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3 years ago

A squirrel runs from its nest in a tree

Physics

1 answer:

Debora [2.8K]3 years ago

3 0

I need more information to further answer this question.

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A tube with a cap on one end, but open at the other end, has a fundamental frequency of 130.8 Hz. The speed of sound is 343 m/s

sergey [27]

Answer:

Y = V / f where Y equals wavelength

4 Y1 = V / f1 for a closed pipe the wavelength is 1/4 the length of the pipe

2 Y2 = V / f2 for the open pipe the wavelength is 1/2 the length of the pipe

Y1 / Y2 = 2 = f2 / f1 dividing equations

f2 = 2 f1

the new fundamental frequency is 2 * 130.8 = 261.6

(The new wavelength is 1/2 the original wavelength so the frequency must double to produce the same speed.

8 0

3 years ago

What is the relationship between balancing equations and the law of conservation of matter

harkovskaia [24]

Every chemical equation adheres to the law of conservation of mass, which states that matter cannot be created or destroyed. Therefore, there must be the same number of atoms of each element on each side of a chemical equation.

7 0

2 years ago

Displacement vectors of 4km north, 2km south, 5km north, 5km south combine to a total displacement of

goldfiish [28.3K]

<u>Answer</u>

The combined displacement is 2km north

<u>Explanation</u>

Since displacement is a vector quantity, we take into account the direction.

Good for us all the displacement vectors are in the same dimension, so we can make north positive and south negative or vice-versa.

We now add to obtain,

$4+-2+5+-5$

This will simplify to

$=4-2+5-5=2$

Therefore the combined displacement is 2km north

5 0

3 years ago

Define couple and give 2 examples

Elodia [21]

Answer:

Two equal and opposite parallel forces not acting along the same line, form a couple. A couple is always needed to produce the rotation.

For example, turning a key in a lock and turning a steering wheel.

3 0

2 years ago

Read 2 more answers

A wheel initially spinning at wo = 50.0 rad/s comes to a halt in 20.0 seconds. Determine the constant angular acceleration and t

svetlana [45]

Answer:

(I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

Explanation:

Given that,

Initial spinning = 50.0 rad/s

Time = 20.0

Distance = 2.5 m

Mass of pole = 4 kg

Angle = 60°

We need to calculate the angular acceleration

Using formula of angular velocity

$\omega=-\alpha t$

$\alpha=-\dfrac{\omega}{t}$

$\alpha=-\dfrac{50.0}{20.0}$

$\alpha=-2.5\ rad/s^2$

The angular acceleration is -2.5 rad/s²

We need to calculate the number of revolution

Using angular equation of motion

$\theta=\omega_{0}t+\dfrac{1}{2}\alpha t$

Put the value into the formula

$\theta=50\times20-\dfrac{1}{2}\times2.5\times20^2$

$\theta=500\ rad$

The number of revolution is 500 rad.

(II). We need to calculate the torque

Using formula of torque

$\vec{\tau}=\vec{r}\times\vec{f}$

$\tau=r\times f\sin\theta$

Put the value into the formula

$\tau=2.5\times4\times 9.8\sin60$

$\tau=84.87\ N-m$

Hence, (I). The angular acceleration and number of revolution are -2.5 rad/s² and 500 rad.

(II). The torque is 84.87 N-m.

8 0

3 years ago