-- The vertical component of the ball's velocity is 14 sin(<span>51°) = 10.88 m/s
-- The acceleration of gravity is 9.8 m/s².
-- The ball rises for 10.88/9.8 seconds, then stops rising, and drops for the
same amount of time before it hits the ground.
-- Altogether, the ball is in the air for (2 x 10.88)/(9.8) = 2.22 seconds
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-- The horizontal component of the ball's velocity is 14 cos(</span><span>51°) = 8.81 m/s
-- At this speed, it covers a horizontal distance of (8.81) x (2.22) = <em><u>19.56 meters</u></em>
before it hits the ground.
As usual when we're discussing this stuff, we completely ignore air resistance.
</span>
Answer:
50% of it .
Explanation:
50% of it is illuminated by the Sun.
Answer:
the value of force, F=4.0N
Explanation:
Firstly, recall velocity-time equation
- v=u+at
- (4)=(2)+a(5)
- a=0.4m/s²
Secondly, recall the Newton's 2nd Law
- <em>F</em><em>=</em><em>ma</em>
- <em>F</em><em>=</em><em>(</em><em>1</em><em>0</em><em>)</em><em>(</em><em>0</em><em>.</em><em>4</em><em>)</em>
- <em>F</em><em>=</em><em>4</em><em>.</em><em>0</em><em>N</em>
Answer:
1.7 seconds
Explanation:
To clear the intersection, the total distance to be covered = 59.7 + 25 =84.7m
first we need to find the initial speed to just enter the intersection by using the third equation of motion
v^2 - u^2 = 2*a*s
45^2 - u^2 = 2 * -5.7 * 84.7
u^2 = 45^2 +965.58
u^2 = 2990.58
u = 54.7 m/s
Now for time we apply the first equation of motion
v-u =a * t
t = (v-u)/a = (45 - 54.7)/-5.7 = 1.7seconds