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2 years ago

If the frequency of wave is 400 Hz and its wavelength is 2.5 m, what is the velocity?

1 answer:

6 0

1000 m/s

You have the wavelength and frequency, you just need to solve for velocity. You can do this by multiplying each side of the equation by frequency.

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What is the weight of a 20 kg box on the Earth?<br> A) 2 N <br> B) 20 N<br> C) 200 N

jarptica [38.1K]

C 200 Newtons ........

6 0

3 years ago

g A rotating wheel requires 5.00 s to rotate 28.0 revolutions. Its angular velocity at the end of the 5.00-s interval is 96.0 ra

Setler [38]

Answer:

The angular acceleration of the wheel is 15.21 rad/s².

Explanation:

Given that,

Time = 5 sec

Final angular velocity = 96.0 rad/s

Angular displacement = 28.0 rev = 175.84 rad

Let $\alpha$ be the angular acceleration

We need to calculate the angular acceleration

Using equation of motion

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

Put the value in the equation

$175.84=\omega_{i}\times 5+\dfrac{1}{2}\times\alpha\times(5)^2$

$175.84=\omega_{i}\times 5+12.5\alpha$ ......(I)

Again using equation of motion

$\omega_{f}=\omega_{i}+\alpha t$

Put the value in the equation

$96.0=\omega_{i}+\alpha \times 5$

On multiply by 5 in both sides

$480=\omega_{i}\times 5+\alpha\times 25$ ....(II)

On subtract equation (I) from equation (II)

$480-175.84=\alpha(25-5)$

$304.16=\alpha\times20$

$\alpha=\dfrac{304.16}{20}$

$\alpha=15.21\ rad/s^2$

Hence, The angular acceleration of the wheel is 15.21 rad/s².

5 0

2 years ago

Christopher conducts an experiment in which he tests how much sugar dissolves at different temperatures of water. One step requi

sveticcg [70]

305.5 i just took the quiz

5 0

3 years ago

Read 2 more answers

A 1.0 kg mass is attached to the end of a vertical ideal spring with a force constant of 400 N/m. The mass is set in simple harm

Marina86 [1]

Answer:

$v(0)=2m/s$

Explanation:

The instantaneous velocity of a point mass that executes a simple harmonic movement is given by:

$v(t)=\omega *A*cos(\omega t + \phi)$

Where:

$\omega=Angular\hspace{3}frequency\\A=Amplitude\\\phi=Initial\hspace{3}phase$

Express the amplitude in meters:

$10cm*\frac{1m}{100cm} =0.1m$

The angular frequency can be found using the next equation:

$\omega=\sqrt{\frac{k}{m} }$

Using the data provided:

$\omega=\sqrt{\frac{400}{1} } =20$

At the equilibrium position:

$\phi=0$

$v(0)=20*(0.1)cos(20*0+0)=2*cos(0)=2*1=2m/s$

6 0

3 years ago

A jet transport has a weight of 2.25 x 106 N and is at rest on the runway. The two rear wheels are 16.0 m behind the front wheel

Rudik [331]

Answer:

Explanation:

Given that,

Weight of jet

W = 2.25 × 10^6 N

It is at rest on the run way.

Two rear wheels are 16m behind the front wheel

Center of gravity of plane 10.6m behind the front wheel

A. Normal force entered on the ground by front wheel.

Taking moment about the the about the real wheel.

Check attachment for better understanding

So,

Clock wise moment = anti-clockwise moment

W × 5.4 = N × 16

2.25 × 10^6 × 5.4 = 16•N

N = 2.25 × 10^6 × 5.4 / 16

N = 7.594 × 10^5 N

B. Normal force on each of the rear two wheels.

Using the second principle of equilibrium body.

Let the rear wheel normal be Nr and note, the are two real wheels, then, there will be two normal forces

ΣFy = 0

Nr + Nr + N — W = 0

2•Nr = W—N

2•Nr = 2.25 × 10^6 — 7.594 × 10^5

2•Nr = 1.491 × 10^6

Nr = 1.491 × 10^6 / 2

Nr = 7.453 × 10^5 N

6 0

3 years ago