Answer:
The pulling force that the man should apply to create an upward acceleration of 
is 
Explanation:
Hi
As it shows in the drawing at the end, we have that the total mass of the man plus the platform is 
, then the force downward 
is 
.
Due the man needs to do a pulling force upward capable of lifting himself and the platform with an acceleration of , this force should create an acceleration greater than gravity by . then 
. Therefore the force should be 
.
Finally, as we have a pulley arrangement connected to the platform, it gives the man a mechanical advantage, so he has to do only half of that upward force
, therefore 
.
Answer:
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.
Assumptions:
- The object is dropped in a free fall.
- There's no air resistance.
- The downward acceleration due to gravity is
Explanation:
Consider the "SUVAT" equation
,
where
is the final velocity,
is the initial velocity,
is the acceleration of the object, and
is the change in the object's position.
For example, if the bottle needs to achieve a speed of 
by the time it reaches the ground,
since the statement that the bottle is "dropped" implies a free fall.
.
Apply the previous equation to find the minimum height, :
.
Replace the value and apply the formula to find the minimum height required to reach different final speeds.
Answer: They’d fall at the same speed, because air resistance is the only thing that makes an object fall faster than another. There’s a video somewhere on the internet of a bowling ball and a feather falling at the same speed in a vacuum, if you look for it. Hope this helps!
Wave speed = frequency * wavelength
Wave speed = 686 * 2.00
Wave speed = 1,372 (m/s)
Answer: a) E= 6.63x10^-19J
E= 3.97×10^2KJ/mol
b) E = 3.31×10^-19J
E= 18.8×10^4 KJ/mol
C) E = 1.32×10^-33J
E= 8.01×10^-10KJ/mol
Explanation:
a) E = h ×f
h= planks constant= 6.626×10^-34
E=(6.626×10^-34)×(1.0×10^15)
E=6.63×10^-19J
1mole =6.02×10^23
E=( 6.63×10^-19)×(6.02×10^23)
E=3.97×10^2KJ/mol
b) E =(6.626×10^-34)/(1.0×10^15)
E=3.13×10^-19J
E= 3.13×10^-19) ×(6.02×10^23)
E= 18.8×10^3KJ/MOL
c) E= (6.626×10^-34) /0.5
E= 1.33×10^-33J
E= (1.33×10^-33) ×(6.02×10^23)
E= 8.01×10^-10KJ/mol
