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Iteru [2.4K]
3 years ago
7

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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viva [34]

Answer:

the acceleration during the collision is: - 5  \frac{m}{s^2}

Explanation:

Using the formula:

a=\frac{\Delta\,v}{\Delta\,t}

we get:

a=\frac{-0.4-0.6}{0.2} \,\frac{m}{s^2} =\frac{-1}{0.2} \,\frac{m}{s^2} =-5\,\,\frac{m}{s^2}

4 0
3 years ago
You lift a 25-kg child 0.80 m, slowly carry him 10 m to the playroom, and finally set him back down 0.80 m onto the playroom flo
maksim [4K]

To solve this problem we will apply the work theorem which is expressed as the force applied to displace a body. Considering that body strength is equivalent to weight, we will make the following considerations

\text{Mass of the child} = m = 25kg

\text{Acceleration due to gravity} = g = 9.81m/s^2

\text{Height lifted} = h = 0.80m (Upward)

Work done to upward the object

W = mgh

W = (25)(9.81)(0.8)

W = 196.2J

Horizontal Force applied while carrying 10m,

F = 0N

W = 0J

Height descended in setting the child down

h' = -0.8m (Downwoard)

W = mgh'

W = (25)(9.81)(-0.80)

W = -196.2J

For full time, assuming that the total value of work is always expressed in terms of its symbol, it would be zero, since at first it performs the same work that is later complemented in a negative way.

6 0
3 years ago
How long does it take a cougar to run 170 meter if it run 5m/s?
Igoryamba

Answer: 34 seconds

Explanation:

170m/5m=34seconds

3 0
3 years ago
Read 2 more answers
10 solve the following humeric problem - S. a What is the lift height of salu? T. object energy required to 25 kg to mass of a 1
Salsk061 [2.6K]
  • Mass=m=25kg
  • Height=h=10m
  • Acceleration due to gravity=g=10m/s^2

\boxed{\sf E_P=mgh}

\\ \sf\longmapsto E_P=25(10)(10)

\\ \sf\longmapsto E_P=25(100)

\\ \sf\longmapsto E_P=2500J

6 0
2 years ago
A 1kg pool ball travels to the right at 2m/s and hits another stopped pool ball bounces back to the left at 1m/s what is the vel
Burka [1]
Ptotal=Ptotal —> m1v1+m2v2=m1v1’+m2v2’ —> (1kg)(2m/s)+(1kg)(0m/s)=(1kg)(-1m/s)+(1kg)(v2’) —> v2’=3m/s

answer: v=3m/s
5 0
3 years ago
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