Answer:
The cooling time will not be reduced.
Explanation:
The time to cook is virtually the same in both types, vigorously and gently boiling water.
The reason cooking of spaghetti calls for vigorously boiling water is to keep the pasta agitated so that they do not stick to one another.
The temperature of boiling water is the same for both vigorously boiling water and gently boiling water, therefore there will be little time difference in when the potatoes will cook when it is done with vigorously boiling water than when it is cooked with gently boiling water.
However cooking potatoes in vigorously boiling water may cause the water to dry up on time and the potatoes get burnt.
Answer:
15193.62 m/s
Explanation:
t = Time taken = 6.5 hours
u = Initial velocity = 0 (Assumed)
m = Mass of rocket = 1380 kg
F = Thrust force = 896 N
v = Final velocity
a = Acceleration of the rocket
Force
Equation of motion
The velocity of the rocket after 6.5 hours of thrust is 15193.62 m/s
Answer:
(a) 0.017m/s^2
(b) 17/100,000
(c) 0.17m, 0.558ft
Explanation:
(a) speed = 60mph = 60m/1h × 1h/3600s = 0.017m/s, time = 10s
Acceleration (a) = speed ÷ time = 0.017m/s ÷ 10s = 0.0017m/s^2
(b) g = 9.8m/s^2, a = 0.0017m/s^2
a/g = 0.0017/9.8 = 0.00017 = 17/100,000
(c) Distance = speed × time = 0.017m/s × 10s = 0.17m
Distance in foot = 0.17 × 3.2808ft = 0.558ft
Answer:
H(max) = (v²/2g)
Explanation:
The maximum height the ball will climb will be when there is no friction at all on the surface of the hill.
Normally, the conservation of kinetic energy (specifically, the work-energy theorem) states that, the change in kinetic energy of a body between two points is equal to the work done in moving the body between the two points.
With no frictional force to do work, all of the initial kinetic emergy is used to climb to the maximum height.
ΔK.E = W
ΔK.E = (final kinetic energy) - (initial kinetic energy)
Final kinetic energy = 0 J, (since the body comes to rest at the height reached)
Initial kinetic energy = (1/2)(m)(v²)
Workdone in moving the body up to the height is done by gravity
W = - mgH
ΔK.E = W
0 - (1/2)(m)(v²) = - mgH
mgH = mv²/2
gH = v²/2
H = v²/2g.
