Ask question

Ask question

All categories

2 years ago

8

A balance accurate to one-hundredth of a gram measures the mass of a rock to be 56.10 grams. How many significant digits are in

this value?

2 answers:

natita [175]2 years ago

7 0

By definition we have to:

The significant figures are the digits of a number that we consider not null.

We have different rules to consider significant figures depending on the number given.

Among some of the rules are:

1) All non-zero digits are significant.

2) For numbers greater than 1, the zeros to the right of the comma are significant.

For this case we have the following number:

56.10 grams

According to the rules, the number zero is counted as a significant figure.

Therefore, the number has four significant figures.

Answer:

56.10 grams = four significant figures.

natali 33 [55]2 years ago

3 0

There are 3 significant didgits

You might be interested in

A siphon pumps water from a large reservoir to a lower tank that is initially empty. The tank also has a rounded orifice 20 ft b

trasher [3.6K]

Answer:

height of the water rise in tank is 10ft

Explanation:

Apply the bernoulli's equation between the reservoir surface (1) and siphon exit (2)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} + z_1= \frac{P_2}{pg} + \frac{V_2^2}{2g} +z_2$

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$ -------(1)

substitute $P_a_t_m for P_1, (P_a_t_m +pgh) for P_2$

0ft/s for V₁, 20ft for (z₁ - z₂) and 32.2ft/s² for g in eqn (1)

$\frac{P_1}{pg} + \frac{V^2_1}{2g} +( z_1-z_2)= \frac{P_2}{pg} + \frac{V_2^2}{2g}$

$\frac{P_1}{pg} + \frac{0^2_1}{2g} +( 20)= \frac{(P_a_t_m+pgh)}{pg} +\frac{V^2_2}{2\times32.2} \\\\V_2 = \sqrt{64.4(20-h)}$

Applying bernoulli's equation between tank surface (3) and orifice exit (4)

$\frac{P_3}{pg} + \frac{V^2_3}{2g} + z_3= \frac{P_4}{pg} + \frac{V_4^2}{2g} +z_4$

substitute

$P_a_t_m for P_3, P_a_t_m for P_4$

0ft/s for V₃, h for z₃, 0ft for z₄, 32,2ft/s² for g

$\frac{P_a_t_m}{pg} + \frac{0^2}{2g} +h=\frac{P_a_t_m}{pg} + \frac{V_4^2}{2\times32.2} +0\\\\V_4 =\sqrt{64.4h}$

At equillibrium Fow rate at point 2 is equal to flow rate at point 4

Q₂ = Q₄

A₂V₂ = A₃V₃

The diameter of the orifice and the siphon are equal , hence there area should be the same

substitute A₂ for A₃

$\sqrt{64.4(20-h)}$ for V₂

$\sqrt{64.4h}$ for V₄

A₂V₂ = A₃V₃

$A_2\sqrt{64.4(20-h)} = A_2\sqrt{64.4h}\\\\20-h=h\\\\h= 10ft$

Therefore ,height of the water rise in tank is 10ft

3 0

2 years ago

A wooden piece is made in different shapes take length (l) = radius (r) = 2m Calculate its volume as a:

In-s [12.5K]

This question deals with the volume of different shapes.

a) volume of the sphere is "33.51 m³".

b) volume of the cylinder is "25.13 m³".

a)

The volume of a sphere is given by the following formula:

$Volume = \frac{4}{3}\pi r^3\\\\Volume = \frac{4}{3}\pi (2\ m)^3$

<u>Volume = 33.51 m³</u>

<u />

b)

The volume of a cylinder is given by the following formula:

$Volume = \pi r^2l\\\\Volume =\pi (2\ m)^2(2\ m)$

<u>Volume = 25.13 m³</u>

<u />

Learn more about <em>volume </em>here:

brainly.com/question/16686115?referrer=searchResults

The attached picture shows the formulae of the <em>volume</em> of different shapes.

7 0

2 years ago

With your hand parallel to the floor and your palm upright, you lower a 3-kg book downward. If the force exerted on the book by

Stells [14]

For Newton's second law, the resultant of the forces acting on the book is equal to the product between the mass of the book and its acceleration:
$\sum F = ma$ (1)

There are only two forces acting on the book:
- its weight, directed downward: mg
- the force exerted by the hand on the book, of 20 N, directed upward

so, equation (1) becomes
$mg - F = ma$
from which we can calculate the book's acceleration, a:
$a= g - \frac{F}{m}= 9.81 m/s^2 - \frac{20 N}{3 kg}=3.14 m/s^2$

7 0

2 years ago

Read 2 more answers

The air in a pipe resonates at 150 Hz and 750 Hz, one of these resonances being the fundamental. If the pipe is open at both end

Xelga [282]

Answer:

Explanation:

Two frequencies with magnitude 150 Hz and 750 Hz are given

For Pipe open at both sides

fundamental frequency is 150 Hz as it is smaller

frequency of pipe is given by

$f=\frac{nv}{2L}$

where L=length of Pipe

v=velocity of sound

$f=150\ Hz$ for n=1

and f=750 is for n=5

thus there are three resonance frequencies for n=2,3 and 4

For Pipe closed at one end

frequency is given by

$f=\frac{(2n+1)}{4L}\cdot v$

for n=0

$f_1=\frac{v}{4L}$

$f_1=150\ Hz$

for n=2

$f_2=\frac{5v}{4L}$

Thus there is one additional resonance corresponding to n=1 , between $f_1$ and $f_2$

8 0

2 years ago

A man walks 30 m to the west, then 5 to the east in 45 seconds

exis [7]

His average speed is (35m/45s) = 7/9 meters per second.

His average velocity is (30m W + 5m E) / (45s) = 25 m/s West .

8 0

2 years ago