Worn tyres will make the braking distance longer. True or false. Why?

Standing waves can be established within a cylindrical volume. However, it is not known if the volume is closed at both ends, op

Evgen [1.6K]

The frequencies are missing in the question. The three successive resonance frequencies within the volume are $f_1, f_2 \text{ and}\ f_3$ .

Solution :

Let the volume be : v

The frequency for one end open and one end closed is given by :

So, $f_n = \frac{(2n-1)v}{4L}$

Therefore,

$f_1 = \frac{v}{4L}$ , $f_2 = \frac{3v}{4L}$ , $f_3 = \frac{5v}{4L}$

So, $f_2 - 3f_1$ and $f_3=\frac{5}{3}f_2$

Therefore, the ratio of $\frac{f_3}{f_2}=\frac{5}{3}$ which is not a whole number.

Now the frequency for open volume and closed at both end

$f_n=\frac{nv}{4L}$

So, $f_1=\frac{v}{4L}$ , $f_2 = 2f_1$ , $f_2=3f_1$

From above formulae we can see that ratio of is not a whole number that is a identification for the frequency of the volume at one end open and one end closed.

Also ratio of the consecutive frequency of the volume at open from both side and closed from both side is always a whole number.

4 0

3 years ago

A student presses a book between his hands, as the drawing indicates. The forces that he exerts on the front and back covers of

Aleksandr-060686 [28]

Answer: 39.5 N

Explanation:

2 x force x static friction coefficient = weight (weight=2μsF)

force = ?

static friction coefficient = 0.416

weight = 32.9N

substituting the values above into the equation we get

2 x F x 0.416 = 32.9

F = 32.9 / (2 x 0.416)

F = 39.5N

4 0

3 years ago

Sand falls from a conveyor belt at a rate of 30 m3m3/min onto the top of a conical pile. The height of the pile is always 3535 o

Reil [10]

Explanation:

As the given data is as follows.

h = $\frac{3}{5}d$

= $\frac{3}{5} \times (2r)$

Also, we know that r = $\frac{4}{3}h$

and Volume (V) = $\frac{1}{3} \pir^{2}h$

= $\frac{1}{3} \pi (\frac{4}{3}h)^{2} h$

= $\frac{16}{27} \pi h^{3}$

And, $\frac{dV}{dt} = \frac{3 \times 16}{27} \pi h^{2} \frac{dh}{dt}$

$\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}$

Putting the given values into the above formula as follows.

$\frac{dV}{dt} = \frac{16}{9} \pi h^{2} \frac{dh}{dt}$

$30 m^{3}/min = \frac{16}{9} \pi (2)^{2} \frac{dh}{dt}$

[tex]\frac{dh}{dt} = 1.343 m/min

or, = 134.3 cm/min (as 1 m = 100 cm)

thus, we can conclude that the height changing at 134.3 cm/min when the pile is 2 m high.

4 0

3 years ago

One of the dangers of tornados and hurricanes is the rapid drop in air pressure that is associated with such storms. Assume that

Ierofanga [76]

Answer:

Force = 22508.86 N

Explanation:

Given:

Area of the window, A = 2.02 m²

Air pressure inside the house, P₁ = 1.02 atm

External air pressure, P₂ = 0.910 atm

Now, The difference in the pressure, ΔP = P₁ - P₂ = 1.02 - 0.910 = 0.11 atm

or

ΔP = 0.11 atm × (101300 Pa / 1 atm )

ΔP = 11143 Pascals

Now,

the Force is given as:

Force = Pressure × area

on substituting the values, we get

Force = 11143 Pa × 2.02 m²

or

Force = 22508.86 N

4 0

3 years ago

According to the Second Law of Thermodynamics, a. any process during which the entropy of the universe increases will be product

Iteru [2.4K]

Answer:

The correct option is;

a. Any process in which the entropy of the universe increases will be product-favored

Explanation:

According to the second law of thermodynamics, the change in entropy of a closed system with time is always positive. That is the entropy of the entire universe, considered as an isolated system, always increases with time, hence the entropy change in the universe will always be positive.

$\Delta S_{universe} = \Delta S_{system} + \Delta S_{surroundings} >0$

Therefore, any process in which the entropy of the universe increases will be product favored.

4 0

3 years ago