You haven't told us anything about the detectors being used. We don't know how the sensitivity of the detector is related to the total number of photons absorbed, and we don't even know whether you and your friend are both using the same type of detector.
All we can do, in desperation, is ASSUME that the minimum time required to just detect a star is inversely proportional to the total number of its photons that strike the detector. That is, assume . . .
(double the number of photons) ===> (detect the source in half the time) .
-- The intensity of light delivered to the prime focus of a telescope is directly proportional to the AREA of its objective lens or mirror, which in turn is proportional to the square of its radius or diameter.
So your telescope gathers (0.18/0.05)² = 12.96 times as much light as your friends telescope does.
-- So we'd expect your instrument to detect the same star in
(119.5 min) / (12.96) = <em>9.22 minutes .</em>
We're simply comparing the performance of two different telescopes as they observe the same object, so the star's magnitude doesn't matter.
Answer:
True
Explanation:
The statement is true because According to Schrödinger it's not possible to get the exact position of where an electron is located. However, he showed from an equation that we can at determine the most likely area where an electron will be and he said they are most likely to be in the most dense area of the clouds.
Answer:
Using the range formula R = v^2 sin 2 theta / g
or v^2 = R * g / sin 86.4
v^2 = 3.14 m * 9.81 m/s2 / .998
v^2 = 30.9 m^2 / s^2
v = 5.56 m/s
This hasn't really proved the question - this would give
vy = 5.56 * sin 43.2 = 3.81 m/s
vx = 5.56 * cos 43.2. = 4.05 m/s
t = 1.57 / 4.05 = .387 sec to reach the waterfall
h = 3.81 * .387 - 4.9 (.387)^2 = .74 m well above the height of the falls
There seems another way to do this
vy / vx = tan 43.2 vy = .939 vx
h = vy t - 1/2 g t^2 and t = 1.57 / vx
h = 1.57 tan 43.2 - 4.9 (1.57 / vx)^2
Solving for vx I get vx = 3.26 m/s vy = 3.06 m/s v = 4.47 m/s
In Newton's cannonball experiment, if the velocity is equal to the orbital velocity then the cannonball will stay in Orbit.
Newtons cannonball experiment stated that the distance that a cannonball will travel, before being drawn into the Earth by the forces of gravity, is dependent on the initial velocity.
Therefore, if the cannonball is launched at a velocity that matches the orbital velocity, then it will not be able to be drawn in by gravity due to the Earth moving away from the cannonball at the same speed at which the cannonball itself is falling.
This means that the cannonball will continue to fall without reaching the Earth, therefore staying in orbit, much like that of the moon or planets around the sun.
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