consider the motion in Y-direction
v₀ = initial velocity = 29 Sin62 = 25.6 m/s
a = acceleration = - 9.8 m/s²
t = time of travel
Y = vertical displacement = - 0.89 m
using the equation
Y = v₀ t + (0.5) a t²
- 0.89 = (25.6) t + (0.5) (- 9.8) t²
t = 5.3 sec
consider the motion along the horizontal direction :
v₀ = initial velocity = 29 Cos62 = 13.6 m/s
a = acceleration = 0 m/s²
t = time of travel = 5.3 sec
X = horizontal displacement =?
using the equation
X = v₀ t + (0.5) a t²
X = (13.6) (5.3) + (0.5) (0) t²
X = 72.1 m
d = distance traveled by the center fielder to catch the ball = 107 - x = 107 - 72.1 = 34.9 m
t = time taken = 5.3 sec
v = speed of center fielder
using the equation
v = d/t
v = 34.9/5.3
v = 6.6 m/s
Answer:
<u>We are given: </u>
initial velocity (u) = 0 m/s
final velocity (v) = 10 m/s
displacement (s) = 20 m
acceleration (a) = a m/s/s
<u>Solving for 'a'</u>
From the third equation of motion:
v² - u² = 2as
replacing the variables
(10)² - (0)² = 2(a)(20)
100 = 40a
a = 100 / 40
a = 2.5 m/s²
Answer:
1807.56 kJ
Explanation:
Parameters given:
Current, I = 8.9A
Time, t = 4.7hrs = 4.7 * 3600 = 16920 secs
Voltage, V = 12V
Electrical energy is given as:
E = I*V*t
Where I = Current
V = Voltage/Potential differenxe
t = time in seconds.
E = 8.9 * 12 * 16920
E = 1807056 J = 1807.056 kJ
Explanation:
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