4. Table 2.4 shows how the displacement of a runner changed
during a sprint race. Draw a displacement–time graph to show
this data, and use it to deduce the runner’s speed in the middle
of the race.
Table 2.4 Data for a sprinter during a race
Displacement
(m)
0 4 10 20 50 80 105
Time (s) 1 2 3 6 9 12
Answer:
Smaller refractive power
Explanation:
The refractive power of an eye is the extent to which it can converge or diverge the light rays.
Near point is the the closest point for an eye such that when an object is placed at that point the image it forms is sharp and clearly visible to the eye.
A the person ages, the ciliary muscles of the eyes weakens and as a result the lens contracts and the formation of the image takes place behind the retina instead of forming at the retina.
Thus the near point also increases and the refractive power becomes smaller.
consider the right direction as positive and left direction as negative.
M = mass of the ball = 5 kg
m = mass of stone = 1.50 kg
= initial velocity of the ball before collision = 0 m/s
= initial velocity of the stone before collision = 12 m/s
= final velocity of the ball after collision = ?
= final velocity of the stone after collision = - 8.50 m/s
using conservation of momentum
M + m = M + m
(5) (0) + (1.5) (12) = 5 + (1.50) (- 8.50)
= 6.15 m/s
h = height gained by the ball
using conservation of energy
Potential energy gained by ball at Top = kinetic energy at the bottom
Mgh = (0.5) M
(9.8) h = (0.5) (6.15)²
h = 1.93 m
Answer:
Where is the text?
Explanation:
If you refer to the short sentence you wrote as text, I believe the answer is probably the word "crashes" because it shows how the momentum was transferred.
Ok the velocity of an object in free fall is given by the equation :
v=v0-gt, where v0 is the original velocity, g is the gravitational constant (9.8 m/s^2) and t is the time.
so, we substitute values into this equation. v=35.8-9.8*2.5; v=11.3 m/s
