All categories

3 years ago

Would the value of "g" be affected if the size of the Bob is varied ?

Physics

1 answer:

marshall27 [118]3 years ago

6 0

Answer:

Any variable would be affected if the product is varied.

Explanation:

If you have any questions feel free to ask in the comments - Mark

ALSO Happy Valentines Day ❤

You might be interested in

What is not changed when work is done by a machine?

Irina18 [472]

B) The amount of work done

8 0

3 years ago

Read 2 more answers

A man performed 500 j of work moving a couch in his living room

Sergio039 [100]

Incomplete question. Missing part:

"it took him 10 seconds to move the couch. What was his power?"

Answer:

50 W

Explanation:

The power developed is given by:

$P=\frac{W}{t}$

where

W is the work done

t is the time taken

In this problem,

W = 500 J

t = 10.0 s

Substituting into the equation, we find

$P=\frac{500}{10}=50 W$

6 0

4 years ago

the solubility product of lead fluoride is 3.6 x 10–8. what is its solubility in 0.10M NaF solution, in grams per liter

BartSMP [9]

Answer:

8.8 × 10⁻³ g/L

Explanation:

NaF is a strong electrolyte that ionizes according to the following reaction.

NaF(aq) → Na⁺(aq) + F⁻(aq)

Then, the concentration of F⁻ will also be 0.10 M.

In order to find the solubility of PbF₂ (S), we will use an ICE Chart.

PbF₂(s) ⇄ Pb²⁺(aq) + 2 F⁻(aq)

I 0 0.10

C +S +2S

E S 0.10 + 2S

The solubility product (Kps) is:

Kps = 3.6 × 10⁻⁸ = [Pb²⁺].[F⁻]² = S . (0.10 + 2S)²

In the term 0.10 + 2S, 2S is negligible in comparison with 0.10 and we can omit it to simplify calculations.

Kps = 3.6 × 10⁻⁸ = S . (0.10)²

S = 3.6 × 10⁻⁵ M

The molar mass of PbF₂ is 245.20 g/mol. The solubility of PbF₂ in g/L is:

3.6 × 10⁻⁵ mol/L × 245.20 g/mol = 8.8 × 10⁻³ g/L

4 0

3 years ago

How does the observed change in the sea floor age support he theory of sea floor spreading

RUDIKE [14]

Dna is the way to answer this questions

6 0

3 years ago

After a fall, a 90 kg rock climber finds himself dangling from the end of a rope that had been 16 m long and 7.8 mm in diameter

Nesterboy [21]

Explanation:

Given that,

Mass of the rock climber, m = 90 kg

Original length of the rock, L = 16 m

Diameter of the rope, d = 7.8 mm

Stretched length of the rope, $\Delta L=3.1\ cm$

(a) The change in length per unit original length is called strain. So,

$\text{strain}=\dfrac{\Delta L}{L}\\\\\text{strain}=\dfrac{3.1\times 10^{-2}}{16}\\\\\text{strain}=0.00193$

(b) The force acting per unit area is called stress.

$\text{stress}=\dfrac{mg}{A}\\\\\text{stress}=\dfrac{90\times 10}{\pi (3.9\times 10^{-3})^2}\\\\\text{stress}=1.88\times 10^7\ Pa$

$Y=\dfrac{\text{stress}}{\text{strain}}\\\\Y=\dfrac{1.88\times 10^7}{0.00193}\\\\Y=9.74\times 10^9\ N/m^2$

Hence, this is the required solution.

8 0

3 years ago

Would the value of "g" be affected if the size of the Bob is varied ?​

Would the value of "g" be affected if the size of the Bob is varied ?