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3 years ago

How do you open a door if its not locked?

1 answer:

4 0

If you're trying to get through a door with a standard spring lock—not a deadbolt—you can use a credit card or other flexible plastic card to open it. Just wedge the card into the gap between the door and the frame and bend it away from the knob. If it's the right kind of lock, it should come right open.

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Biological systems use free energy based on empirical data that all organisms require a constant energy input. The first law of

dem82 [27]

Answer:

Explanation:

7 0

3 years ago

On the basis, a given figure, calculate rfficiency of the lever.

daser333 [38]

Explanation:

here,load=100N

effort=200N

distance travelled by load=2cm

distance travelled by effort=10cm

Now, output work=L*Ld

=100*2

=200g

Again,

input work =E*Ed

200*10

2000j

again, efficiency = output work/input work*100%

200/2000*100%

=10%

please don't forget to write symbol

4 0

3 years ago

The first thing that made Trestail curious about the Crippen case was that __________.

Anna35 [415]

Possibly the third one down. I don't think that DNA was around in Crippen's time. I'm not a criminologist or associated fun and games, though.

7 0

3 years ago

If we are viewing the atom in such a way that the electron's orbit is in the plane of the paper with the electron moving clockwi

Anastaziya [24]

Answer:

The magnitude of the electric field is $5.1 \times 10^{11} \frac{N}{C}$

Explanation:

Given:

Charge of electron $q = 1.6 \times 10^{-19}$ C

Separation between two charges $r = 5.3 \times 10^{-11}$ m

For finding the magnitude of the electric field,

$E= \frac{kq}{r^{2} }$

Where $k = 9 \times 10^{9}$

$E = \frac{9 \times 10^{9} \times 1.6 \times 10^{-19} }{(5.3 \times 10^{-11} )^{2} }$

$E = 5.1 \times 10^{11}$ $\frac{N}{C}$

Therefore, the magnitude of the electric field is $5.1 \times 10^{11} \frac{N}{C}$

5 0

4 years ago

Steam at 400C has a specific volume of 0.02m3/kg. Determine the pressure of the steam based on a) the ideal gas equation b) the

nikklg [1K]

Answer:

by ideal gas pressure = 15529.475 kPa

by compressibility chart pressure = 12576 kPa

by steam tables Pressure = 12517 kPa

Explanation:

given data

temperature T = 400°C = 673 K

volume v = 0.02 m³/kg

to find out

pressure by ideal gas, compressibility chart and steam tables

solution

we know here by table

gas constant R is 0.4615 kJ/ kg-K

and critical temp Tc = 647.1 K

and critical pressure Pc = 22064 kPa

so by ideal gas pressure is

pressure = R×T / v

pressure = 0.4615 × 673 / 0.02

pressure = 15529.475 kPa

and

by compressibility chart

temperature reduce is = T/ Tc

temperature reduce Tr = 673 / 647.1

Tr = 1.040 K

so pseudo reduce volume is here

reduce volume Vr = v / ( RTc/Pc)

reduce volume Vr = $\frac{0.02}{\frac{461.5(647.1)}{22064*10^{3} } }$

0.02 / ( 461.5(647.1) / 22064×10³)

reduce volume = 1.48

and we know by compressibility chart

reduce pressure Pr is 0.57

pressure = Pr × Pc

pressure = 0.57 × 22064 × 10³

pressure = 12576 kPa

and

from steam table

pressure is 12.5 MPa at 673 K and 0.020030 m³/kg

pressure is 15 MPa at 673 K and 0.015671 m³/kg

pressure P is

$\frac{0.02 - 0.020030}{0.015671 - 0.020030}$ = $\frac{ P - 12.5}{15 - 12.5}$

Pressure = 12517 kPa

4 0

3 years ago