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3 years ago

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A 20-coil spring has a spring constant of 100 N/m. Suppose you cut this spring in half. That is, there are two 10-coil springs.

Find the spring constant of the 10-coil spring.

1 answer:

Tom [10]3 years ago

5 0

Just need the points lol idk

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If a certain shade of blue has a frequency of 7.33 x 10^14Hz what is the energy of exactly one photon of this light. Is this as

geniusboy [140]

In calculating the energy of a photon of light, we need the relationship for energy and the frequency which is expressed as:

E=hv

where h is the Planck's constant (6.626 x 10-34 J s)and v is the frequency.
E = 6.626 x 10-34 J s (<span>7.33 x 10^14 /s) = 4.857 x 10^-19 J</span>

6 0

3 years ago

- What are (a) the x component, (b) the y component, and (c) the z component of r = a - b +c if a= 7.8 + 6.6 - 7.1 , b= -2.9 + 7

frez [133]

Answer:

Explanation:

a= 7.8i + 6.6j - 7.1k

b= -2.9 i+ 7.4 j+ 3.9k , and

c = 7.6i + 8.8j + 2.2k

r = a - b +c

=7.8i + 6.6j - 7.1k - ( -2.9i + 7.4j+ 3.9k )+ ( 7.6i + 8.8j + 2.2k)

= 7.8i + 6.6j - 7.1k +2.9i - 7.4j- 3.9k )+ 7.6i + 8.8j + 2.2k

= 18.3 i +18.3 j - k

the angle between r and the positive z axis.

cosθ = 18.3 / √18.3² +18.3² +1

the angle between r and the positive z axis.

= 18.3 / 25.75

cos θ= .71

45 degree

6 0

3 years ago

Five hundred million years ago, basaltic lava flowed in an area now known as Monticello, the historic home of Thomas Jefferson.

aev [14]

Explanation:

Igneous - metamorphic - sedimentary

A rock cycle provides the cyclic transformation of one rock type to another in nature.

There are three main types of rock involved in the rock cycle;

igneous rocks are derived from the cooling and solidification of molten magma
metamorphic rocks are changed rocks subjected to intense pressure and temperature
sedimentary rocks are derived from rock sediments that have been lithified.

The history of the rock in Monticello begins with igneous rock formation. Basalt is an igneous rock that forms from the cooling and solidification of molten magma. Under intense pressure and temperature regimes, they are changed to metamorphic rocks.

Agents of denudation such as wind, water and glacier weathers the rock and disintegrates it. They are then carried into basins where they are deposited. Here they form sedimentary rock.

The process still goes on as the sedimentary rock gets taken into depth, they can either melt to form igneous rock or be changed to metamorphic rocks.

learn more:

metamorphic process brainly.com/question/869769

sedimentary rocks brainly.com/question/9131992

#learnwithBrainly

5 0

3 years ago

Calculate the magnitude of the electric field inside the solid at a distance of 9.50 cm from the center of the cavity. Express y

WITCHER [35]

Question:

A point charge of -2.14uC is located in the center of a spherical cavity of radius 6.55cm inside an insulating spherical charged solid. The charge density in the solid is 7.35×10−4 C/m^3.

a) Calculate the magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity.

Express your answer using two significant figures.

Answer:

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity $3.65\times 10^5N/C$

Explanation:

A point charge ,q = $-2.14\times 10^{-6} C$ is located in the center of a spherical cavity of radius , $r =6.55\times 10^{-2}$ m inside an insulating spherical charged solid.

The charge density in the solid , d = $7.35 \times 10^{-4}C/m^3.$

Distance from the center of the cavity,R = $9.5\times 10^{-2 }m$

Volume of shell of charge= V = $(\frac{4\pi}{3})[ R^3 - r^3 ]$

Charge on the shell ,Q = $V \times d'$

$Q =(\frac{4\pi}{3})[ R^3 - r^3 ] \times d$

$Q = 4.1888*\times 10^{-4 }[8.57375 - 2.81011 ]\times 7.35\times 10^{-4}$

$Q = 4.1888\times 10^{-4} [5.76364 ] \times 7.35 \times 10^{-4}$

$Q =2.4143 \times 10^{-4} \times 7.35 \times 10^ { -4}$

$Q =1.7745 \times 10^{-6 }C$

Electric field at $9.5\times 10^{-2}$ m due to shell $E1 = \frac{k Q}{R^2}$

E1 = $\frac{ 9 \times 10^9\times 1.7745\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E1 =1.769\times 10^6 N/C$

Electric field at $9.5\times 10^{-2}$ due to 'q' at center $E2 = \frac{kq}{R^2}$

E2 = $\frac{ - 9 \times 10^9\times 2.14\times 10^{-6 }}{ 90.25\times 10^{-4}}$

$E2 =2.134\times 10^6 N/C$

The magnitude of the electric field inside the solid at a distance of 9.50cm from the center of the cavity

= E2- E1

$=[ 2.134 - 1.769 ]\times 10^6$

$= 3.65\times 10^5 N/C$

8 0

3 years ago

1. Space probes have not landed on Pluto yet. Describe three types of information you would collect if you were designing the pr

konstantin123 [22]

Answer:

Space probes are made to conduct science experiments. They do not have people on them. Space probes have helped scientists get information about our solar system. Most probes are not designed to return to Earth. Some have landed on other planets! Others have flown past the planets and taken pictures of them for scientists to see. There are even some space probes that go into orbit around other planets and study them for a long time. The information they gather is used to help us understand the weather and other changes which happen on planets other than the Earth. This information is important in helping to plan other space missions such as ones to Mars and to Saturn.

Explanation:

7 0

3 years ago