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bixtya [17]
2 years ago
7

What are some guidelines for preventing quackery and fraud when shopping? Select

Physics
1 answer:
artcher [175]2 years ago
6 0

The steps/guidelines for preventing quackery and fraud when shopping are :

  • Be suspicious of a sales pitch that promises results too good to be true
  • Be cautious of mail-order and internet sales
  • Be wary of product claims
  • Be wary of advisors who sell products

<h3>What is Quackery and Fraud </h3><h3 />

Quackery involves the claim of knowledge and provision of a fake and dishonest service to unsuspecting customers or clients and this is very common in medicine.

Fraud is the false representation of a material or service in order to cause loss of money or pain to an unsuspecting client or customer. some steps to prevent fraud are as follows :

  • If a sales pitch is too good to be true one should keep off from such product/service.
  • when making internet sales and mail-order customers should be careful as there are a lot of imposters on the internet.

Hence we can conclude that The steps/guidelines for preventing quackery and fraud when shopping are as listed above

Learn more about Fraud : brainly.com/question/1260589

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The density of ice can help preserve the habitats of aquatic organisms, but it can also cause the death of an organism. Which st
Paul [167]

Answer:

I think it's D. The expansion of water as it freezes increases the amount of nutrients that can dissolve in liquid water, but it could cause fluid in cells to dissolve more harmful substances.

Explanation:

I know when water freezes, it expands and between the two answers that discuss the expansion of water, D sounds the most logical to me lol.

6 0
3 years ago
A disk is uniformly accelerated from rest with angular acceleration α. The magnitude of the linear acceleration of a point on th
solniwko [45]

Answer:

a = R\alpha\sqrt{1 + \alpha^2t^4}

Explanation:

As we know that the acceleration of a point on the rim of the disc is in two directions

1) tangential acceleration which is given as

a_t = R\alpha

2) Centripetal acceleration

a_c = \omega^2 R

here we know that

\omega = \alpha t

a_c = (\alpha t)^2 R

now we know that net linear acceleration is given as

a = \sqrt{a_c^2 + a_t^2}

so we have

a = \sqrt{R^2\alpha^2 + R^2(\alpha t)^4}

a = R\alpha\sqrt{1 + \alpha^2t^4}

4 0
3 years ago
Determine the energy required to accelerate an electron between each of the following speeds. (a) 0.500c to 0.900c MeV (b) 0.900
Aleonysh [2.5K]

Answer:

The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

Explanation:

We know that,

Mass of electron m_{e}=9.11\times10^{-31}\ kg

Rest mass energy for electron = 0.511 Mev

(a). The energy required to accelerate an electron from 0.500c to 0.900c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.500c)^2}{c^2}}}

E=0.582\ Mev

(b). The energy required to accelerate an electron from 0.900c to 0.942c Mev

Using formula of rest,

E=\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{f}^2}{c^2}}}-\dfrac{E_{0}}{\sqrt{1-\dfrac{v_{i}^2}{c^2}}}

E=\dfrac{0.511}{\sqrt{1-\dfrac{(0.942c)^2}{c^2}}}-\dfrac{0.511}{\sqrt{1-\dfrac{(0.900c)^2}{c^2}}}

E=0.350\ Mev

Hence, The energy required to accelerate an electron is 0.582 Mev and 0.350 Mev.

4 0
3 years ago
The waste product of photosynthesis is:
zalisa [80]

Answer:

The waste product of photosynthesis is oxygen

6 0
3 years ago
The rocket now has a thruster that malfunctions and is now pushing the rocket in the wrong direction. What is the new net force
Lana71 [14]

Answer:

Daddy Dr Dr Dr Dr

Explanation:

3 0
3 years ago
Read 2 more answers
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