Refer to the diagram shown below.
m = the mass of the object
x = the distance of the object from the equilibrium position at time t.
v = the velocity of the object at time t
a = the acceleration of the object at time t
A = the amplitude ( the maximum distance) of the mass from the equilibrium
position
The oscillatory motion of the object (without damping) is given by
x(t) = A sin(ωt)
where
ω = the circular frequency of the motion
T = the period of the motion so that ω = (2π)/T
The velocity and acceleration are respectively
v(t) = ωA cos(ωt)
a(t) = -ω²A sin(ωt)
In the equilibrium position,
x is zero;
v is maximum;
a is zero.
At the farthest distance (A) from the equilibrium position,
x is maximum;
v is zero;
a is zero.
In the graphs shown, it is assumed (for illustrative purposes) that
A = 1 and T = 1.
Answer:
50 N/m
Explanation:
Elastic energy = kinetic energy
EE = KE
½ kx² = ½ mv²
½ k (4 m)² = ½ (8.0 kg) (10.0 m/s)²
k = 50 N/m
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Answer: On December 20th, 1951 in Idaho, United States.
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The world's first experimental nuclear power plant was the Experimental Breeder Reactor Number One (EBR-I), which was built in a desert in Idaho, United States.
This reactor made history when, on December 20th, 1951, four 200-watt light bulbs were illuminated by means of atomic energy, specifically by nuclear fission reaction.
Answer:
1.082 mm
Explanation:
From the question, we can see that we were given The following
Wavelength of the atoms, λ = 502 nm = 502*10^-9 m
Radius of the screen away from the double slit, r = 1.1 m
We know that Y(20) = 10.2 mm = 10.2*10^-3 m
d = (20 * R * λ) / Y(20)
d = (20 * 1.1 * 502*10^-9)/10.2*10^-3
d = 1.1*10^-5 / 10.2*10^-3
d = 1.082 mm
Therefore, we can say that the distance of separation between the two slits is 1.082 mm
Answer:
(a). The average daily demand of this subdivision is 2444.44 gallon/min.
(b). The design-demand used to design the distribution system is 2444.44 gallon/min.
Explanation:
Given that,
Area = 1100 acres
Number of house in 1 acres = 4


Per house water demand = 800 g/day/house
(a). We need to calculate the average daily demand of this subdivision
Using formula for average daily demand





The average daily demand of this subdivision is 2444.44 gallon/min.
(b). We need to calculate the design-demand used to design the distribution system
Using formula for the design-demand



Hence, (a). The average daily demand of this subdivision is 2444.44 gallon/min.
(b). The design-demand used to design the distribution system is 2444.44 gallon/min.