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Alenkinab [10]
3 years ago
12

You are designing an office building. In order to reduce noise from high-frequency sounds between offices above and below each o

ther, would you choose carpet on concrete or wood floors? Explain your answer.
Chemistry
2 answers:
Monica [59]3 years ago
8 0

Answer:

concrete has more mass and it will block more noise than than the wood plus the wood is thiner the thicker the substance is the better u won't hear noise

Explanation:

miskamm [114]3 years ago
7 0
Carpet for sure. Have you ever heard how loud a dog walking on wood compared to carpet is.
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Which type of electromagnetic waves are dangerous enough to be used to kill cancerous cells?
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Gamma rays

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2 years ago
Compare What do infrared technologies, fiber optic
LuckyWell [14K]

Answer:

ptic fiber communication and satellite communication are the leading technologies which are revolutionizing the world of telecommunications. Both technologies have their advantages and limitations which make them suitable for certain type of applications. This article will provide an overview of optic fiber and satellite communication technologies and present a comparison of the features and related issues.

Optic Fiber Communication

Optic Fiber communication transmits information by sending pulses of light (using laser) through an optic fiber. The low signal loss in optic fibers and high data rate of transmission systems, allow signals with high data rates (exceeding several Gbps) to travel over long distances (more than 100 km) without a need of repeater or amplifier. Moreover, using wavelength division multiplexing (WDM) allows a single fiber to carry multiple signals (upto 10 different signals) of multi-Gbps transmissions. Optic Fiber communication offers extremely high bandwidth, immunity to electromagnetic interference, non-existent delays and immunity from interception by external means. In the 1980s and 1990s, the continents were linked together using undersea optic fiber bringing about a paradigm shift in the global telecommunications.

These advancements in optic fiber communication has resulted in decrease of satellite communications for several types of communications. For instance, transmission between fixed locations or point-to-point communications, where large bandwidths are required (such as transoceanic telephone systems) are made through optic fiber instead of using satellite communication. Optic Fiber communication is also used to transmit telephone signals, Internet communication, LAN (Gigabit LAN) and cable television signals.

Satellite Communication

Satellite communications use artificial satellites as relays between a transmitter and a receiver at different locations on Earth. Satellite systems allow users to bypass typical carrier offices and to broadcast information to multiple locations. Communications satellites are used for radio, TV, telephone, Internet, military and other applications. There are more than 2,000 satellites around Earth’s orbit, being used for communication by both government and private organizations.

Communication Satellites are LOS (line-of-sight) microwave systems with a repeater. These satellites rotate around the earth with the speed of earth and are known as geostationary satellites. The limitations of antenna size also limits focusing capability making the coverage for a single satellite transmitter very large. This makes satellite communication ideal for TV and radio services as the signal has to flow from a single point to many points in a single direction. The large distance of satellites from the earth (about 22,300 miles) results in delays which adversely effects two-way communication like mobile conversations. Low earth orbit satellites can be used for two-way mobile communication because less power is required to reach those satellites.

Explanation:

6 0
3 years ago
A 52.0 mL portion of a 1.20 M solution is diluted to a total volume of 278 mL. A 139 mL portion of that solution is diluted by a
RoseWind [281]

Answer:

C_3=0.125M

Explanation:

Hello!

In this case, we can divide the problem in two steps:

1. Dilution to 278 mL: here, the initial concentration and volume are 1.20 M and 52.0 mL respectively, and a final volume of 278 mL, it means that the moles remain the same so we can write:

V_1C_1=V_2C_2

So we solve for C2:

C_2=\frac{C_1V_1}{V_2}=\frac{52.0mL*1.20M}{278mL}\\\\C_2=0.224M

2. Now, since 111 mL of water is added, we compute the final volume, V3:

V_3=139+111=250mL

So, the final concentration of the 139 mL portion is:

C_3=\frac{139 mL*0.224M}{250mL}\\\\C_3=0.125M

Best regards!

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