Answer: 50π m ≈ 157 m
Explanation:
100 rev/min (2π rad/rev) / (60 sec/min) = 3⅓π rad/s
d = ωrt = 3⅓π(0.50)(30) = 50π m ≈ 157 m
Answer:
The punnet chart indicate the cross between a black cat and a spotted cat.
Explanation:
As the punnet chart was not given with the question, the question is searched and found online which is as attached herewith.
As seen from the punned table the column head has two capital H which indicate that the cat is a black cat.
White the row heads indicate a Capital and a small h thus this is a spotted cat
Thus the punnet chart indicate the cross between a black cat and a spotted cat.
To solve the problem it is necessary to take into account the concepts related to simple pendulum, i.e., a point mass that is suspended from a weightless string. Such a pendulum moves in a harmonic motion -the oscillations repeat regularly, and kineticenergy is transformed into potntial energy and vice versa.
In the given problem half of the period is equivalent to 1 second so the pendulum period is,

From the equations describing the period of a simple pendulum you have to

Where
g= gravity
L = Length
T = Period
Re-arrange to find L we have

Replacing the values,


In the case of the reduction of gravity because the pendulum is in another celestial body, as the moon for example would happen that,




In this way preserving the same length of the rope but decreasing the gravity the Period would increase considerably.
The answer is C nucleic acids
Answer:
696.83 m/s
Explanation:
m = mass of water = 565 g = 0.565 kg
c = specific heat of water = 4186 J/(kg⁰C)
ΔT = Change in temperature = T₂ - T₁ = 80 - 22 = 58 ⁰C
v = speed gained by water
Using conservation of energy
Kinetic energy gained by water = heat required to warm water
(0.5) m v² = m c ΔT
(0.5) v² = c ΔT
(0.5) v² = (4186) (58)
v = 696.83 m/s