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mixas84 [53]
2 years ago
8

A bob of mass m is suspended from a fixed point with a massless string of length L (i.e., it is a pendulum). You are to investig

ate the motion in which the string moves in a cone with half-angle θ. What tangential speed, v , must the bob have so that it moves in a horizontal circle with the string always making an angle θ from the vertical?Express your answer in terms of some or all of the variables m, L, and θ, as well as the acceleration due to gravity g.v = Lsin(θ) / sqrt(Lcos(θ) / g)v = Lsin(θ) / sqrt(Lcos(θ) / g)
Physics
1 answer:
Brrunno [24]2 years ago
7 0

Answer:

b

Explanation:

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1.An elevator is ascending with constant speed of 10 m/s. A boy in the elevator throws a ball upward at 20 m/ a from a height of
laiz [17]

(a) The maximum height reached by the ball from the ground level is 75.87m

(b) The time taken for the ball to return to the elevator floor is 2.21 s

<u>The given parameters include:</u>

  • constant velocity of the elevator, u₁ = 10 m/s
  • initial velocity of the ball, u₂ = 20 m/s
  • height of the boy above the elevator floor, h₁ = 2 m
  • height of the elevator above the ground, h₂ = 28 m

To calculate:

(a) the maximum height of the projectile

total initial velocity of the projectile = 10 m/s + 20 m/s  = 30 m/s (since the elevator is ascending at a constant speed)

at maximum height the final velocity of the projectile (ball), v = 0

Apply the following kinematic equation to determine the maximum height of the projectile.

v^2 = u^2 + 2(-g)h_3\\\\where;\\\\g \ is \ the \ acceleration \ due \ to\  gravity = 9.81 \ m/s^2\\\\h_3 \ is \ maximum \ height \ reached \ by \ the \ ball \ from \ the \ point \ of \ projection\\\\0 = u^2 -2gh_3\\\\2gh_3 = u^2 \\\\h_3 = \frac{u^2}{2g} \\\\h_3 = \frac{(30)^2}{2\times 9.81} \\\\h_3 = 45.87 \ m

The maximum height reached by the ball from the ground level (h) = height of the elevator from the ground level + height of he boy above the elevator + maximum height reached by elevator from the point of projection

h = h₁ + h₂ + h₃

h = 28 m + 2 m  +  45.87 m

h = 75.87 m

(b) The time taken for the ball to return to the elevator floor

Final height of the ball above the elevator floor = 2 m + 45.87 m = 47.87 m

Apply the following kinematic equation to determine the time to return to the elevator floor.

h = vt + \frac{1}{2} gt^2\\\\where;\\\\v \ is \ the \ initial \ velocity \ of \ the \ ball \ at \ the \ maximum \ height = 0\\\\h = \frac{1}{2} gt^2\\\\gt^2 = 2h\\\\t^2 = \frac{2h}{g} \\\\t = \sqrt{\frac{2h}{g}} \\\\t = \sqrt{\frac{2\times 47.87}{9.81}} \\\\t = 2.21 \ s

To learn more about projectile calculations please visit: brainly.com/question/14083704

6 0
2 years ago
A 25.0 g marble sliding to the right at 20.0 cm/s overtakes and collides elastically with a 10.0 g marble moving in the same dir
ikadub [295]
In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,

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The kinetic energy of the system before the collision is solved below.

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This value should also be equal to KE2, which can be calculated using the conditions after the collision.

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ikadub [295]

Answer:

no idea

Explanation:

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