Answer: The expected population that wants flexible working hours in ranging from 58% to 62 %.
Explanation:
Percentage of sample respondents agreed for flexible working hours = 60%
The margin of the error = ± 2%
The expected population that wants flexible working hours in ranging from:
lower range
upper range
that is 58% to 62 %.
The expected population that wants flexible working hours in ranging from 58% to 62 %.
Would presume the energy as kinetic energy.
E = (1/2)*mv²
But m = 0.05kg, velocity here = 0.70c, where c is the speed of light ≈ 3* 10⁸ m/s
Ke = (1/2)*mv² = 0.5*0.05*(0.7*<span>3* 10⁸)</span>² = 1.1025 * 10¹⁵ Joules
There is no exact match from the options.
Note:
The height of a high bar from the floor is h = 2.8 m (or 9.1 ft).
It is not provided in the question, so the standard height is assumed.
g = 9.8 m/s², acceleration due to gravity.
Note that the velocity and distance are measured as positive upward.
Therefore the floor is at a height of h = -2.8 m.
First dismount:
u = 4.0 m/s, initial upward velocity.
Let v = the velocity when the gymnast hits the floor.
Then
v² = u² - 2gh
v² = 16 - 2*9.8*(-2.8) = 70.88
v = 8.42 m/s
Second dismount:
u = -3.0 m/s
v² = (-3.0)² - 2*9.8*(-2.8) = 63.88 m/s
v = 7.99 m/s
The difference in landing velocities is 8.42 - 7.99 = 0.43 m/s.
Answer:
First dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 8.42 m/s downward
Second dismount:
Acceleration = 9.8 m/s² downward
Landing velocity = 7.99 m/s downward
The landing velocities differ by 0.43 m/s.
Answer:
(a). The initial velocity is 28.58m/s
(b). The speed when touching the ground is 33.3m/s.
Explanation:
The equations governing the position of the projectile are
where 
is the initial velocity.
(a).
When the projectile hits the 50m mark, 
; therefore,
solving for 
we get:
Thus, the projectile must hit the 50m mark in 1.75s, and this condition demands from equation (1) that
which gives
(b).
The horizontal velocity remains unchanged just before the projectile touches the ground because gravity acts only along the vertical direction; therefore,
the vertical component of the velocity is
which gives a speed 
of
