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anzhelika [568]
3 years ago
10

A small sphere with mass m is attached to a massless rod of length L that is pivoted at the top, forming a simple pendulum. The

pendulum is pulled to one side so that the rod is at an angle θ from the vertical, and released from rest.
a. Show the pendulum just after it is released. Draw vectors representing the forces acting on the small sphere and the acceleration of the sphere. Accuracy counts! At this point, what is the linear acceleration of the sphere?
b. Repeat part (a) for the instant when the pendulum rod is at an angle 9/2 from the vertical.
c. Repeat part (a) for the instant when the pendulum rod is vertical. At this point, what is the linear speed of the sphere?

Physics
1 answer:
USPshnik [31]3 years ago
4 0

Answer:

a) see attached, a = g sin θ

b)

c)   v = √(2gL (1-cos θ))

Explanation:

In the attached we can see the forces on the sphere, which are the attention of the bar that is perpendicular to the movement and the weight of the sphere that is vertical at all times. To solve this problem, a reference system is created with one axis parallel to the bar and the other perpendicular to the rod, the weight of decomposing in this reference system and the linear acceleration is given by

          Wₓ = m a

          W sin θ = m a

          a = g sin θ

b) The diagram is the same, the only thing that changes is the angle that is less

                θ' = 9/2  θ

             

c) At this point the weight and the force of the bar are in the same line of action, so that at linear acceleration it is zero, even when the pendulum has velocity v, so it follows its path.

The easiest way to find linear speed is to use conservation of energy

Highest point

            Em₀ = mg h = mg L (1-cos tea)

Lowest point

          Emf = K = ½ m v²

          Em₀ = Emf

          g L (1-cos θ) = v² / 2

              v = √(2gL (1-cos θ))

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andrey2020 [161]

Answer:

39.26 s

Explanation:

From the question given above, the following data were obtainedb

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Distance travelled (s) = 1700 m

Time (t) =?

Next, we shall determine the final velocity of the plane. This can be obtained as follow:

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Distance travelled (s) = 1700 m

Final velocity (v) =?

v² = u² + 2as

v² = 6² + (2 × 1.9 × 1700)

v² = 36 + 6460

v² = 6496

Take the square root of both side

v = √6496

v = 80.6 m/s

Finally, we shall determine the time taken before the plane lifts off. This can be obtained as follow:

Initial velocity (u) = 6 m/s

Acceleration (a) = 1.9 m/s²

Final velocity (v) = 80.6 m/s

Time (t) =?

v = u + at

80.6 = 6 + 1.9t

Collect like terms

80.6 – 6 = 1.9t

74.6 = 1.9t

Divide both side by 1.9

t = 74.6 / 1.9

t = 39.26 s

This, it will take 39.26 s before the plane lifts off.

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3 years ago
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melamori03 [73]

Examples of student-led organizations are:

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8 0
2 years ago
À stone is thrown
masya89 [10]
A = -9.8

v = -9.8t -8

s = -4.9 t2 -8t +25

So… -5t^2 -8t + 25 =0, we’ll rearrange to 5t^2 + 8t - 25. We get two roots, one is positive and is 1.59 seconds

V = -9.8(1.59) - 8 = -23.6

So… it takes 1.59 seconds to hit the ground at -23.6 m/s.
3 0
3 years ago
A 20 cm-long wire carrying a current of 6 A is immersed in a uniform magnetic field of 3 T. If the magnetic field is oriented at
SOVA2 [1]

Answer:

the  magnitude of the force that the wire will experience = 1.8 N

Explanation:

The force on a current carrying wire placed in a magnetic field is :

F = Idl × B

where:

I = current flowing through the wire

dl = length of the wire

B = magnetic field

We can equally say that :

|F| = IdlBsin \theta

where : sin θ is the angle at which the orientation from the magnetic field  to the wire occurs = 30°

Then;

|F| = B\ I \ L \ sin \theta

Given that:

L = 20 cm = 0.2 m

I = 6 A

B = 3 T

θ = 30°

Then:

F = 3 × 6 × 0.2 sin 30°

F = 1.8 N

Therefore, the  magnitude of the force that the wire will experience = 1.8 N

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