Before the cells in a person's body can use the food that the person eats, the food must be A. chemically broken down into sugar
2 answers:
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Answer:
Explanation:
The change in potential energy can be expressed as:
where K is a constant with a value of , q1 and q2 are the charges of the proton and the electron and r is the distance between them.
The charge for the proton is and the charge for the electron is
.
Converting r=1.0nm to m:
Replacing values:
(a) 0.96 m/s
The period of the wave corresponds to the time taken for one complete oscillation of the boat, from the highest point to the highest point again. Since the time between the highest point and the lowest point is 2.5 s, the period is twice this time:
The frequency of the waves is the reciprocal of the period:
The wavelength instead is just the distance between two consecutive crests, so
And the wave speed is given by:
(b) 0.265 m
The total distance between the highest point of the wave and its lowest point is
d = 0.53 m
The amplitude is just the maximum displacement of the wave from the equilibrium position, so it is equal to half of this distance. So, the amplitude is
(c) Amplitude: 0.15 m, wave speed: same as before
In this case, the amplitude of the wave would be lower. In fact,
d = 0.30 m
So the amplitude would be
Instead, the wave speed would not change, since neither the frequency nor the wavelength of the wave have changed.
Answer:
For situation (a)
net charge E = E₊₂ + E₋₅ + E₋₃
E = K(q/d²)
where K = 8.99e9
d = 5.7cm = 5.7e-2m
Therefore,
E₊₂(x) = K(q/d²) = (8.99e9)× ((2.0e-6)÷(5.7e-2)) = 3.15e5(+x)
E₋₅(y) = K(q/d²) = (8.99e9)× ((5.0e-6)÷(5.7e-2)) = 7.88e5(+y)
E₋₃(x) = K(q/d²) = (8.99e9)× ((3.0e6)÷(5.7e-2)) = 4.73e5(+x)
thus
E = E₊₂ + E₋₅ + E₋₃
= 3.15e5(x) + 7.88e5(y) + 4.73e6(x)
= 7.88e6(x) + 7.88e6(y)
use Pythagorean theorem
I <em>E </em>I = = 1.242e6
∅ = =
= 45°
Thus for (a) net magnitude = 1.115e6 @ 45° above +x axis
for situation (b)
net charge E = E₊₄ + E₊₁ + E₋₁ + E₊₆
E₊₄(x) = K(q/d²) = (8.99e9)× ((4.0e-6)÷(5.7e-2)) = 6.30e5(+x)
E₊₁(y) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(-y)
E₋₁(x) = K(q/d²) = (8.99e9)× ((1.0e-6)÷(5.7e-2)) = 1.58e5(+x)
E₊₆(y) = K(q/d²) = (8.99e9)× ((6.0e-6)÷(5.7e-2)) = 9.46e5(+y)
thus,
E = E₊₄ + E₊₁ + E₋₁ + E₊₆
= 6.30e5(x) - 1.58e5(y) + 1.58e5(x) + 9.46e5(y)
= 7.88e5(x) + 7.88e5(y)
use Pythagorean theorem
I <em>E </em>I = = 1.242e6
∅ = =
= 45°
Thus for (a) and (b) the net magnitude = 1.242e6 @ 45° above +x axis
Explanation:
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