Answer:a)45cos40 b)45sin40 c) 2.95 d)102
Explanation:
a) 
b) 
c) Since air resistance is negligible, the only force acting on the golf ball is gravity. Thus, its vertical acceleration is -<em>g</em>. We know the final velocity must be 0 m/s, because this will be when the golf ball reaches the maximum height and starts to change direction (it falls back to the ground). We also know initial velocity in the vertical direction (see part b). Thus, we can use this equation: 
.
d) The horizontal distance traveled is dependent on (1) how long the ball is in the air and (2) what the horizontal velocity is. (1) was found in part c, and (2) was found in part a.
(BELOW YOU CAN FIND ATTACHED THE IMAGE OF THE SITUATION)
Answer:
Explanation:
For this we're going to use conservation of mechanical energy because there are nor dissipative forces as friction. So, the change on mechanical energy (E) should be zero, that means:
(1)
With 
the initial kinetic energy, 
the initial potential energy, 
the final kinetic energy and 
the final potential energy. Note that initialy the masses are at rest so 
, when they are released the block 2 moves downward because m2>m1 and finally when the mass 2 reaches its maximum displacement the blocks will be instantly at rest so 
. So, equation (1) becomes:
(2)
At initial moment all the potential energy is gravitational because the spring is not stretched so 
and at final moment we have potential gravitational energy and potential elastic energy so 
, using this on (2)
(3)
Additional if we define the cero of potential gravitational energy as sketched on the figure below (See image attached), 
and we have by (3) :
(4)
Now when the block 1 moves a distance d upward the block 2 moves downward a distance d too (to maintain a constant length of the rope) and the spring stretches a distance d, so (4) is:
dividing both sides by d
, with k the constant of the spring and g the gravitational acceleration.
Answer:
Explanation:
We put the charges in the ascending order as follows
1.53 P
3.26 P
4.66 P
5.09 P
6.39 P
where P is equal to 10⁻¹⁹
we round off given charges as follows
1.53 P → 1.6 P
3.26 P → 3.2 P
4.66 P → 4.8 P
5.09 P → 4.8 P
6.39 P → 6.4 P
We see that 2 nd to 4 th charges are integral multiples of first charge . That means these charges are supposed to be made of combination of first charge . So first charge appears to be minimum possible charge .
Hence this charge may exist on single electron.
