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miss Akunina [59]

2 years ago

6

What problem did the crew of Biosphere 2 face?

1 answer:

konstantin123 [22]2 years ago

3 0

Answer: A loss of oxygen

Explanation: There wasn’t enough reliable oxygen for the researchers to inhale

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An ideal vapor-compression refrigeration cycle that uses refrigerant-134a as its working fluid maintains a condenser at 800 kpa

AlladinOne [14]

Answer:

P= 30.87 kw

Explanation:

find the solution below

7 0

4 years ago

Read 2 more answers

A 4.55 nF parallel-plate capacitor contains 27.5 μJ of stored energy. By how many volts would you have to increase this potentia

aivan3 [116]

Answer:

$\Delta V=V_{2}-V_{1}=45.4V$

Explanation:

The energy, E, from a capacitor, with capacitance, C, and voltage V is:

$E=\frac{1}{2} CV^{2}$

$V=\sqrt{2E/C}$

If we increase the Voltage, the Energy increase also:

$V_{1}=\sqrt{2E_{1}/C}$

$V_{2}=\sqrt{2E_{2}/C}$

The voltage difference:

$V_{2}-V_{1}=\sqrt{2E_{2}/C}-\sqrt{2E_{1}/C}$

$V_{2}-V_{1}=\sqrt{2*55*10^{-6}/4.55*10^{-9}}-\sqrt{2*27.5*10^{-6}/4.55*10{-9}}=45.4V$

5 0

3 years ago

Two large parallel conducting plates carrying opposite charges of equal magnitude are separated by 2.20 cm. If the surface charg

bekas [8.4K]

Answer:

$5.3\times 10^3 N/C$

Explanation:

We are given that

Distance between plates=d=2.2 cm= $2.2\times 10^{-2} m$

$1 cm=10^{-2} m$

$\sigma=47nC/m^2=47\times 10^{-9}C/m^2$

Using $1 nC=10^{-9} C$

We have to find the magnitude of E in the region between the plates.

We know that the electric field for parallel plates

$E=\frac{\sigma}{2\epsilon_0}$

$E_1=\frac{\sigma}{2\epsilon_0}$

$E_2=\frac{\sigma}{2\epsilon_0}$

$E=E_1+E_2$

$E=\frac{\sigma}{2\epsilon_0}+\frac{\sigma}{2\epsilon_0}=\frac{\sigma}{\epsilon_0}$

Where $\epsilon_0=8.85\times 10^{-12}C^2/Nm^2$

Substitute the values

$E=\frac{47\times 10^{-9}}{8.85\times 10^{-12}}$

$E=5.3\times 10^3 N/C$

Hence, the magnitude of E in the region between the plates= $5.3\times 10^3 N/C$

5 0

3 years ago

A box of mass 20kg is pulled up an inclined plane by a force of 285N. Given that the value of the incline angle is 30 degrees an

docker41 [41]

Given :

Mass of box , m = 250 kg.

Force applied , F = 285 N.

The value of the incline angle is 30°.

the coefficient of dynamic friction is $\mu=0.72$ .

To Find :

The speed with which the box is moving with, assuming it takes 4 seconds to reach the top of the incline.

Solution :

Net force applied in box is :

$F=285 - mgsin\ \theta - \mu mg cos \ \theta\\ \\F=285-mg( sin \ \theta - \mu cos\ \theta)\\\\F=285 - 20\times 10( \dfrac{1}{2}+0.72\times \dfrac{\sqrt{3}}{2})\\\\F=60.29\ N$

Acceleration , $a=\dfrac{F}{m}=\dfrac{60.29}{20}=3.01\ m/s^2$ .

By equation of motion :

$v=u+at\\\\v=0+3.01\times 4\\\\v=12.04\ m/s$

Therefore, the speed of box is 12.04 m/s.

Hence, this is the required solution.

6 0

3 years ago

If you had 6,000 milligrams of rock to move how many kilograms is that?

True [87]

Answer:

0.006

Explanation:

100mg= 0.0001kgghsjsslslkdn d

5 0

3 years ago