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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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There is strong evidence that Europa, a satellite of Jupiter, has a liquid ocean beneath its icy surface. Many scientists think

dangina [55]

Answer:

4.44 rpm

Explanation:

$\omega$ = Angular speed

G = Gravitational constant = 6.67 × 10⁻¹¹ m³/kgs²

r = Radius of Europa = $\frac{3138000}{2}\ m$

R = Radius of arm = 6 m

The acceleration due to gravity is given by

$g=\frac{GM}{r^2}\\\Rightarrow g=\frac{6.67\times 10^{-11}\times 4.8\times 10^{22}}{\left(\frac{3138000}{2}\right)^2}\\\Rightarrow g=1.3\ m/s^2$

Here the centripetal acceleration of the arm and acceleration due to gravity are equal

$a_c=\omega^2R$

$a_c=g\\\Rightarrow \omega^2R=1.3\\\Rightarrow \omega^2\times 6=1.3\\\Rightarrow \omega=\sqrt{\frac{1.3}{6}}\\\Rightarrow \omega=0.46547\ rad/s$

Converting to rpm

$1\ rad/s=\frac{60}{2\pi}\ rpm$

$0.46547\ rad/s=0.46547\times \frac{60}{2\pi}\ rpm=4.44\ rpm$

The angular speed of the arm is 4.44 rpm

8 0

2 years ago

What is your operational definition for a fast reaction time?

LuckyWell [14K]

Answer:

The ability to react to a certain stimulus with a speedy and effective manner

Explanation:

7 0

3 years ago

Plz help me with question 3

choli [55]

Answer:

C. 110 - 140

Explanation:

5 0

2 years ago

Read 2 more answers

A hoodlum throws a stone vertically downward with an initial speed v0 from the roof of a building, a height h above the ground.

Vaselesa [24]

V2=u2+2as
v2=144+600
v2=744
v=√744

6 0

3 years ago

Block A of mass M is on a horizontal surface of negligible friction. An identical block B is attached to block A by a light stri

miv72 [106K]

Answer:

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

Explanation:

For this problem let's use Newton's second law applied to each body

Body A

X axis

T = m_A a

Axis y

N- W_A = 0

Body B

Vertical axis

W_B - T = m_B a

In the reference system we have selected the direction to the right as positive, therefore the downward movement is also positive. The acceleration of the two bodies must be the same so that the rope cannot tension

We write the equations

T = m_A a

W_B –T = M_B a

We solve this system of equations

m_B g = (m_A + m_B) a

a = m_B / (m_A + m_B) g

In this initial case

m_A = M

m_B = M

a = M / (1 + 1) M g

a = ½ g

Let's find the tension

T = m_A a

T = M ½ g

T = ½ M g

Now we change the mass of the second block

m_B = 2M

a = 2M / (1 + 2) M g

a = 2/3 g

We seek tension for this case

T’= m_A a

T’= M 2/3 g

Let's look for the relationship between the tensions of the two cases

T’/ T = 2/3 M g / (½ M g)

T’/ T = 4/3

T’= 4/3 T

The new tension is 4/3 = 1.33 of the previous tension the answer e

4 0

3 years ago