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2 years ago

A form of oxygen with three atoms of

Physics

1 answer:

lesya692 [45]2 years ago

5 0

Answer:

ozone, letter b

ozone is basically an allotrope of oxygen having three atoms instead of two.

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A parked car's horn (belonging to a musician) emits a concert A of frequency 440 on a day when the speed of sound is 342 m/s. Yo

julsineya [31]

Answer:

19.08 m/s

Explanation:

f = actual frequency emitted by the parked car's horn = 440 Hz

V = speed of sound = 342 m/s

f' = frequency of the horn observed by you = 466 Hz

v = speed of your car moving towards the parked car = ?

frequency of the horn observed by you is given as

$f' = \frac{Vf}{V - v}$

$466 = \frac{(342)(440)}{342 - v}$

v = 19.08 m/s

3 0

2 years ago

A student examines a 20-meter long rectangular stream channel and takes the following measurements: width of stream = 4 meters,

Scrat [10]

Answer:

The discharge of the stream at this location is 40 cubic meters per second.

Explanation:

The discharge is the volume flow rate of the water in the stream. For this purpose we can use the following formula:

Discharge = Volume Flow Rate = (Cross-Sectional Area)(Velocity of Stream)

Volume Flow Rate = (Width of Stream)(Depth of Stream)(Velocity of Stream)

Volume Flow Rate = (4 meters)(2 meters)(5 meters per second)

<u>Volume Flow Rate = 40 cubic meters per second</u>

Therefore, the discharge of the stream at this location is found to be <u>40 cubic meters per second</u>

This result shows that 40 cubic meters volume of water passes or discharges through this point in a time of one second. Hence, this is called the volume flow rate or the discharge of the stream.

3 0

2 years ago

A round pipe of varying diameter carries petroleum from a wellhead to a refinery. At the wellhead, the pipe's diameter is 50.9 c

seraphim [82]

Answer:

Flow rate 2.34 m3/s

Diameter 0.754 m

Explanation:

Assuming steady flow, the volume flow rate along the pipe will always be constant, and equals to the product of flow speed and cross-section area.

The area at the well head is

$A = \pi r_w^2 = \pi (0.509/2)^2 = 0.203 m^2$

So the volume flow rate along the pipe is

$\dot{V} = Av = 0.203 * 11.5 = 2.34 m^3/s$

We can use the similar logic to find the cross-section area at the refinery

$A_r = \dot{V}/v_r = 2.34 / 5.25 = 0.446 m^2$

The radius of the pipe at the refinery is:

$A_r = \pi r^2$

$r^2 =A_r/\pi = 0.446/\pi = 0.141$

$r = \sqrt{0.141} = 0.377m$

So the diameter is twice the radius = 0.38*2 = 0.754m

6 0

2 years ago

How does a arrow represent displacement

Lera25 [3.4K]

Answer:

An arrow represents displacement. Because it's a vector quantity.

Vector quantity= Direction+Magnitude

When an arrow is put above a unit like

acceleration
displacement

it means that it's a vector quantity.

Hope this helps ;) ❤❤❤

4 0

3 years ago

Objects with masses of 165 kg and a 465 kg are separated by 0.340 m. (a) Find the net gravitational force exerted by these objec

Dmitrij [34]

Answer:

(a) $F_{net}$ = 4.19 x $10^{-5}$ N, and its direction is towards $m_{2}$ .

(b) It must be placed inside a hollow shell.

Explanation:

Let, $m_{1}$ = 165 kg, $m_{2}$ = 465 kg, $m_{3}$ = 60 kg, and the distance between $m_{1}$ and $m_{2}$ is 0.340 m.

(a) Since $m_{3}$ is placed midway between $m_{1}$ and $m_{2}$ , then its distance to both masses is 0.170 m.

From the Newton's law of universal gravitation,

F = $\frac{Gm_{1}m_{2} }{r^{2} }$

Where all variables have their usual meaning.

Then,

a. $F_{net}$ = $F_{23}$ - $F_{13}$

$F_{13}$ = $\frac{6.67*10^{-11}*165*60 }{(0.17)^{2} }$

= 2.25 x $10^{-5}$ N

$F_{23}$ = $\frac{6.67*10^{-11}*465*60 }{(0.17)^{2} }$

= 6.44 x $10^{-5}$ N

∴ $F_{net}$ = = 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

The net force exerted by the two masses on the 60 kg object is 4.19 x $10^{-5}$ N.

(ii) / $F_{net}$ / = / $F_{23}$ / - / $F_{13}$ /

= 6.44 x $10^{-5}$ - 2.25 x $10^{-5}$

= 4.19 x $10^{-5}$ N

(iii) The direction of the net force is to the right i.e towards $m_{2}$ .

(b) For the net force experienced by the 60 kg object to be zero, it must be placed inside a hollow shell.

7 0

2 years ago