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1 year ago

13

if a car travels around a circular track with a radius of 100 m once every 10s what will its average angular velocity around the

track be?

1 answer:

noname [10]1 year ago

7 0

Answer:

0.628 radians per second

Explanation:

Angular velocity

= 2pi/T

= 2(3.14159)/10

= 0.628 radians per second

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What is the fate of solar radiation that reaches the earth?

expeople1 [14]

When solar radiation reaches the Earth it quickly dissipates as most of the radiation and UV rays are blocked by ozone layer, but more radiation and UV rays are able to get through because of global warming.

6 0

2 years ago

How do I solve such problem???

pashok25 [27]

As far as I'm concerned, this is a bogus question, or at least a severely corrupted one.

The three numbers given can NOT all be true on Earth.

-- It rolled off the table at 7.6 m/s . By golly, there you are! Its initial horizontal velocity is 7.6 m/s, and it has no vertical velocity until it leaves the table.

-- There are no horizontal forces that we're aware of acting on the object. So it maintains the same horizontal velocity for the rest of the story. It's 10.5m away from the table in (10.5 m) / (7.8 m/s) = 1.35 second .

-- Vertically, it's just an object dropped from 17.6m off the floor. Shockingly, the distance it falls in time 'T' is (1/2 g) T². In 1.35 second, that's 8.88 meters ! . . . only about halfway to the floor !

-- In order to fall 17.6 m to the floor, it would need 1.89 seconds. In <u>that</u> length of time, however, it would travel (7.8 m/s) x (1.89 s) = 14.78 m away from the base of the table.

So you see, either . . .

-- the table is NOT 17.6m tall, or

-- the object does NOT roll off of the table at 7.8 m/s, or

-- it does NOT land 10.5 m away from the base of the table.

OR . . .

-- the table is not on Earth, and gravity is not 9.8 m/s² !

We often see questions posted on Brainly with not enough given information, OR with some information given that's not needed because it's not involved the answer.

THIS one is different, and it's unusual. In this one, we have<em> too much</em> given information, we can't ignore any of it because it's all related, but it's inconsistent and it CAN't all be true.

(Unless the whole story takes place on a mystery planet that is not Earth. Which I'm not going to take the time and effort right now to figure out what the acceleration of gravity has to be in order to make all of the given information compatible.)

7 0

2 years ago

Using a density of air to be 1.21kg/m3, the diameter of the bottom part of the filter as 0.15m (assume circular cross-section),

salantis [7]

Answer:

The drag coefficient is $D_z = 1.30512$

Explanation:

From the question we are told that

The density of air is $\rho_a = 1.21 \ kg/m^3$

The diameter of bottom part is $d = 0.15 \ m$

The power trend-line equation is mathematically represented as

$F_{\alpha } = 0.9226 * v^{0.5737}$

let assume that the velocity is 20 m/s

Then

$F_{\alpha } = 0.9226 * 20^{0.5737}$

$F_{\alpha } = 5.1453 \ N$

The drag coefficient is mathematically represented as

$D_z = \frac{2 F_{\alpha } }{A \rho v^2 }$

Where

$F_{\alpha }$ is the drag force

$\rho$ is the density of the fluid

$v$ is the flow velocity

A is the area which mathematically evaluated as

$A = \pi r^2 = \pi \frac{d^2}{4}$

substituting values

$A = 3.142 * \frac{(0.15)^2}{4}$

$A = 0.0176 \ m^2$

Then

$D_z = \frac{2 * 5.1453 }{0.0176 * 1.12 * 20^2 }$

$D_z = 1.30512$

3 0

2 years ago

A wave is produced from a vibrating string on a violin. Two observers are standing 20 m and 40 m away from the musician. The obs

max2010maxim [7]

Well the one that is closer can see and hear more

3 0

2 years ago

What would be the answer for this and how?

Veronika [31]

Answer:

B. 6 cm

Explanation:

First, we calculate the spring constant of a single spring:

$k = \frac{F}{\Delta x}\\$

where,

k = spring constant of single spring = ?

F = Force Applied = 10 N

Δx = extension = 4 cm = 0.04 m

Therefore,

$k = \frac{10\ N}{0.04\ m}\\k = 250\ N/m\\$

Now, the equivalent resistance of two springs connected in parallel, as shown in the diagram, will be:

$k_{eq} = k + k\\k_{eq} = 2k = 2(250\ N/m)\\k_{eq} = 500\ N/m\\$

For a load of 30 N, applying Hooke's Law:

$\Delta x = \frac{F}{k_{eq}}\\\\\Delta x = \frac{30\ N}{500\ N/m}\\\\\Delta x = 0.06\ m = 6\ cm\\$

Hence, the correct option is:

<u>B. 6 cm</u>

7 0

2 years ago