Answer:
time to fall is 3.914 seconds
Explanation:
given data
half distance time = 1.50 s
to find out
find the total time of its fall
solution
we consider here s is total distance
so equation of motion for distance
s = ut + 0.5 × at² .........1
here s is distance and u is initial speed that is 0 and a is acceleration due to gravity = 9.8 and t is time
so for last 1.5 sec distance is 0.5 of its distance so equation will be
0.5 s = 0 + 0.5 × (9.8) × ( t - 1.5)² ........................1
and
velocity will be
v = u + at
so velocity v = 0+ 9.8(t-1.5) ..................2
so first we find time
0.5 × (9.8) × ( t - 1.5)² = 9.8(t-1.5) + 0.5 ( 9.8)
solve and we get t
t = 3.37 s
so time to fall is 3.914 seconds
Answer:
Approximately 
(given that the magnitude of this charge is 
.)
Explanation:
If a charge of magnitude 
is placed in an electric field of magnitude 
, the magnitude of the electrostatic force on that charge would be 
.
The magnitude of this charge is 
. Apply the unit conversion 
:
.
An electric field of magnitude 
would exert on this charge a force with a magnitude of:
.
Note that the electric charge in this question is negative. Hence, electrostatic force on this charge would be opposite in direction to the the electric field. Since the electric field points due south, the electrostatic force on this charge would point due north.
Vertical forces:
There is a force of 579N acting upward, and a force of 579N
acting downward.
The vertical forces are balanced ... they add up to zero ...
so there's no vertical acceleration.
Not up, not down.
Horizontal forces:
There is a force of 487N acting to the left, and a force of 632N
acting to the right.
The net horizontal force is
(487-left + 632-right) - (632-right - 487-right) = 145N to the right.
The net force on the car is all to the right.
The car accelerates to the right.
Answer:
True
Explanation:
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