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Rufina [12.5K]
3 years ago
7

Plzz answer this question correctly

Physics
1 answer:
Leno4ka [110]3 years ago
6 0

Answer:

by reducing friction.....

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In what way are all sounds alike?
Anna35 [415]

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D. all sounds are caused by vibrations

Explanation:

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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
Read 2 more answers
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