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myrzilka [38]
3 years ago
6

Electricity seems like a fundamental element in the construction of any home. However, to have electricity run into and out of a

home, there must be a complete circuit. Write a brief essay describing the minimum requirements for any electric circuit. Within your essay, describe the task for each of these elements using your lessons and experience within this virtual lab. 150 words max
Physics
1 answer:
olganol [36]3 years ago
8 0

Answer:

•Most of the components that make up curcuits are conductors. Electricity flows easily thought conductors such as metal or water.Electricity does not flow easily thought insulators such as wood or rubber.

•When a battery is used to power a circuit, the circuit the current must travel from one pole of the battery to the opposite battery.

•Resisotors show the flow of Electricity throught a circuit this can prevent components of the circuit from 'brning out'. In a curcuit diagram, resistors are represented by bending lines.

Explanation:

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A satellite is in a circular orbit about the earth (ME = 5.98 x 10^24 kg). The period of the satellite is 1.26 x 10^4 s. What is
Soloha48 [4]

Answer: V=5839.051m/s  

Explanation:

According to the <u>Third Kepler’s Law</u> of Planetary motion:

T^{2}=\frac{4\pi^{2}}{GM}a^{3}   (1)

Where;:

T=1.26(10)^{4}s is the period of the satellite

G is the Gravitational Constant and its value is 6.674(10)^{-11}\frac{m^{3}}{kgs^{2}}

M=5.98(10)^{24}kg is the mass of the Earth

a  is the semimajor axis of the orbit the satllite describes around the Earth (as we know it is a circular orbit, the semimajor axis is equal to the radius of the orbit).

On the other hand, the orbital velocity V is given by:

V=\sqrt{\frac{GM}{a}}   (2)

Now, from (1) we can find a, in order to substitute this value in (2):

a=\sqrt[3]{\frac{T^{2}GM}{4\pi}^{2}}   (3)

a=\sqrt[3]{\frac{(1.26(10)^{4}s)^{2}(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{4\pi}^{2}}   (4)

a=11705845.57m   (5)

Substituting (5) in (2):

V=\sqrt{\frac{(6.674(10)^{-11}\frac{m^{3}}{kgs^{2}})(5.98(10)^{24}kg)}{11705845.57m}}   (6)

V=5839.051m/s   (7)  This is the speed at which the satellite travels

6 0
3 years ago
Galileo, Giotto, Magellan, Mariner 2 These are all _________________ used to travel out of Earth's orbit to explore places that
kykrilka [37]
Probes,and engineered unmanned space technology..
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3 years ago
What does gravity do to objects as they:<br><br> a) are thrown up:<br><br> b) fall toward earth:
tangare [24]

Answer:

B

Explanation:

Gravity is when objects are pulled down to earth

7 0
3 years ago
A 24 kg child slides down a 3.3-m-high playground slide. She starts from rest, and her speed at the bottom is 3.0 m/s.a. What en
Gelneren [198K]

Answer:

(a) Potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

(b) U=668.16\ J

Explanation:

Given:

  • mass of the child, m=24\ kg
  • height of the slide, h=3.3\ m
  • initial velocity of the child at the slide, v_i=0 m.s^{-1}
  • final velocity of the child at the bottom of slide, v_f=3\ m.s^{-1}

(a)

∴The initial potential energy of the child is converted into the kinetic energy at the bottom off the slide and a part of which is lost into friction generating heat between the contact surfaces.

Initial potential energy:

PE=m.g.h

PE=24\times 9.8\times 3.3

PE=776.16\ J

Kinetic energy at the bottom of the slide:

KE=\frac{1}{2} m.v^2

KE= 0.5\times 24\times 3^2

KE= 108\ J

(b)

Now, the difference in the potential and kinetic energy is the total change in the thermal energy of the slide and the seat of her pants.

This can be given as:

U=PE-KE

U=776.16-108

U=668.16\ J

4 0
3 years ago
If it requires 4.5 J of work to stretch a particular spring by 2.3 cm from its equilibrium length, how much more work will be re
saveliy_v [14]

Answer:

\Delta W=24.1162\ J

Explanation:

Given:

  • work done to stretch the spring, W=4.5\ J
  • length through which the spring is stretched beyond equilibrium, \Delta x=2.3\ cm=0.023\ m
  • additional stretch in the spring length, \delta x=3.5\ cm=0.035\ m

<u>We know the work done in stretching the spring is given as:</u>

W=\frac{1}{2} \times k.\Delta x^2

where:

k = stiffness constant

4.5=0.5\times k\times 0.023^2

k=17013.2325\ N.m^{-1}

Now the work done in stretching the spring from equilibrium to (\Delta x+\delta x):

W'=0.5\times k.(\Delta x+\delta x)^2

W'=0.5\times 17013.2325\times 0.058^2

W'=28.6162\ J

So, the amount of extra work done:

\Delta W=W'-W

\Delta W=28.6162-4.5

\Delta W=24.1162\ J

4 0
3 years ago
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