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2 years ago

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 water is the greatest at a temperature of

bija089 [108]

You anwser is B. 277 k

4 0

3 years ago

Read 2 more answers

Gold has a specific heat of 0.130 J/g*C. If 195 joules of heat are added to 15 grams of gold how much does the temperature of th

Anna71 [15]

The correct answer to the question is : $100\ ^0C$

EXPLANATION :

As per the question, the specific heat of gold is given as c = $0.130\ J/g^0C$

The heat given to the gold dQ = 195 J

The mass of the gold is given as m = 15 gram.

We are asked to calculate the change in temperature.

Let the change in temperature is dT.

We know that dQ = mcdT

$dT=\ \frac{dQ}{mc}$

$=\ \frac{195}{15\times 0.130}$

$=\ 100\ ^0C$ [ANS]

Hence, the change in temperature is 100 degree celsius.

5 0

3 years ago

A spring with a spring constant of 350 N/m pulls a door closed. How much work is done as the spring pulls the door at a constant

german

The work done to stretch the spring will be 112 J.

<h3>What is spring force?</h3>

The force required to extend or compress a spring by some distance scales linearly with respect to that distance is known as the spring force. Its formula is

F = kx

The given data in the problem is;

F is the spring force =?

K is the spring constant= 8.5 N/m

x is the length by which spring got stretched = 1.2m

The work is done to stretch the spring is;

$\rm W= \frac{1}{2} kx^2 \\\\ W=\frac{1}{2} \times 350 \times (0.850-0.050)^2 \\\\ W=0.5 \times 350 \times (0.80)^2 \\\\W=112 \ J$

To learn more about the spring force refer to the link;

brainly.com/question/4291098

#SPJ1

3 0

2 years ago

When Simon grows (let’s say he doubles his mass) what happens to his GPE when he is at the balcony?

DerKrebs [107]

Answer:

His gravitational potential energy will increase as well.

Explanation:

Let gpe represent gravitational potential energy.

gpe = mass × gravitational field strength × height

From the formula above, we can conclude that as the mass of a body increases, it's gpe increases too.

5 0

2 years ago

The height of a typical playground slide is about 1.8 m and it rises at an angle of 30 ∘ above the horizontal.

Salsk061 [2.6K]

Answer:

$5.94\ \text{m/s}$

$1.7$

$0.577$

Explanation:

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

$\theta$ = Angle of slope = $30^{\circ}$

v = Velocity of child at the bottom of the slide

$\mu_k$ = Coefficient of kinetic friction

$\mu_s$ = Coefficient of static friction

h = Height of slope = 1.8 m

The energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2\\\Rightarrow v=\sqrt{2gh}\\\Rightarrow v=\sqrt{2\times 9.81\times 1.8}\\\Rightarrow v=5.94\ \text{m/s}$

The speed of the child at the bottom of the slide is $5.94\ \text{m/s}$

Length of the slide is given by

$l=h\sin\theta\\\Rightarrow l=1.8\sin30^{\circ}\\\Rightarrow l=0.9\ \text{m}$

$v=\dfrac{1}{2}\times5.94\\\Rightarrow v=2.97\ \text{m/s}$

The force energy balance of the system is given by

$mgh=\dfrac{1}{2}mv^2+\mu_kmg\cos\theta l\\\Rightarrow \mu_k=\dfrac{gh-\dfrac{1}{2}v^2}{gl\cos\theta}\\\Rightarrow \mu_k=\dfrac{9.81\times 1.8-\dfrac{1}{2}\times 2.97^2}{9.81\times 0.9\cos30^{\circ}}\\\Rightarrow \mu_k=1.73$

The coefficient of kinetic friction is $1.7$ .

For static friction

$\mu_s\geq\tan30^{\circ}\\\Rightarrow \mu_s\geq0.577$

So, the minimum possible value for the coefficient of static friction is $0.577$ .

8 0

3 years ago