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zhannawk [14.2K]

3 years ago

8

what term describes the situation where the frequency of the wind match the frequency of the building? a. velocity b. resonance

C. amplification d. force

1 answer:

saul85 [17]3 years ago

3 0

The answer is B) resonance

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Assuming 70% of Earth's surface is covered in water at an average depth of 2.5 mi, estimate the mass of the water on Earth in Ki

saw5 [17]

This is an excellent question that i do not have the answer to.

7 0

2 years ago

How much heat is required to evaporate 0.15 kg of lead at 1750°C, the boiling point for lead? The heat of vaporization for lead

lara [203]

Answer:

Heat required = mass× latent heat Q = 0.15 × 871 ×

3 0

2 years ago

A speaker emits sound waves in all directions, and at a distance of 28 m from it the intensity level is 73 db. What is the total

Oduvanchick [21]

Answer:

P=0.197 Watt

Explanation:

Given That

B=73dB

Io= 1 * 10^-12

$I=Io * 10^{\frac{B}{10} } \\I=2 * 10^{-5} Wm^{-2} \\A=4 * pi * r^{2}\\ A=9852.03 m^{2}\\ Power= I * A\\P= 0.197 Watt$

6 0

2 years ago

A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 1

Bumek [7]

Answer:

202.8m

Explanation:

Given that A pirate fires his cannon parallel to the water but 3.5 m above the water. The cannonball leaves the cannon with a velocity of 120 m/s. He misses his target and the cannonball splashes into the briny deep.

First calculate the total time travelled by using the second equation of motion

h = Ut + 1/2gt^2

Let assume that u = 0

And h = 3.5

Substitute all the parameters into the formula

3.5 = 1/2 × 9.8 × t^2

3.5 = 4.9t^2

t^2 = 3.5/4.9

t^2 = 0.7

t = 0.845s

To know how far the cannonball travel, let's use the equation

S = UT + 1/2at^2

But acceleration a = 0

T = 2t

T = 1.69s

S = 120 × 1.69

S = 202.834 m

Therefore, the distance travelled by the cannon ball is approximately 202.8m.

4 0

2 years ago

(a) What is the minimum width of a single slit (in multiples of λ ) that will produce a first minimum for a wavelength λ ? (b) W

Kitty [74]

Answer:

The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

Explanation:

Given that,

Wavelength = λ

For D to be small,

We need to calculate the minimum width

Using formula of minimum width

$D\sin\theta=n\lambda$

$D=\dfrac{n\lambda}{\sin\theta}$

Where, D = width of slit

$\lambda$ = wavelength

Put the value into the formula

$D=\dfrac{n\lambda}{\sin\theta}$

Here, $\sin\theta$ should be maximum.

So. maximum value of $\sin\theta$ is 1

Put the value into the formula

$D=\dfrac{1\times\lambda}{1}$

$D=\lambda$

(b). If the minimum number is 50

Then, the width is

$D=\dfrac{50\times\lambda}{1}$

$D=50\lambda$

(c). If the minimum number is 1000

Then, the width is

$D=\dfrac{1000\times\lambda}{1}$

$D=1000\lambda$

Hence, The minimum value of width for first minima is λ

The minimum value of width for 50 minima is 50λ

The minimum value of width for 1000 minima is 1000λ

4 0

3 years ago