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3 years ago

What is the difference between methane and fluorinated gases

1 answer:

6 0

Methane is emitted during the production and transport of coal,natural gas,and oil,while fluorinated gases are emitted from a variety industrial processes.

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Water displacement is used to find what? a. volume b. density

Alex

Answer:

Volume

Explanation:

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4 0

2 years ago

Astronauts often undergo special training in which they are subjected to extremely high centripetal accelerations. One device ha

egoroff_w [7]

The centripetal acceleration of an object is given by the relation,

$Ac =V^2/R$

where Ac = centripetal acceleration = $98 m/s^2$

R = radius of rotation = 15 m

V = speed of astronaut

Hence, $\frac{V^2}{15} =98$

solving this we get, V = 38.34 m/s

3 0

3 years ago

Read 2 more answers

A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor

Allisa [31]

Answer:

the impedance of the circuit is 25.7 ohms.

Explanation:

It is given that,

Voltage, V = 50 volts

Frequency, f = 60 Hz

Resistance, R = 25 ohms

Capacitive resistance, $X_C=18\ ohms$

Inductive resistance, $X_L=24\ ohms$

We need to find the impedance of the circuit. It is given by :

$Z=\sqrt{R^2+(X_L-X_C)^2}$

$Z=\sqrt{25^2+(24-18)^2}$

Z = 25.7 ohms

So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.

6 0

3 years ago

Which adaptation is likely to increase the chances of survival of an animal in a rainforest?

34kurt

Black-spotted skin coat as camouflage while stalking prey.
Survival = avoiding predators or capturing prey successfully

6 0

3 years ago

Read 2 more answers

A heat pump has a coefficient of performance that is 60% of the Carnot heat pump coefficient of performance. The heat pump is us

Nataliya [291]

Answer:

$T_C$ =118.8 K= 154.2°C

Explanation:

COP_max of carnot heat pump= $\frac{T_{H} }{T_{H}-T_{C} }$

where T_H and T_C are temperatures of hot and cold reservoirs

Also COP= $\frac{Q_H}{W}$

in the question $COP= \frac{60}{100} \times COP_{max}$

⇒ $\frac{Q_H}{W} =\frac{60}{100}\times\frac{T_H}{T_H-T_C}$

heat is added directly to be as efficient as via heat pump

$Q_H= W$

and T_H= 24° C= 297 K

$1=\frac{60}{100}\times \frac{297}{297-T_C}$

on calculating the above equation we get

$T_C$ =118.8 K

the outdoor temperature for efficient addition of heat to interior of home

$T_C$ =118.8 K= 154.2°C

6 0

3 years ago