Answer:
ω = 0.05 rad/s
Explanation:
We consider the centripetal force acting as the weight force on the surface of the cylinder. Therefore,

where,
ω = angular velocity of cylinder = ?
g = required acceleration = 9.8 m/s²
r = radius of cylinder = diameter/2 = 5.9 mi/2 = 2.95 mi = 4023.36 m
Therefore,

<u>ω = 0.05 rad/s</u>
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:
<h3><u>ELECTRIC POTENTIAL</u></h3>
• the amount of work needed to move a unit charge from a reference point to a specific point against an electric field.
Answer:
- 5436 J
Explanation:
mass of car, m = 120 kg
radius of loop, r = 12 m
velocity at the bottom (A) = Va = 25 m/s
Velocity at the top(B) = Vb = 8 m/s
Vertical distance from A to B = diameter of loop, h = 2 x 12 = 24 m
by use of Work energy theorem
Work done by all the forces = change in kinetic energy of the body
Work done by the force + Work done by the friction = Kinetic energy at B - kinetic energy at A
- m x g x h + Work done by friction = 0.5 x 120 x (Vb^2 - Va^2)
- 120 x 9.8 x 24 + Work done by friction = 60 x (64 - 625)
- 28224 + Work done by friction = - 33660
Work done by friction = -33660 + 28224 = - 5436 J
Answer:
= 0.331 J / g ° C
Explanation:
We have a calorimetry exercise where all the heat yielded by one of the components is absorbed by the other.
Heat ceded Qh = m1 ce1 (
-
)
Heat absorbed Qc = m2 ce2 (
- T₀)
Body 1 is metal and body 2 is water
. Where m are the masses of the two bodies, ce their specific heat and T the temperatures
Qh = Qc
m₁
(
-
) = m₂
(
- T₀)
we clear the specific heat of the metal
= m₂
(
- T₀) / (m₁ (
-
))
= 50.00 4.184 (20.15 -10.79) / (75.00 (99.0-20.15))
= 209.2 (9.36) / (75 78.85)
= 1958.11 / 5913.75
= 0.331 J / g ° C