Plzz answer this question correctly

In what way are all sounds alike?

Anna35 [415]

Answer:

D. all sounds are caused by vibrations

Explanation:

5 0

3 years ago

A car has a mass of 900 kg and a truck has a mass of 1800 kg. In which of the following situations would they have the same mome

Kaylis [27]

a) since force = mass * acceleration

f= 900 * 0 (because constant speed = 0 acceleration)

similarly b) f = 0

7 0

3 years ago

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A 40-cmcm-long tube has a 40-cmcm-long insert that can be pulled in and out. A vibrating tuning fork is held next to the tube. A

Licemer1 [7]

Answer:

1070 Hz

Explanation:

First, I should point out there might be a typo in the question or the question has inconsistent values. If the tube is 40 cm long, standing waves cannot be produced at 42.5 cm and 58.5 cm lengths. I assume the length is more than the value in the question then. Under this assumption, we proceed as below:

The insert in the tube creates a closed pipe with one end open and the other closed. For a closed pipe, the difference between successive resonances is a half wavelength $\frac{\lambda}{2}$ .

Hence, we have

$\dfrac{\lambda}{2}=58.5-42.5=16 \text{ cm}$

$\lambda=32\text{ cm}=0.32 \text{ m}$ .

The speed of a wave is the product of its wavelength and its frequency.

$v=f\lambda$

$f=\dfrac{v}{\lambda}$

$f=\dfrac{343}{0.32}=1070 \text{ Hz}$

7 0

3 years ago

Is an example of a physical process that a physical geographer would study.

diamong [38]

D. The environment
Is the right answer

8 0

3 years ago

Question 1 of 4 Attempt 4 The acceleration due to gravity, ???? , is constant at sea level on the Earth's surface. However, the

Evgen [1.6K]

Answer:

g(h) = g ( 1 - 2(h/R) )

<em>*At first order on h/R*</em>

Explanation:

Hi!

We can derive this expression for distances h small compared to the earth's radius R.

In order to do this, we must expand the newton's law of universal gravitation around r=R

Remember that this law is:

$F = G \frac{m_1m_2}{r^2}$

In the present case m1 will be the mass of the earth.

Additionally, if we remember Newton's second law for the mass m2 (with m2 constant):

$F = m_2a$

Therefore, we can see that

$a(r) = G \frac{m_1}{r^2}$

With a the acceleration due to the earth's mass.

Now, the taylor series is going to be (at first order in h/R):

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R}$

a(R) is actually the constant acceleration at sea level

and

$a(R) =G \frac{m_1}{R^2} \\ \frac{da(r)}{dr}_{r=R} = -2 G\frac{m_1}{R^3}$

Therefore:

$a(R+h) \approx G\frac{m_1}{R^2} -2G\frac{m_1}{R^2} \frac{h}{R} = g(1-2\frac{h}{R})$

Consider that the error in this expresion is quadratic in (h/R), and to consider quadratic correctiosn you must expand the taylor series to the next power:

$a(R+h) \approx a(R) + h \frac{da(r)}{dr}_{r=R} + \frac{h^2}{2!} \frac{d^2a(r)}{dr^2}_{r=R}$

6 0

4 years ago

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