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gtnhenbr [62]
3 years ago
6

What’s gravitational pull of the earth

Physics
2 answers:
DENIUS [597]3 years ago
8 0

Explanation:

9.807 m/s squared is the gravitational pull of planet Earth.

OLga [1]3 years ago
3 0

Answer:

9.8 m/s^2

Explanation:

9.8 meters per second squared. A common misconception is that heavier objects fall faster, but gravity affects all things equally.

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What is the correct reason for blinking / flickering of stars? Explain it.
Rzqust [24]

Answer:

d. changing refractive index of gases in the atmosphere

Explanation:

light from a star passes through the atmosphere to reach our eyes. however, the layers of gases in the atmosphere are in constant motion, so the light gets bent and appears to be flickering.

i hope this helps! :D

6 0
3 years ago
When n liters of fuel were added to a tank that was already 1 3 full, the tank was filled to 7 9 of its capacity. In terms of n,
Zarrin [17]

Answer: 9/4n

Explanation:

Let's take T to represent tank

When n litres of fuel was added to tank that was 1/3 full = n + 1/3T

The tank was 7/9T

n+1/3T = 7/9T

n =7/9T - 1/3T

n = 4/9

If a full tank is taken to be 1 = 9/9

Hence, we will have 1 = x × n

1 = x × 4/9

x = 9/4

Hence 9/4 of n = 9/4n

8 0
3 years ago
A list of heat transfer examples are listed below. Which is NOT an example of convection?
Alex777 [14]

Answer:

C. Hot pavement burning your feet

Explanation:

Convection is a type of process where heat travels upwards and cold travels downwards.

7 0
3 years ago
A positron is an elementary particle identical to an electron except that its charge is . An electron and a positron can rotate
denis23 [38]

Explanation:

According to the energy conservation,

          F_{centripetal} = F_{electric}

            \frac{mv^{2}}{r} = \frac{kq^{2}}{d^{2}}

           v^{2} = \frac{kq^{2}r}{d^{2}m}

                 = \frac{9 \times 10^{9} N.m^{2}/C^{2} \times 1.6 \times 10^{-19} C \times 0.75 \times 10^{-9} m}{(1.50 \times 10^{-9}m)^{2} \times 9.11 \times 10^{-31} kg}

                = 8.430 \times 10^{10} m^{2}/s^{2}

             v = \sqrt{8.430 \times 10^{10} m^{2}/s^{2}}

                = 2.903 \times 10^{5} m/s

Formula for distance from the orbit is as follows.

               S = 2 \pi r

                  = 2 \times 3.14 \times 0.75 \times 10^{-9} m

                  = 4.71 \times 10^{-9} m

Now, relation between time and distance is as follows.

                T = \frac{S}{v}

       \frac{1}{f} = \frac{S}{v}

or,           f = \frac{v}{S}          

                = \frac{2.903 \times 10^{5} m/s}{4.71 \times 10^{-9} m}      

                = 6.164 \times 10^{13} Hz

Thus, we can conclude that the orbital frequency for an electron and a positron that is 1.50 apart is 6.164 \times 10^{13} Hz.

7 0
3 years ago
The vertical normal stress increase caused by a point load of 10 kN acting on the ground surface at a point 1m vertically below
Tomtit [17]

Answer:

(b) 4.775 kN

Explanation:

see the attached file

8 0
3 years ago
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