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Anna11 [10]
3 years ago
12

What are significant figures?

Physics
2 answers:
vekshin13 years ago
8 0
“The term significant figures refers to the number of important single digits (0 through 9 inclusive) in the coefficient of an expression in scientific notation . The number of significant figures in an expression indicates the confidence or precision with which an engineer or scientist states a quantity.”
11Alexandr11 [23.1K]3 years ago
4 0
You can check this out for further understanding

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An atom’s mass number equals the number of
steposvetlana [31]
An atom's mass number equals the number of protons plus the number of neutrons.
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An ion has unequal numbers of which two particles?
Scrat [10]
Protons and electrons
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a car initially at 39.40 m/s accelerates at -3.00 m/s^2 how far has a car travel when it comes to a stop 9.35 seconds later
Kipish [7]

Answer:

499.523. meter

<em>I</em><em> hope</em><em> it's</em><em> helps</em><em> you</em>

6 0
3 years ago
a large sphere is on a horizontal field on a sunny day. at a certain time the shadow reaches out a distance of 10 m from the poi
Mars2501 [29]

The radius of the sphere in meters is ,r = 10\sqrt{5} -20

Think about the angle the ground and the shadow make. Since the sun's beams are parallel, the angle created by the stick's shadow is also equal. Since the stick is 1 m high and its shadow is 2 m long, we know that the stick's angle is arctan 1/2. Therefore, by thinking of a right-angled triangle,

r/10 = tan [arctan(1/2)] = tan (1/2)

Since, tan (θ/2) = 1-cos(θ) / sin(θ)

we find that,

r/10 = \sqrt{5} -2

Hence, r = 10\sqrt{5} -20

So, the radius of the sphere in meters is ,r = 10\sqrt{5} -20

Learn more about radius (r) of the sphere here;

brainly.com/question/14100787

#SPJ4

5 0
1 year ago
A 1.0-cm-tall object is 13 cm in front of a converging lens that has a 40 cm focal length.
kicyunya [14]

A) Image position: -19.3 cm

B) Image height: 1.5 cm, upright

Explanation:

A)

In order to calculate the image position, we can use the lens equation:

\frac{1}{p}+\frac{1}{q}=\frac{1}{f}

where

p is the distance of the object from the lens

q is the distance of the image from the lens

f is the focal length

In this problem, we have:

p = 13 cm (object distance)

f = 40 cm (focal length, positive for a converging lens)

So the image distance is

\frac{1}{q}=\frac{1}{f}-\frac{1}{p}=\frac{1}{40}-\frac{1}{13}=-0.0519\\q=\frac{1}{-0.0519}=-19.3 cm

The negative sign means that the image is virtual.

B)

In order to calculate the image height, we use the magnification equation:

\frac{y'}{y}=-\frac{q}{p}

where

y' is the image height

y is the object height

In this problem, we have:

y = 1.0 cm (object height)

p = 13 cm

q = -19.3 cm

Therefore, the image heigth is

y'=-\frac{qy}{p}=-\frac{(-19.3)(1.0)}{13}=1.5 cm

And the positive sign means the image is upright.

6 0
3 years ago
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