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Katen [24]
2 years ago
14

A pendulum consists of a small object hanging from the ceiling at the end of a string of negligible mass. The string has a lengt

h of 0.79 m. With the string hanging vertically, the object is given an initial velocity of 1.1 m/s parallel to the ground and swings upward in a circular arc. Eventually, the object comes to a momentary halt at a point where the string makes an angle θ with its initial vertical orientation and then swings back downward. Find the angle θ.
Physics
1 answer:
valkas [14]2 years ago
6 0

Answer:

θ = 21.7 °.

Explanation:

Let the length of the pendulum be L. . Initial kinetic energy of pendulum

is 1/2 mv² . Initial  potential energy is zero .

The distance by which  the bob of pendulum is raised

= L - Lcosθ

= L (1 - cos θ )

Increase in potential energy = mgL(1- cosθ )

According to conservation of mechanical energy

1/2 mv² = mgL ( 1 - cos θ)

v² = 2gL(1-cosθ)

Putting the values of different variables given

( 1.1 )² = 2 x 9.8 x .79 (1-cosθ)

(1-cosθ) = .071

cosθ = .929

θ = 21.7 °.

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2 years ago
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Answer:

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putting in M =0.420kg, g = 9.8m/s^2 and solving for F_t we get:

F_t = Mg\:tan(5^o)

F_t = (0.420kg)(9.8m/s^2)\:tan(5^o)

\boxed{F_t = 0.3601N.}

Now, this thrust produced is related to to the air ejection speed v by the relation

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where dM/dt is the rate of air ejection which we know is

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and since 1m^3 = 1.2kg,

0.03m^3/s \rightarrow 0.036kg/s

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putting this into equation (2) and the value of F_t we get:

0.3601N = 0.036v

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\boxed{v =10m/s.}

which is the speed of the air ejected.

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2 years ago
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Answer:

Explanation:

given that

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The electric field located inside the sphere is zero.

b. The electric field located 0.25m beneath the sphere.

Since the radius is 0.75m

Then, the total distance of the electric field from the centre of the circle is 0.75+0.25=1m

Then

E=kq/r2

K=9e9Nm2/C2

q=0.13e-9C

r=1m

Then,

E= 9e9×0.13e-9/1^2

E=1.17N/C. Q.E.D

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2 years ago
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Answer:

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Subject: physics

Level: High School

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