
Reading the temperature of a solution by using a thermometer is an example of a(n) ________.
blsea [12.9K]

Answer:

B. Observation

Explanation:

Using a thermometer to read the temperature of a solution is tantamount to the making an observation.

Observation are recorded using our senses of sight, taste, earing, feeling etc or by the use of instrument.

  • Through observation, data is usually collected to make inferences about an experiment.
  • An observation leads to the formulation of a hypothesis which is scientific guess that leads to experimental designs.
  • Conclusions are drawn from the information of data obtained from an experiment.
4 0
2 years ago
Question 1 of 10
Novay_Z [31]

Answer:

C. 5.6 × 10^11 N/C

Explanation:

The electric field E at a distance R from a charge Q is given by

E = k\dfrac{Q}{R^2}

where k = 9*10^9Nm/C is the coulomb's constant.

Now, in our case

R = 0.0075m

Q = 0.0035C;

therefore,

E = (9*10^9)\dfrac{0.0035C}{(0.0075m)^2}

\boxed{E = 5.6*10^{11}N/C.}

which is choice C from the options given<em> (at least it resembles it).</em>

6 0
3 years ago
What is the total resistance in this circuit?
saul85 [17]

Answer:B

Explanation:

6 0
3 years ago
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