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3 years ago

THIS IS ME BEWARE UGLINESS

Physics

2 answers:

Gemiola [76]3 years ago

8 0

Answer:

u aint ugly, but careful there are creeps on the internet

Explanation:

Nadusha1986 [10]3 years ago

3 0

Answer:

DONT POST PICS OF YOURSELF ON THIS!!!!!!!!!!!!! WHAT IF I WAS A CREEPY OLD MAN? STAY SAFE ONLINE!!!!!!!!!!!!!

Explanation:

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The pressure at the top of a liquid

myrzilka [38]

suman ek din maxi pahankar peshab ma hath dali thi

6 0

3 years ago

A model airplane with a mass of 0.741kg is tethered by a wire so that it flies in a circle 30.9 m in radius. The airplane engine

Tju [1.3M]

Answer:

$_T}=24.57Nm$

ω = 0.0347 rad/s²

a ≅ 1.07 m/s²

Explanation:

Given that:

mass of the model airplane = 0.741 kg

radius of the wire = 30.9 m

Force = 0.795 N

The torque produced by the net thrust about the center of the circle can be calculated as:

$_T } = Fr$

where;

F represent the magnitude of the thrust

r represent the radius of the wire

Since we have our parameters in set, the next thing to do is to replace it into the above formula;

So;

$_T}=(0.795)*(30.9)$

$_T}=24.57Nm$

(b)

Find the angular acceleration of the airplane when it is in level flight rad/s²

$_T}=I \omega$

where;

I = moment of inertia

ω = angular acceleration

The moment of inertia (I) can also be illustrated as:

$I = mr^2$

I = ( 0.741) × (30.9)²

I = 0.741 × 954.81

I = 707.51 Kg.m²

$_T}=I \omega$

Making angular acceleration the subject of the formula; we have;

$\omega = \frac{_T}{I}$

ω = $\frac{24.57}{707.51}$

ω = 0.0347 rad/s²

(c)

Find the linear acceleration of the airplane tangent to its flight path.m/s²

the linear acceleration (a) can be given as:

a = ωr

a = 0.0347 × 30.9

a = 1.07223 m/s²

a ≅ 1.07 m/s²

5 0

3 years ago

Find the distance along an arc on the surface of the earth that subtends a central angle of 1 minutes (1 minute = 1/60 degree).

-Dominant- [34]

Answer:

1.152 miles

Explanation:

Given: central angle = 1 minute = $(\frac{1}{60}) ^{o}$

radius of the earth = 3960 miles

The length of an arc = $\frac{\alpha }{360^{o} }$ 2 $\pi$ r

where: $\alpha$ is the central angle, and r is the radius.

Thus,

Distance along the arc = $\frac{\alpha }{360^{o} }$ 2 $\pi$ r

Distance along the arc = $\frac{(\frac{1}{60}) ^{o} }{360^{o} }$ x 2 x $\frac{22}{7}$ x 3960

= $\frac{(\frac{1}{60}) ^{o} }{360^{o} }$ x 24891.4286

= 1.1524

The required distance along an arc is 1.152 miles.

4 0

4 years ago

Which is an example of fraudulent bookkeeping on the income side?

Nadya [2.5K]

A customer is billed for a service or product, but it gets recorded in the ledger at a lower amount is an example of fraudulent bookkeeping on the income side. This fraud would likely happen in the account receivable department of a company which generates sales for its function<span>. The sale is the primary source of company's operational income.</span>

3 0

3 years ago

At a beach the light is generallypartially polarized owing to reflections off sand and water. At a particular beach on a particu

Ne4ueva [31]

Answer:

a) 0.159

b) 0.84

Explanation:

The Horizontal component is 2.3 times the vertical component

Let the horizontal electric field component = $E_{h}$

Let the vertical electric field component = $E_{v}$

The formula for light intensity is given by:

$I = \frac{E_{m} ^{2} }{2c \mu}$ ..............................(1)

$E_{m}$ is the resolution of the vertical and horizontal components, $E_{h} and E_{v}$

$E_{m} ^{2} = E_{h} ^{2} + E_{v} ^{2}$ ..................(2)

Light intensity before the glasses were put on:

$I_{1} = \frac{E_{m} ^{2} }{2c \mu_{1} }$ .............................(3)

Put equation (2) into equation (3)

$I_{1} = \frac{E_{h} ^{2} + E_{v} ^{2}}{2c \mu_{1} }$ .............................(4)

After the glasses were put on the horizontal component vanishes, i.e. $E_{h} = 0$

$I_{2} = \frac{ E_{v} ^{2}}{2c \mu_{2} }$ ...................................(5)

Divide equation (5) by equation (4)

$\frac{I_{2} }{I_{1} } = \frac{E_{v} ^{2} }{E_{h} ^{2} + E_{v} ^{2}}$ ...............................(6)

But $E_{h} = 2.3E_{v}$ ......................(7)

Insert equation (7) into (6)

$\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{(2.3E_{v})^{2} + E_{v} ^{2} } \\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{5.29E_{v}^{2} + E_{v} ^{2} }\\\frac{I_{2} }{I_{1} } = \frac{E_{v}^{2} }{6.29E_{v}^{2} }\\\frac{I_{2} }{I_{1} } =\frac{1}{6.29} \\$