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3 years ago

a stockbroker "talks up" a piece of stock that he wants you to buy because it was a high return rate. This is an example of: ?

Physics

2 answers:

kozerog [31]3 years ago

8 0

I believe the answer is manipulation of information. Hope I helped!

Margaret [11]3 years ago

7 0

<h2>Answer:</h2>

This is an example of "Marketing".

There different strategies that stockbroker used to sell their products. They try to engage customers with this kind of lavish offers. So, that trapped the attention and can sell their products.

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The generator at a power plant produces AC at 20,000 V. A transformer steps this up to 355,000 V for transmission over power lin

Masja [62]

Answer:

Number of coil in the output is 39938

Explanation:

We have given a step up transformer

Input voltage of transformer, that is primary voltage $v_p=20000volt$

Output voltage, that is secondary voltage $v_s=355000volt$

Number of turns in primary $N_p=2250$

For transformer we know that $\frac{V_p}{V_s}=\frac{N_p}{N_s}$

$\frac{20000}{355000}=\frac{2250}{N_s}$

$N_s=39937.5$

As the number of turns can not be in fraction so number of turns in the output coil is 39938

7 0

3 years ago

Why are the coldest places on earth found near the poles

Nata [24]

Polar regions do not receive direct sunlight during the winter months due to the tilt in the Earth's<span> axis. Hence, polar regions can get very cold. Antarctica is the </span>coldest place on Earth. <span>The </span>coldest places on Earth<span> tend to be located </span>near the poles<span>. Hope this answers the question.</span>

6 0

3 years ago

A block of mass 0.1 kg is attached to a spring of spring constant 21 N/m on a frictionless track. The block moves in simple harm

bogdanovich [222]

Answer:

A) 2.75 m/s B) 0.1911 m C) 0.109 s

Explanation:

mass of block = M =0.1 kg

spring constant = k = 21 N/m

amplitude = A = 0.19 m

mass of bullet = m = 1.45 g = 0.00145 kg

velocity of bullet = vᵇ = 68 m/s

as we know:

Angular frequency of S.H.M = ω₀ = $\sqrt\frac{k}{M}$

= $\sqrt\frac{21}{0.1}$

= 14.49 rad/sec

<h3>A) Speed of the block immediately before the collision:</h3>

displacement of Simple Harmonic Motion is given as:

$x = A sin (\omega t + \phi)\\$

Differentiating this to find speed of the block immediately before the collision:

$v=\frac{dx}{dt}= A\omega_{o} cos (\omega_{o}t =\phi}\\$

As bullet strikes at equilibrium position so,

φ = 0

t= 2nπ

⇒ cos (ω₀t + φ) = 1

⇒ $v= A\omega_{o}$

$v=(.19)(14.49)\\v= 2.75 ms^{-1}$

<h3>B) If the simple harmonic motion after the collision is described by x = B sin(ωt + φ), new amplitude B:</h3>

S.H.M after collision is given as :

$x= Bsin(\omega t + \phi)$

To find B, consider law of conservation of energy

$K.E = P.E\\K.E= \frac{1}{2}(m+M)v^{2} \\P.E = \frac{1}{2} kB^{2}$

$\frac{m+M}{k} v^{2} = B^{2} \\B =\sqrt\frac{m+M}{k} v\\B = \sqrt\frac{.00145+0.1}{21} (2.75)\\B = .1911m$

<h3>C) Time taken by the block to reach maximum amplitude after the collision:</h3>

Time period S.H.M is given as:

$T=2\pi \sqrt\frac{m}{k}\\ for given case\\m= m=M\\then\\T=2\pi \sqrt\frac{m+M}{k}$

Collision occurred at equilibrium position so time taken by block to reach maximum amplitude is equal to one fourth of total time period

$T=\frac{\pi }{2}\sqrt\frac{m+M}{k} \\T=0.109 sec$

5 0

3 years ago

A polo player hits a ball with a mass of 0.42 kg with a force of 7.35 Newtons.

Serga [27]

Due to the law of conservation of momentum, the force exerted on the mallet is equal and opposite to the force exerted on the ball, so the answer is C.

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liraira [26]

The correct answer is A <span />

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