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3 years ago

What did scientists predict the big bang should have left?

Physics

2 answers:

snow_tiger [21]3 years ago

7 0

I believe the answer is 'A'

VARVARA [1.3K]3 years ago

4 0

the answer is A

hope this helps b!

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A proposed space elevator would consist of a cable stretching from the earth's surface to a satellite, orbiting far in space, th

NISA [10]

To solve this problem we will apply the concepts related to energy conservation. Here we will use the conservation between the potential gravitational energy and the kinetic energy to determine the velocity of this escape. The gravitational potential energy can be expressed as,

$PE= \frac{GMm}{d}$

The kinetic energy can be written as,

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

Where,

$G = 6.67*10^{-11}m^3/kg\cdot s^2$ Gravitational Universal Constant

$m = 5.972*10^{24}kg$ Mass of Earth

$h = 56*10^6m$ Height

$r = 6.378*10^6m$ Radius of Earth

From the conservation of energy:

$\frac{1}{2} mv^2 = \frac{GMm}{d}$

Rearranging to find the velocity,

$v = \sqrt{\frac{2Gm}{d}} \rightarrow$ Escape velocity at a certain height from the earth

If the height of the satellite from the earth is h, then the total distance would be the radius of the earth and the eight,

$d = r+h$

$v = \sqrt{\frac{2Gm}{r+h}}$

Replacing the values we have that

$v = \frac{2(6.67*10^{-11})(5.972*10^{24})}{6.378*10^6+56*10^6}$

$v = 3.6km/s$

Therefore the escape velocity is 3.6km/s

3 0

3 years ago

Two balloons are charged with an identical quantity and type of charge: -4 nc They are held apart at a separation distance of 70

blagie [28]

Answer:

29.4 uN

Explanation:

The electric force between two charges can be calculated using Coulomb's Law. According to this law the force between two point charges is given as:

$F=k\frac{q_{1} q_{2} }{r^{2}}$

where k is a proportionality constant known as the Coulomb's law constant. Its value is $9 \times 10^{9}$ Nm²/C²

r = distance between charges = 70 cm = 0.7 m

q1 = q2 = 4nC = $4 \times 10^{-9}$ C

The negative sign indicates that the charges are negative. In the formula we will only use the magnitude of the charges.

Using these values in the formula, we get:

$F=9 \times 10^{9} \times \frac{4 \times 10^{-9} \times 4 \times 10^{-9}}{0.7^{2}}\\\\ F=2.94\times10^{-7} N\\\\ F=29.4 \times 10^{-6} N\\\\ F=29.4 \mu N$