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1 year ago

The weight of an object never changes. A. True B. False

Chemistry

2 answers:

Shalnov [3]1 year ago

6 0

Answer: false .

Explanation: Weight is the measure of the force of gravity on an object. The mass of an object will never change, but the weight of an item can change based on its location.

For example, you may weigh 100 pounds on Earth, but in outer space you would be weightless.Mass never changes. Weight, which is a product of mass times gravity, may change due to a difference in gravity

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Alla [95]1 year ago

3 0

Answer:

$\huge\boxed{\textbf{False}}$

Explanation:

Definition of weight:

The gravitational force acting on a body is called weight.

Explanation:

The weight of the body changes with different factors concerned e.g. height h, acceleration due to gravity g etc.

We know that,

W = mg

Mass is a constant quantity, however g changes with the altitude or place. Consequently, weight also does not remain constant.

Example:

An object having mass 1 kg will have a weight of 9.8 N on Earth while its weight on moon will be 1.6 N due to the difference in the values of g.

$\rule[225]{225}{2}$

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What is the electron configuration of aluminum?

Bond [772]

What is the electron configuration of aluminum is 1s^2 2s^2 2p^6 3s^2 3p^1.

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2 years ago

In a Dam, If we doubled the depth of the dam the hydrostatic force will be ?

AnnZ [28]

Answer: Option (A) is the correct answer.

Explanation:

Force acting on a dam is as follows.

F = $\frac{1}{2}\rho g\omega H^{2}$ .......... (1)

Now, when we double the depth then it means H is increasing 2 times and then the above relation will be as follows.

F' = $\frac{1}{2}\rho g\omega (2)^{2}$

F' = $\frac{1}{2}\rho g\omega 4$ ........... (2)

Now, dividing equation (1) by equation (2) as follows.

$\frac{F}{F'}$ = $\frac{\frac{1}{2}\rho g\omega H^{2}}{\frac{1}{2}\rho g\omega 4}$

Cancelling the common terms we get the following.

$\frac{F}{F'}$ = $\frac{1}{4}$

4F = F'

Thus, we can conclude that if doubled the depth of the dam the hydrostatic force will be 4F.

4 0

2 years ago

An environmental scientist developed a new analytical method for the determination of cadmium (cd^2+) in mussels. To validate th

Anettt [7]

Answer:

The method is accurate in the calculation of the $Cu^+2$

Explanation:

As a first step we have to calculate the average concentration of $Cu^+2$ find it by the method.

$\frac{0.782+0.762+0.825+0.838+0.761 }{5} =0.79 ppm$

Then we have to find the standard deviation:

$s=\sqrt{\frac{1}{N-1}\sum_{i=1}^N(x_i-\bar{x})^2}=0.0359$

For the confidence interval we have to use the formula:

μ=Average± $\frac{t*s}{\sqrt{n} }$

Where:

t=t student constant with 95 % of confidence and 5 data=2.78

μ= $0.79$ ± $\frac{2.78*0.0359}{\sqrt{5} }$

upper limit: 0.84

lower limit: 0.75

If we compare the limits of the value obtanied by the method (Figure 1 Red line) with the reference material (Figure 1 blue line) we can see that the values obtained by the method are within the values suggested by the reference material. So, it's method is accurate.