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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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Forces applied in

Anni [7]

Answer:

A -Added when in the same direction

Subtracted when in opposite directions.

Explanation:

8 0

2 years ago

a person dives off the edge of aa cliff 33m above the surface of the sea. assuming that air resistance is negligible, how long d

USPshnik [31]

Answer:

The time is $t = 2.595 \ s$

The speed is $v = 25.43 \ m/s$

Explanation:

From the question we are told that

The height of the cliff is $h = 33 \ m$

Generally from kinematic equation we have that

$h = ut + \frac{1}{2} gt^2$

before the jump the persons initial velocity is u = 0 m/s

$33 = 0 * t + \frac{1}{2} 9.8 * t^2$

=> $t = 2.595 \ s$

Generally from kinematic equation

$v= u + gt$

=> $v= 0 + 9.8 * 2.595$

=> $v = 25.43 \ m/s$

3 0

3 years ago

A ball is thrown up onto a roof, landing 4 sec later at height of 20m above the release level. The balls path just before landin

Kobotan [32]

c) what is the angle (relative to the horizontal) of the balls initial velocity?

8 0

3 years ago

The drawing shows three particles far away from any other objects and located on a straight line. The masses of these particles

belka [17]

Answer:

$F_a=5.67\times 10^{-5}\ N$

$F_b=3.49\times 10^{-5}\ N$

$F_c=9.16\times 10^{-5}\ N$

Explanation:

Given:

mass of particle A, $m_a=363\ kg$

mass of particle B, $m_b=517\ kg$

mass of particle C, $m_c=154\ kg$

All the three particles lie on a straight line.

Distance between particle A and B, $x_{ab}=0.5\ m$

Distance between particle B and C, $x_{bc}=0.25\ m$

Since the gravitational force is attractive in nature it will add up when enacted from the same direction.

Force on particle A due to particles B & C:

$F_a=G. \frac{m_a.m_b}{x_{ab}^2} +G. \frac{m_a.m_c}{(x_{ab}+x_{bc})^2}$

$F_a=6.67\times 10^{-11}\times (\frac{363\times 517}{0.5^2}+\frac{363\times 154}{(0.5+0.25)^2})$

$F_a=5.67\times 10^{-5}\ N$

Force on particle C due to particles B & A:

$F_c=G.\frac{m_c.m_b}{x_{bc}^2} +G.\frac{m_c.m_a}{(x_{ab}+x_{bc})^2}$

$F_c=6.67\times 10^{-11}\times (\frac{154\times 517}{0.25^2}+\frac{154\times 363}{(0.25+0.5)^2} )$

$F_c=9.16\times 10^{-5}\ N$

Force on particle B due to particles C & A:

$F_b=G.\frac{m_b.m_c}{x_{bc}^2} -G.\frac{m_b.m_a}{x_{ab}^2}$

$F_b=6.67\times 10^{-11}\times (\frac{517\times 154}{0.25^2}-\frac{517\times 363}{0.5^2} )$

$F_b=3.49\times 10^{-5}\ N$

3 0

3 years ago

How much heat is contained in 100 kg of water at 60.0 °C?

yulyashka [42]

Answer:

so how much heat is there at 0 C? That's zero. But for every degree above that you have 4.184 J. You take it from there. Remember q = mc*delta T.

8 0

3 years ago