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Tcecarenko [31]
3 years ago
10

The ejection seat has an acceleration of 8gees (8xgravity or ~80m/s/s). He has a mass of 70kg. The total force on him from the c

hair/rocket would be ?
(80m/s/s)(70kg)=5600N
(80m/s/s)(70kg)=5600N + Fg = 5600N+(70kg)(9.8N/kg)~5600N+700N=6300N
(80m/s/s)(70kg)=5600N - Fg = 5600N+(70kg)(9.8N/kg)~5600N-700N=4900N
I need the time

please explain need this ASAP
Physics
1 answer:
Georgia [21]3 years ago
5 0

I assume you're talking about a pilot. If the ejection seat has an acceleration of 8<em>g</em>, then it would exert a normal force of 8<em>g</em> (70 kg) ≈ 5600 N.

(This is assuming the pilot is flying horizontally at a constant speed, and the seat is ejected vertically upward.)

To reiterate, this is *only* the force exerted by the seat on the pilot. Contrast this with the <em>net</em> force on the pilot, which would be the normal force minus the pilot's weight, 5600 N - (70 kg)<em>g</em> ≈ 4900 N.

If instead the seat ejects the pilot directly downward, the force exerted by the seat would have the same magnitude of 5600 N, but its direction would be reversed to point downward, making it negative. But the <em>net</em> force would change to -5600 N - (70 kg)<em>g</em> ≈ -6300 N

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Which pair below describes isotopes of the same element? A) an atom with 6 protons and 8 neutrons - an atom with 8 protons and 6
iogann1982 [59]

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It BEGINS to turn into KE immediately, when Darwin lets go of the ball, and it starts to fall.

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Calculation:

When an observer changes their position, an apparent change in the object's position takes place. This change can be calculated using the angle ( or semi-angle) made by the observer and object i.e. the angle made between the two lines of observation from the object to the observer.

Thus from the relation of parallax of a celestial body we get:

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Converting the actual distance into light years we get:

23.81 parsecs = 23.81 × (3.26 light yrs) = 77.658 light-years

Therefore, the actual distance of Regulus from Earth is 23.81 parsecs which is 77.658 in light years.

Learn more about astronomical units here:

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