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3 years ago

What is the period of a pendulum that takes 5 seconds to make a complete back and forth vibration

Physics

1 answer:

dezoksy [38]3 years ago

8 0

Answer:

Explanation:

-A period is the time it takes to make one full back-and-forth swing.

-A period is calculated as the time taken for a complete swing divided by one cycle:

$T=t/1c\\\\=5s/1\\\\=5s$

Hence, the period of the pendulum is 5s

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A water balloon is thrown at 20 m/s from the top of a 20 m high building, what is its speed when it hits the ground? Does the an

Oksana_A [137]

Answer:

The final velocity is 28.14 m/s

Yes the angle of projection matters

Explanation:

Given;

initial velocity of the water balloon, u = 20 m/s

height of the building, h = 20 m

let the final speed of the ball when it hits the ground = v

The final speed is calculated as follows;

v² = u² + 2gh

v² = (20)² + 2(9.8)(20)

v² = 400 + 392

v² = 792

v = √792

v = 28.14 m/s

Yes the angle matters, if the balloon had been dropped at a certain angle, the final velocity would have been estimated using the following formula;

$v_y^2 = u_y^2 sin^2(\theta) + 2gh_y$

where;

θ is the angle of projection, which accounts for the vertical component of the velocity.

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3 years ago

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Natasha_Volkova [10]

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3 years ago

Read 2 more answers

Consider the displacement vectors Ā=(i +6j)m, B = (3i– 7j)m,

andrezito [222]

Answer:

Dx = -0.5

Dy = -0.25

Explanation:

Two vectors are given in rectangular components form as follows:

A = i + 6j

B = 3i - 7j

It is also given that:

A - B - 4D = 0

so, we solve this to find D vector:

(i + 6j) - (3i - 7j) - 4D = 0

- 2i - j = 4D

D = - (2/4)i - (1/4)j

D = - (1/2)i - (1/4)j

D = - 0.5i - 0.25j

Therefore,

Dx = -0.5

Dy = -0.25

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3 years ago

A car travels around a level, circular track that is 750m across. What coefficient of friction is required to ensure the car can

Crank

The coefficient of friction must be 0.196

Explanation:

For a car moving on a circular track, the frictional force provides the centripetal force needed to keep the car in circular motion. Therefore, we can write:

$\mu mg = m\frac{v^2}{r}$

where the term on the left is the frictional force acting between the tires of the car and the road, while the term on the right is the centripetal force. The various terms are:

$\mu$ is the coefficient of friction between the tires and the road