What is 2 plus 2 and 1 plus 1

Physics

2 answers:

Stels [109]3 years ago

8 0

Answer: 6

Explanation:

Have an awesome amazing day

stiks02 [169]3 years ago

4 0

Answer:

2 plus 2 = 4 AND 1 plus 1 = 2

Explanation:

I guess combine both equations if your detailed instructions allow.

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Satellite A is orbiting Earth at an altitude of 500 km and Satellite B is orbiting 800 km above the surface . How does the veloc

sp2606 [1]

The velocity of Satellite A is 2% greater than velocity of satellite B.

The given parameters;

<em>Altitude of Satellite A = 500 km</em>
<em>Altitude of Satellite B = 800 km</em>

The forces acting on the Satellites are given as follows;

$F_c = \frac{mv^2}{r} \\\\F_g = \frac{GMm}{r^2} \\\\\frac{v^2}{r} = \frac{GM}{r^2}\\\\v^2 = \frac{GM}{r} \\\\v^2 r = GM\\\\v_1^2 r _1 = v_2^2 r_2\\\\v_A^2r_A = v_B^2 r_B\\\\(\frac{v_A}{v_B} )^2 = \frac{r_B}{r_A} \\\\\frac{v_A}{v_B} = \sqrt{\frac{r_B}{r_A} } \\\\\frac{v_A}{v_B} = \sqrt{\frac{(800,000\ + \ 6.4 \times 10^6}{(500,000\ + \ 6.4 \times 10^6} )} \\\\\frac{v_A}{v_B} = 1.02 \\\\v_A = 1.02 \ v_B$

$v_A = v_B( 100\% \ + 2\%)\\\\v_A = 100\%v_B \ + \ \ 2\% v_B\\\\v_A = v_B \ \ + \ 2\% v_B$

Thus, the velocity of Satellite A is 2% greater than velocity of satellite B.

Learn more about velocity of satellite here: brainly.com/question/13981089

4 0

3 years ago

A person scuffing her feet on a wool rug on a dry day accumulates a net charge of -52 µc. how many excess electrons does this pe

LenKa [72]

Using q=1.60217662 E-19 C we get:
$\frac{52 \times 10^-^6}{1.60217662 \times 10^-^1^9}=3.24558 \times 10^1^4 electrons$
Then using 9.10938356E-31 kg as the mass, we get
$(3.24558 \times 10^1^4)(9.10938356\times 10^-^3^1)=2.95653\times 10^-^1^6kg$

6 0

3 years ago

Read 2 more answers

Concept map for kinetic energy, work and power

Helen [10]

Kinetic energy: the energy of motion

Work: the change in kinetic energy

Power: the rate of work done

Explanation:

The kinetic energy of an object is the energy possessed by the object due to its motion. Mathematically, it is given by:

$K=\frac{1}{2}mv^2$

where

m is the mass of the object

v is its speed

The work done an object is the amount of energy transferred; according to the energy-work theorem, it is equal to the change in kinetic energy of an object:

$W=K_f - K_i$