As the pendulum swings through the origin, it has a speed of v0 = 1.1 m/s. If the mass on the end of the pendulum is 15 g, and t
he pendulum is 30 cm long, how high does the pendulum swing before coming back down? What is the angle corresponding to that height?
1 answer:
Answer:
1: 6.18 cm
2: 52.5609 degrees
Explanation:
We have the pendulum speed at the origin, and in that moment, all energy is kinetic, so we can calculate the pendulum energy by:
Ec = 0.5*m*v^2 = 0.5*0.015*1.1^2 = 0.0091 J
Now with that energy, we can calculate the height the pendulum will reach, as in that moment, the kinetic energy is totally converted to gravitational potencial energy:
Eg = m*g*h = 0.0091
0.015 * 9.81 * h = 0.0091
h = 0.0091 / (0.015 * 9.81 ) = 0.0618 m = 6.18 cm
Looking at the image attached, we can see that the pendulum will form a triangle, and one of the cathetus will be the length of the pendulum minus the height it went up, and the hypotenusa will be the pendulum length.
So, we know that the sine of the angle will be the division between the opposite cathetus and the hypotenusa:
sin(angle) = (30-6.18)/30 = 23.82/30 = 0.794 -> angle = 52.5609 degrees
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Answer:
The correct answer is B
Explanation:
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Φ = ∫ E. dA =
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∫ E dA = / ε₀
The area of a sphere is
A = 4π r²
E 4π r² = / ε₀
E = (1 /4πε₀ ) q / r²
Having the solution of the problem let's analyze the points:
A ) r = 3R / 4 = 0.75 R.
In this case there is no charge inside the Gaussian surface therefore the electric field is zero
E = 0
B) r = 5R / 4 = 1.25R
In this case the entire charge is inside the Gaussian surface, the field is
E = (1 /4πε₀ ) Q / (1.25R)²
E = (1 /4πε₀ ) Q / R2 1 / 1.56²
E₀ = (1 /4π ε₀ ) Q / R²
= Eo /1.56
²
= 0.41 Eo
C) r = 2R
All charge inside is inside the Gaussian surface
=(1 /4π ε₀
) Q 1/(2R)²
= (1 /4π ε₀
) q/R² 1/4
= Eo 1/4
= 0.25 Eo
D) False the field changes with distance
The correct answer is B