Answer:
Acceleration = 192.3 m/s² (Approx.)
Explanation:
Given:
Force = 125 N
Mass of ball = 0.65 kg
Find:
Acceleration
Computation:
We know that;
Acceleration = Force / Mas
So,
Acceleration = 125 / 0.65
Acceleration = 192.3 m/s² (Approx.)
d=vi*t+(1/2)gt²
d=11 m
g=9.8 m/s²
vi=0 m/s
11 m=0 m/s*t+(1/2)9.8 m/s²t²
11 m=4.9 m/s²t²
t²=11 m / 4.9 m/s²
t=√(11 m / 4.9 m/s²)=1.489... s≈1.5 s
Answer: the time the sone is in flight is 1.5 s
Answer:
10.23m/s^2
Explanation:
GIven data
mass of elevator = 2125 kg
Force= 21,750 N
Required
The maximum acceleration upward
F= ma
a= F/m
a=21,750/2125
a= 10.23m/s^2
Hence the acceleration is 10.23m/s^2
Answer:
W ≅ 292.97 J
Explanation:
1)What is the work done by tension before the block goes up the incline? (On the horizontal surface.)
Workdone by the tension before the block goes up the incline on the horizontal surface can be calculated using the expression;
W = (Fcosθ)d
Given that:
Tension of the force = 62 N
angle of incline θ = 34°
distance d =5.7 m.
Then;
W = 62 × cos(34) × 5.7
W = 353.4 cos(34)
W = 353.4 × 0.8290
W = 292.9686 J
W ≅ 292.97 J
Hence, the work done by tension before the block goes up the incline = 292.97 J
