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anzhelika [568]
3 years ago
12

A materials density is the same , no matter how large or small the sample is or what it’s shape is as long as it is solid unifor

m piece of the material explain how this is possible and give an example
Physics
1 answer:
Scorpion4ik [409]3 years ago
8 0

Answer:

See the explanation below

Explanation:

Density is defined as the relationship between mass and volume, i.e. the following equation can be used:

density = m/v

where:

density [kg/m^3]

m = mass [kg]

v = volume [m^3]

If we change the volume of a body by reducing its size, its mass will also decrease proportionally with a density as seen in the equation.

m = density*v

To understand this concept more clearly, let's use the following example:

We know that the density of water is equal to 1000 [kg/m^3], that is, 1 cubic meter of water contains 1000 kilograms of water, using the equation.

1000 = m /1

m = 1000*1 = 1000 [kg]

Now if we have 500 kilograms of water, that would pass with the volume so that the density remains constant.

1000 = 500/v

v = 500/1000

v = 0.5 [m^3]

We can see that the volume of water has halved. Since the mass of water was reduced by half. That is, the relationship between mass and volume is proportional to the density of the material or substance.

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If it requires 8.0 J of work to stretch a particular spring by 2.0 cm from its equilibrium length, how much more work will be re
jenyasd209 [6]

Answer:

The amount of work done required to stretch spring by additional 4 cm is 64 J.

Explanation:

The energy used for stretching spring is given by the relation :

E = \frac{1}{2}kx^{2}           .......(1)

Here k is spring constant and x is the displacement of spring from its equilibrium position.

For stretch spring by 2.0 cm or 0.02 m, we need 8.0 J of energy. Hence, substitute the suitable values in equation (1).

8 = \frac{1}{2}\timesk\times k \times(0.02)^{2}

k = 4 x 10⁴ N/m

Energy needed to stretch a spring by 6.0 cm can be determine by the equation (1).

Substitute 0.06 m for x and 4 x 10⁴ N/m for k in equation (1).

E = \frac{1}{2}\times4\times10^{4}\times (0.06)^{2}

E = 72 J

But we already have 8.0 J. So, the extra energy needed to stretch spring by additional 4 cm is :

E = ( 72 - 8 ) J = 64 J

7 0
3 years ago
or Ethernet, if an adapter determines that a frame it has just received is addressed to a different adapter it discards the fram
Artist 52 [7]

For Ethernet, if an adapter determines that a frame it has just received is addressed to a different adapter

a. it discards the frame without sending an error message to the network layer

b. it sends a NACK (not acknowledged frame) to the sending host

c. it delivers the frame to the network layer, and lets the network layer decide what to do

d. it discards the frame and sends an error message to the network layer

Answer:

Option A

Explanation:

The nodal address has to match the signal message address for it to function well but if the it doesn't match the nodal receiver address, it disregards it.

5 0
3 years ago
block of mass 0.5kg on a horizontal surface is attached to a horizontal spring of negligible mass and spring constant 50N/m . Th
Alisiya [41]

Answer:

Explanation:

The mass of the block is 0.5kg

m = 0.5kg.

The spring constant is 50N/m

k =50N/m.

When the spring is stretch to 0.3m

e=0.3m

The spring oscillates from -0.3 to 0.3m

Therefore, amplitude is A=0.3m

Magnitude of acceleration and the direction of the force

The angular frequency (ω) is given as

ω = √(k/m)

ω = √(50/0.5)

ω = √100

ω = 10rad/s

The acceleration of a SHM is given as

a = -ω²A

a = -10²×0.3

a = -30m/s²

Since we need the magnitude of the acceleration,

Then, a = 30m/s²

To know the direction of net force let apply newtons second law

ΣFnet = ma

Fnet = 0.5 × -30

Fnet = -15N

Fnet = -15•i N

The net force is directed to the negative direction of the x -axis

8 0
3 years ago
Why does the surfaces of the moon, mars and earth vary in appearance
Murrr4er [49]

Answer:

due to its distance ..............

4 0
3 years ago
Just a thought
Olin [163]

Answer:

Explanation:

THE FOURTEEN WEEKS' COURSES

IN

NATURAL SCIENCE,

BY

J. DORMAN STEELE, A.M., PHI.D.

Fourteen Weeks iq Natural Philosophy,

Fourteen Weeks iq Ctlenqistry,

Fourteen Weeks iq Descriptive Astroqonqy,

Fourteel Weeks iq Popular Geology,

Fourteeq Weeks iQ2 Human P1ysiology,

Fourteen Weeks iq Zoology,

Fourteeq Weeks iq Botany,

A Key, containing Answers to the Questions

and Problems in Steele's I4 Weeks' Courses,

4 1ISTORIC4L SERIES,

ON THE PLAN OF STEELE'S 14 WEEKS IN THE SCIENCES.

A Brief History of the Urlited States,

A Brief History of France,

The same publishers also offer the following standard scientific

vworks, being more extended or difficult treatises than those of

Prof. Steele, though still of Academic grade.

Peck's Ganot's Natural Philosophy,

Porter's Principles of Chemistry,

Jarvis' Physiology and Laws of Healtl,

Wood's Botanist and Florist,

Clanlbers' Elenments of Zoology,

lcIqtyre's Astroqomy and tle Globes,

Page's. Elen~ents of Geology,

Entered according to Act of Congress, in the year 1869, by

A. S. BARNES & CO.,

In the Clerk's Office of the District Court of the United States

for the Southern District of New York.

sTERLE'S KEY.

6 0
2 years ago
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