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3 years ago

What property is a characteristic of a pure substance that can be observed with out changing the substance into something else

1 answer:

4 0

physical property is a characteristics of a pure substance that can be observed without changing the substance into someone else.

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You are standing on a train station platform as a train goes by close to you. As the train approaches, you hear the whistle soun

adoni [48]

Answer: v = (33320 - 340f2)/( f2 + 98)

Explanation: this question refers to Doppler effect, hence,

f1 = { V / (V - v)} x f2

f1 = 98Hz, f2 = 76 Hz, V = 340m/s, v = ?, fs = fsourcs

For f1,

98 = { 340 / (340 + v)} x fs ...(i)

for f2,

f2 = { 340 / (340 - v)} x fs ... (ii)

Divide (ii) by (i)

98/f2 = [{ 340 / (340 + v)} x fs]÷ [ 340 / (340 - v)} x fs]

98/f2 = {340 / (340 + v)} x fs x (340 - v)} / 340 x fs

98/f2 = (340 + v)/ (340 - v)

Cross multiplying

98(340 - v) = f2 (340 + v)

33320 - 98v = 340f2 + vf2

Collecting like terms

vf2 +98v = 33320 - 340f2

v(f2 +98) = 33320 - 340f2

v = (33320 - 340f2)/( f2 + 98)

5 0

4 years ago

Kelly walks 5 km to school. It takes her 1.5 hours. Calculate her speed in m/s

ratelena [41]

Kelly walks 5km.
1 km = 1000m.
5x1000 = 5000meters
1.5h = 90minutes
90miX60 = 5400s

10m/54s
And so on

7 0

2 years ago

A stone is dropped at t = 0. A second stone, with twice the mass of the first, is dropped from the same point at t = 181 ms. How

lara [203]

Answer:

the center of mass is 316,670m bellow the release point.

Explanation:

First, we must find the distance at which the objects are in time t = 360s

We will use the formula for vertical distance in free fall

$h=v_{0}t+\frac{1}{2} gt^2$

$v_{0}$ is the initial velocity, is 0 for both stones since they were just dropped.

g is acceleration of gravity and t is time ( $g=9.81m/s^2$ )

At $t=360s$ the first stone has been falling for the entire 360 seconds, its position h1 is:

$h_{1}=\frac{1}{2} (9.82m/s^2)(360s^2)=635,688m$

And at 360 seconds the second stone has been fallin for $t= 360s -181 s = 179s$ , so its position h2 is:

$h_{2}=\frac{1}{2} (9.81m/s^2)(179)^2=157,161.1m$

And finally using the equation for the center of mass:

$CM=\frac{m_{1}h_{1}+m_{2}h_{2}}{m_{1}+m_{2}}$

We know that the mass of the second stone is twice the mass of the first stone so:

$m_{1}=m\\m_{2}=2m$

replacing these values in the equation for the center of mass

$CM=\frac{mh_{1}+2mh_{2}}{m+2m}$

$CM=\frac{m(h_{1}+2h_{2})}{3m}=\frac{h_{1}+2h_{2}}{3}$

Finally, replacing the values we found fot h1 and h2:

$CM=\frac{635,688m+2(157,161.1m)}{3}=316,670m$

the center of mass is 316,670m bellow the release point.

8 0

4 years ago

Which of the following statements is correct about the force of gravity between two objects

Aneli [31]

Answer:

i dont know this but please answer before tomarrow please

Explanation:

thanks

4 0

3 years ago

A galloping pony speeds past you at 5 m/s. The frequency of the sound produced by the hooves on the dirt is 221 Hz. Assume the s

kow [346]

Given:
speed of passing pony 5 m/s
frequency of the sound produced: 221 Hz
speed of sound 342 m/s

Let us use the Doppler Shift Formula:
Where the source is moving away from the observer at rest

f' = (v / v+vs) f
Where, vs = Velocity of the Source,
v = Velocity of sound or light in medium,
f = Real frequency,
f' = Apparent frequency.

f'= [342 m/s / (342 m/s+5m/s)] * 221 Hz
f' = 0.9856 * 221Hz
f' = 217.8176 Hz or 218 Hz

The observed frequency of the hooves after the pony has passed your position is 218 Hz.

7 0

3 years ago

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