All categories

3 years ago

When are zeros significant when found to the trailing (to the right) of the decimal point?

Physics

1 answer:

Lelechka [254]3 years ago

7 0

Answer:

Usually, zeroes are found to the right of a decimal point in significant numbers if you have something like $7\frac{1}{100000000000000000000000000}$ (exaggerated example of course), which is when you have a number that is very very close to the integer before it, but isn't that integer.

You might be interested in

What happens to the current in a circuit if a 1.5 volt battery

Alexus [3.1K]

Answer:

the current will improve in performance and tell me if im correct

4 0

3 years ago

Dr. gavin decides that instead of conducting a 2 ´ 4 independent-groups factorial design, he is going to conduct a 2 ´ 4 within-

vampirchik [111]

<span>If Dr. Gavin decides that instead of conducting a 2 ´ 4 independent-groups factorial design, and he is going to conduct a 2 ´ 4 within-subjects factorial design, then the things that will change are the various independent groups which are involved.</span>

8 0

3 years ago

What is N{2}+H{2}=NH{3}

Norma-Jean [14]

Answer:

N2 + 3 H2 = 2 NH3

Explanation:

8 0

3 years ago

Famous scientists, who was credited with perfecting the telescope and saying there is another side to the moon?.

Black_prince [1.1K]

Answer:

Galileo di Vincenzo Bonaiuti de' Galilei

7 0

3 years ago

What could be the possible answer to the question ?<br><br>thankyou ~

Ganezh [65]

The value of the force, F₀, at equilibrium is equal to the horizontal

component of the tension in string 2.

Response:

The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>

<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>

Given:

The weight of the rod = The sum of the vertical forces in the strings

Therefore;

M·g = T₂·cos(37°) + T₁

The weight of the rod is at the middle.

Taking moment about point (2) gives;

M·g × L = T₁ × 2·L

Therefore;

$T_1 = \mathbf{\dfrac{M \cdot g}{2}}$

Which gives;

$M \cdot g = \mathbf{T_2 \cdot cos(37 ^{\circ})+ \dfrac{M \cdot g}{2}}$

$T_2 = \dfrac{M \cdot g - \dfrac{M \cdot g}{2}}{cos(37 ^{\circ})} = \mathbf{\dfrac{M \cdot g}{2 \cdot cos(37 ^{\circ})}}}$

F₀ = T₂·sin(37°)

Which gives;

$F_0 = \dfrac{M \cdot g \cdot sin(37 ^{\circ})}{2 \cdot cos(37 ^{\circ})}} = \dfrac{M \cdot g \cdot tan(37 ^{\circ})}{2} \approx \mathbf{0.377 \cdot M \cdot g}$

F₀ ≈ <u>0.377·M·g</u>

<u />

Learn more about equilibrium of forces here:

brainly.com/question/6995192

3 0

2 years ago

Read 2 more answers