A pendulum is raised to a height of 0.3m above its lowest point and released. What is the velocity of the pendulum at its lowest
1 answer:
Answer:
v = 2,425 m / s
Explanation:
A simple pendulum has anergy stored at the highest point of the path and this energy is conserved throughout the movement.
highest point
Em₀ = U = m g y
lowest point
= K = ½ m v²
Em₀ = Em_{f}
mg y = ½ m v²
v = √ 2gy
let's calculate
v = √ (2 9.8 0.3)
v = 2,425 m / s
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Answer:
a. Fnet =37.67N
b. The direction = 133.4 from the x axis counter clockwise.
c. Option 2
Explanation:
Given that F1 is 53N at 116°, then it will be at a direction of 116-90=26° in the second quadrant.
Given that F2 is 57N at 116°, then it will be at a direction of 217-180=37° in the third quadrant..
Given that F1 is 71N at 20°, then it is in the first quadrant.
a. Fnet= F1+F2+F3
Fnet= -F1sin26i+F2cos26j-F2cos37i-F2sin37j+F3cos20i+F3sin20j
Fnet= 53sin26i+53cos26j-57cos37i-57sin37j+71cos20i+71sin20j
Resolving the vectors into x and y components.
Fnet= -2.04i+37.62j
Magnitude of the vector
Fnet= √((-2.04)^2+(37.62)^2)
Fnet= 37.67N
Fnet is approximately 38N.
b. Direction of the Fnet.
Angle=arctan(y/x)
Angle=arctan(-37.61/2.04)
Angle= -43.37°
The angle is in the negative x axis and positive y axis.
Then the direction becomes 180-43.37
Therefore, the direction of the net force is 133.37°.
c. The instantaneous velocity of a body is always in the direction of the net force at that instant. Option 2 is correct.
Fnet=ma
Fnet= mv/t
So the velocity is in the direction of the Fnet.
To solve this problem it is necessary to apply the principles of conservation of Energy in order to obtain the final work done.
The electric field in terms of the Force can be expressed as
Where,
F = Force
E= Electric Field
q = Charge
Puesto que el trabajo realizado es equivalente al cambio en la energía cinetica entonces tenemos que
KE = W
KE = F*d
In the First Case,
In Second Case,
The total energy change would be subject to,
Therefore the Kinetic Energy change of the charged object is 27.976J