Answer:
prokaryotic cells have no nucleus but contain DNA.
I hope this helps
Answer:
Explanation:
a = 4ms⁻², Vf = 180 m/s & Vi = 140m/s
a = 
4 = 
t = 40/4
t = 10sec
To Measure Distance Use third Equation of Motion:
2aS = Vf²-Vi²
S = 
S = 12800/8 = 1600m
Answer:
Torque = 1191.68 N-m
Explanation:
Given data
mass m = 76 kg
standingdistance r = 1.6 m
Solution
we get here torque that si express as
torque = force × distance ................1
torque = r × F sin(theta)
and we know that
F = mg .........2
and g = 9.8 m/s²
put here value in equation 1 we get
Torque = 76 × 1.6 × 9.8 × sin(90)
Torque = 1191.68 N-m
Answer:
heavy box with no heels on it
Answer:
a) The maximum height the ball will achieve above the launch point is 0.2 m.
b) The minimum velocity with which the ball must be launched is 4.43 m/s or 0.174 in/ms.
Explanation:
a)
For the height reached, we use 3rd equation of motion:
2gh = Vf² - Vo²
Here,
Vo = 3.75 m/s
Vf = 0m/s, since ball stops at the highest point
g = -9.8 m/s² (negative sign for upward motion)
h = maximum height reached by ball
therefore, eqn becomes:
2(-9.8m/s²)(h) = (0 m/s)² - (3.75 m/s²)²
<u>h = 0.2 m</u>
b)
To find out the initial speed to reach the hoop at height of 3.5 m, we again use 3rd eqn. of motion with h= 3.5 m - 2.5m = 1 m (taking launch point as reference), and Vo as unknown:
2(-9.8m/s²)(1 m) = (0 m/s)² - (Vo)²
(Vo)² = 19.6 m²/s²
Vo = √19.6 m²/s²
<u>Vo = 4.43 m/s</u>
Vo = (4.43 m/s)(1 s/1000 ms)(39.37 in/1 m)
<u>Vo = 0.174 in/ms</u>
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