How long does it take to go blind looking at the sun?

2 answers:

8 0

You could do it in just a couple of seconds if you really try. Find a place that's close to the equator, then do it on a day close to March 21 or September 21, at 12:00 noon, when the air is really clear. Keep your eyes wide open and don't squint. Good luck ! I know you can do it !

kondaur [170]3 years ago

7 0

Could be with a click of a finger depending how strong the light of the sun is

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The velocity time graph of a car shown below a) Calculate the magnitude of displacement of the car in 40 seconds. b) During whic

Gekata [30.6K]

Answer:

a) 0 metres

b) From time 0 s to 10 s , the car was accelerated. Its velocity accelerated from 0m/s to 20 m/s

c) 20 m/s

Explanation:

a) <em>Formula of displacement= velocity x time</em>

time=40 s

velocity =0 m/s

∴ displacement= 0 x 40 = 0 m

Magnitude of displacement is 0 m

b) The increase in velocity shows that there has been acceleration.

c) The average velocity of the car is = $\frac{0+40}{2\\}$ {initial velocity + final velocity}

= $\frac{40}{2}$

=20

Therefore, the magnitude of the average velocity of the car is 20 m/s

3 0

3 years ago

A skateboarder flies horizontally off a cement planter. After 3 seconds the skateboarder lands on the ground with a final veloci

evablogger [386]

Given the time, the final velocity and the acceleration, we can calculate the initial velocity using the kinematic equation A:

$v = v_o + a \Delta t$

A skateboarder flies horizontally off a cement planter. After a time of 3 seconds (Δt), he lands with a final velocity (v) of −4.5 m/s. Assuming the acceleration is -9.8 m/s² (a), we can calculate the initial velocity of the skateboarder (v₀) using the kinematic equation A.

$v = v_o + a \Delta t\\\\v_o = v - a \Delta t = (-4.5 m/s) - (-9.8 m/s^{2} ) \times 3 s = 24.9 m/s$