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max2010maxim [7]

2 years ago

12

A 0.25 kg ideal harmonic oscillator has a total mechanical energy of 9.8 J. If the oscillation amplitude is 20.0 cm, what is the

oscillation frequency?a. 4.6 Hz b. 1.4 Hz c. 2.3 Hz d. 3.2 Hz

1 answer:

DanielleElmas [232]2 years ago

6 0

Answer:

7.04 Hz

Explanation:

x(t) = A cos(wt)

v(t) = - wA sin (wt)

Vmax = wA

E = 1/2 * m * (Vmax)^2

E = 1/2*m*(wA)^2

$w = \frac{1}{A} \sqrt{\frac{2E}{m} }$

f = w/ 2*pi

$f = \frac{1}{2*pi*A}*\sqrt{\frac{2E}{m} } \\f = \frac{1}{2 * pi * 0.2}*\sqrt{\frac{2(9.8)}{0.25} } \\f = 7.04 Hz$

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What is the gauge pressure of the water right at the point p, where the needle meets the wider chamber of the syringe? neglect t

Helen [10]

Missing details: figure of the problem is attached.

We can solve the exercise by using Poiseuille's law. It says that, for a fluid in laminar flow inside a closed pipe,

$\Delta P = \frac{8 \mu L Q}{\pi r^4}$

where:

$\Delta P$ is the pressure difference between the two ends

$\mu$ is viscosity of the fluid

L is the length of the pipe

$Q=Av$ is the volumetric flow rate, with $A=\pi r^2$ being the section of the tube and $v$ the velocity of the fluid

r is the radius of the pipe.

We can apply this law to the needle, and then calculating the pressure difference between point P and the end of the needle. For our problem, we have:

$\mu=0.001 Pa/s$ is the dynamic water viscosity at $20^{\circ}$

L=4.0 cm=0.04 m

$Q=Av=\pi r^2 v= \pi (1 \cdot 10^{-3}m)^2 \cdot 10 m/s =3.14 \cdot 10^{-5} m^3/s$

and r=1 mm=0.001 m

Using these data in the formula, we get:

$\Delta P = 3200 Pa$

However, this is the pressure difference between point P and the end of the needle. But the end of the needle is at atmosphere pressure, and therefore the gauge pressure (which has zero-reference against atmosphere pressure) at point P is exactly 3200 Pa.

8 0

1 year ago

Terminal velocity. A rider on a bike with the combined mass of 100kg attains a terminal speed of 15m/s on a 12% slope. Assuming

Firlakuza [10]

Answer:

0.9378

Explanation:

Weight (W) of the rider = 100 kg;

since 1 kg = 9.8067 N

100 kg will be = 980.67 N

W = 980.67 N

At the slope of 12%, the angle θ is calculated as:

$tan \ \theta = \dfrac{12}{100} \\ \\ tan \ \theta = 0.12 \\ \\ \theta = tan^{-1}(0.12) \\\\ \theta = 6.84^0$

The drag force D = Wsinθ

$\dfrac{1}{2}C_v \rho AV^2 = W sin \theta$

where;

$\rho = 1.23 \ kg/m^3$

A = 0.9 m²

V = 15 m/s

∴

Drag coefficient $C_D = \dfrac{2 *W*sin \theta}{\rho *A *V^2}$

$C_D =\dfrac{2 *980.67*sin 6.84}{1.23 *0.9 *15^2}$

$C_D =0.9378$

8 0

1 year ago

A community calendar allows nonprofit organizations to add events and their intended purposes for community members to see.

lbvjy [14]

Out of all given choices, activities and donor list can be found on the given kind of informative website.

Answer: Option A and B

<u>Explanation:</u>

Non-profit need to do the examination and for sake activities that simply aren't successful. And afterwards, they have to look to a portion of these auxiliary changes that to discover increasingly productive approaches to make a feasible money related model for their social change work.

When occasions are crucial, allowed to visit, and concentrated on developing as well as managing present or potential significant donors (people, establishments, corporate pioneers, they can bode well. Yet, only in the event that you catch up with participants on a one-on-one premise to additionally put them in the association and in the long run request that they contribute or reestablish their commitments.

On the off chance, you don't charge those to visit with the goal that can approach them for a greater, and progressively important blessing not far off.

8 0

2 years ago

 A bartender slides a beer mug at 1.50 m/s toward a customer at the end of a frictionless bar that is 1.20 m tall. The customer

Andrew [12]

Answer:

a) the mug hits the floor 0.7425m away from the end of the bar. b) |V|=5.08m/s θ= -72.82°

Explanation:

In order to solve this problem, we must first start by doing a drawing of the situation. (see attached picture).

a)

From the drawing we can see that we are dealing with a two dimensions movement problem. So in order to find out how far away from the bar the mug will fall, we need to start by finding how long it will take the mug to be in the air, so we analyze the vertical movement of the mug.

In order to find the time we need to use the following formula, which contains the data we know:

$y_{f}=y_{0}+v_{y0}t+\frac{1}{2}at^{2}$

we know that $y_{f}=0$ and that $v_{y0}=0$ as well, so the formula is simplified to:

$0=y_{0}+\frac{1}{2}at^{2}$

we can now solve this for t, so we get:

$-y_{0}=\frac{1}{2}at^{2}$

$-2y_{0}=at^{2}$

$\frac{-2y_{0}}{a}=t^{2}$

$t=\sqrt{\frac{-2y_{0}}{a}}$

we know that $y_{0}=1.20m$ and that $a=g=-9.8m/s^{2}$

the acceleration of gravity is negative because the mug is moving downwards. So we substitute them into the given formula:

$t=\sqrt{\frac{-2(1.20m)}{(-9.8m/s^{2})}}$

which yields:

t=0.495s

we can now use this to find the horizontal distance the mug travels. We know that:

$V_{x}=\frac{x}{t}$

so we can solve this for x, so we get:

$x=V_{x}t$

and we can now substitute the values we know:

x=(1.5m/s)(0.495s)

which yields:

x=0.7425m

b) Now that we know the time it takes the mug to hit the floor, we can use it to find the final velocity in the y-direction by using the following formula:

$a=\frac{v_{f}-v_{0}}{t}$

we know the initial velocity in the vertical direction is zero, so we can simplify the formula:

$a=\frac{v_{f}}{t}$

so we can solve this for the final velocity:

$V_{yf}=at$

in this case the acceleration is the same as the acceleration of gravity (which is negative) so we can substitute that and the time we found on the previous part to get:

$V_{yf}=(-9.8m/s^{2})(0.495s)$

which yields:

$V_{yf}=-4.851m/s$

so now we know the components of the final velocity, which are:

$V_{xf}=1.5m/s$ and $V_{yf]=-4.851m/s$

so now we can find the speed by determining the magnitude of the vector, like this:

$|V|=\sqrt{V_{x}^{2}+V_{y}^{2}}$

so we get:

$|V|=\sqrt{(1.5m/s)^{2}+(-4.851m/s)^{2}$

which yields:

|V|=5.08m/s

now, to find the direction of the impact, we can use the following equation:

$\theta = tan^{-1} (\frac{V_{y}}{V_{x}})$

so we get:

$\theta = tan^{-1} (\frac{-4.851m/s}{(1.5m/s)})$

which yields:

$\theta = -72.82^{o}$

4 0

2 years ago

A 0.25-m string, vibrating in its sixth harmonic, excites a 0.96-m pipe that is open at both ends into its second overtone reson

Andrews [41]

Answer:

option D

Explanation:

given,

length of the pipe, L = 0.96 m

Speed of sound,v = 345 m/s

Resonating frequency when both the end is open

$f = \dfrac{nv}{2L}$

n is the Harmonic number

2nd overtone = 3rd harmonic

so, here n = 3

now,

$f = \dfrac{3\times 345}{2\times 0.96}$

f = 540 Hz

The common resonant frequency of the string and the pipe is closest to 540 Hz.

the correct answer is option D

7 0

2 years ago