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2 years ago

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An electric buzzer is activated, then sealed inside a glass chamber. When all of the air is pumped out of the chamber, how is th

e sound from the buzzer affected?

1 answer:

Aleksandr-060686 [28]2 years ago

4 0

The sound is increased because sound waves are in fact mech. waves which means the that they can't travel through empty space and thus need a medium to travel through

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A moving 46.6 kg sled feels a 52.9 N friction force. what is the coefficient of friction

Setler [38]

Answer:

F=UR

52.9=U*46.6

U=52.9/46.6

U=1.135

4 0

2 years ago

Read 2 more answers

A car drives at a constant speed around a banked circular track with a diameter of 136 m . The motion of the car can be describe

galina1969 [7]

Answer:

speed = 44.9m/s

x = 35.5 m, y = 58.0m

Explanation:

A car on a circular track with constant angular velocity ω can be described by the equation of position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}$

The velocity v is given by:

$\overrightarrow {v(t)} = \overrightarrow{\frac{dr}{dt}}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}$

The acceleration a:

$\overrightarrow {a(t)} = \overrightarrow{\frac{dv}{dt}}= -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}$

From the given values we get two equations:

$-\omega^2 Rsin(\omega t)=-15.4\\-\omega^2 Rcos(\omega t)=-25.4$

We also know:

$\overrightarrow {a(t)} = -\omega^2 Rsin(\omega t)\hat{i} - \omega^2 Rcos(\omega t)\hat{j}=-\omega^2\overrightarrow{r(t)}$

The magnitude of the acceleration a is:

$a=\sqrt{(-15.4)^2+(-25.4)^2}=29.7$

The magnitude of position r is:

$r=R=68m$

Plugging in to the equation for a(t):

$\overrightarrow {a(t)} =-\omega^2\overrightarrow{r(t)}$

and solving for ω:

$|\omega|=0.66$

Now solve for time t:

$\frac{sin(0.66t)}{cos(0.66t)}=tan(0.66t)=\frac{15.4}{25.4}\\t=0.83$

Using the calculated values to compute v(t):

$\overrightarrow {v(t)}= \omega Rcos(\omega t)\hat{i} - \omega Rsin(\omega t)\hat{j}\\\overrightarrow {v(t)}=44.88cos(0.55)\hai{i}-44.88sin(0.55)\hat{j}\\\overrightarrow {v(t)}=38.3\hat{i}-23.5\hat{j}$

The speed of the car is:

$\sqrt{38.3^2 + (-23.5)^2} = 44.9$

The position r:

$\overrightarrow {r(t)} = Rsin(\omega t)\hat{i} + Rcos(\omega t)\hat{j}\\\overrightarrow {r(t)} = 68sin(0.55)\hat{i} + 68cos(0.55)\hat{j}\\\overrightarrow {r(t)} = 35.5{i} + 58.0\hat{j}$

5 0

1 year ago

Read 2 more answers

A 6-in-wide polyamide F-1 flat belt is used to connect a 2-in-diameter pulley to drive a larger pulley with an angular velocity

Likurg_2 [28]

Answer:

a) Fc = 4.15 N, Fi = 435.65 N, (F1)a = 640 N, and F2 = 239.6 N,

b) Ha = 1863.75 N, nfs = 1 , length = 11.8 mm

Explanation:

Given that:

γ= 9.5 kN/m³ = 9500N/m3

b = 6 inches = 0.1524 m

t = 0.0013 mm

d = 2 inches = 0.0508 m

n = 1750 rpm

$H_{nom}=2hp=1491.4W$

L = 9 ft = 2.7432 m

Ks = 1.25

g = 9.81 m/s²

a)

$w=\gamma b t = 9500* 0.1524*0.0013=1.88N/m$

$V=\frac{\pi d n}{60} =\pi *0.0508*1750/60=4.65 m/s$

$F_c=\frac{wV^2}{g}=1.88*4.65^2/9.81=4.15N$

$(F_1)_a=bF_aC_pC_v=0.1524*6000*0.7*1=640N$

$T=\frac{H_{nom}n_dK_s}{2\pi n}= \frac{1491*1.25*1}{2*\pi*1750/60}=10.17Nm$

$F_2=(F_1)_a-\frac{2T}{D}= 640-\frac{2*10.17}{0.0508} =239.6N$

$F_i=\frac{(F_1)_a+F_2}{2} -F_c=435.65N$

b)

$H_a=1491*1.25=1863.75W$

$n_f_s=\frac{H_a}{H_{nom}K_S }=1$

dip = $\frac{L^2w}{8F_i} =\frac{2.7432*1.88}{435.65}=11.8mm$

7 0

1 year ago

When you come within 100 ft of the uncontrolled railroad crossing and you cannot see the tracks for 400 ft in both directions, t

Sergio039 [100]

The answer to the blank above is 20. The speed limit given provided that you cannot see the tracks for 400 ft in both directions is always 20 mph. Basically, this 20 mph is decided based upon the traffic laws in order to avoid road casualties. Since it is "uncontrolled railroad crossing", the minimum speed should be implemented to slow down for the purpose of traffic calming measures. Other than this, the 20 mph is also applicable in narrowing roads as well as speed humps.

6 0

2 years ago

If the gas in a container absorbs 275 Joules of heat, has 125 Joules of work done on it, then does 50 Joules of work, what is th

cluponka [151]

Answer:

The increase in the internal energy = 350 J

Explanation:

Given that

Q= 275 J

W= - 125 J

W' = 50 J

W(net)= -125 + 50 = -75 J

Sign -

1.Heat rejected by system - negative

2.Heat gain by system - Positive

3.Work done by system = Positive

4.Work done on the system-Negative

Lets take change in the internal energy =ΔU

We know that

Q= ΔU + W(net)

275 = ΔU -75

ΔU= 275 + 75 J

ΔU=350 J

The increase in the internal energy = 350 J

7 0

2 years ago