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3 years ago

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Standing waves of frequency 68 hz are produced on a string that has mass per unit length 0.0160 kg/m. with what tension must the

string be stretched between two supports if adjacent nodes in the standing wave are to be 0.74 meters apart?

1 answer:

mario62 [17]3 years ago

3 0

0.54 ghost I think probably

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Assume the following values: d1 = 0.880 m , d2 = 1.11 m , d3 = 0.560 m , d4 = 2.08 m , F1 = 510 N , F2 = 306 N , F3 = 501 N , F4

dsp73

Answer:

= 2630.6 N.m

Explanation:

(FR)x = ΣFx = -F4 = -407 N

(FR)y = ΣFy =-F1-F2 -F3 = -510 - 306 - 501 = -1317 N

(MR)B =ΣM + Σ(±Fd)

= MA + F1(d1 +d2) + F2d2 - F4d3

= 1504 + 510(0.880+1.11) +306(1.11) - 407(0.560)

= 2630.64 N.m (counterclockwise)

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4 years ago

An isolated system consists of two masses. The first, m1, has a mass of 1.90 kg, and is initially traveling to the east with a s

Natali [406]

Answer:

m1v1=m2v2, v2=4.3m/s KE=(0.5)(2.94)(4.3)=6.2J

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3 years ago

A sample of an ideal gas has a volume of 2.37 L at 2.80×102 K and 1.15 atm. Calculate the pressure when the volume is 1.68 L and

iogann1982 [59]

Answer:

$p_2 = 1.76 atm$

Explanation:

given data:

v_1 = 2.37 L

v_2 = 1.68 L

p_1 =1.15 atm

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t_1 = 280 K

t_2 = 304 K

from Gas Law Equation

, WE HAVE

$\frac{p_1 v_1}{t_1} =\frac{p_2 v_2}{t_2}$

Putting the values

$\frac{1.15*2.37}{280} =\frac{p_2 *1.68}{304}$

$9.733*10^{-3} = \frac{p_2 *1.68}{304}$

$9.733*10^{-3}*304 = p_2*1.68$

$\frac{9.733*10^{-3}*304}{1.68} =p_2$

$p_2= 1.76 atm$

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4 years ago

A pulley is used to raise a heavy crate. With the pulley it only takes an effort force of 223 N to

RoseWind [281]

Answer:

1784/223 = 8:1

Explanation:

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3 years ago

Please help me Thanks!

Fiesta28 [93]

l will help and if you get it correctly I will say you are welcome

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3 years ago