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3 years ago

7

Alguien me ayuda con calculos estequiometricos?

1 answer:

ruslelena [56]3 years ago

5 0

Answer:

SURE!!!...

But what to calculate!!!....

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RC time constant circuit if R 50 KOC-21 a TOSS c. 1.05 s . what is the expected RC value b. 10.55 d. 0.105 s

Afina-wow [57]

Answer:

Time constant of RC circuit is 0.105 seconds.

Explanation:

It is given that,

Resistance, $R=50\ K\Omega=5\times 10^4\ \Omega$

Capacitance, $C=2.1\ \mu F=2.1\times 10^{-6}\ F$

We need to find the expected time constant for this RC circuit. It can be calculated as :

$\tau=R\times C$

$\tau=5\times 10^4\times 2.1\times 10^{-6}$

$\tau=0.105\ s$

So, the time constant for this RC circuit is 0.105 seconds. Hence, this is the required solution.

7 0

3 years ago

A stone is dropped from the

ICE Princess25 [194]

Height=h=500m
Acceleration=g=10m/s^2
Initial velocity=u=0
Speed of sound=c=340m/s
TIME TAKEN BY STONE TO HIT WATER=t
Time taken by sound to hear back=T

Now

$\\ \sf\longmapsto h=ut+\dfrac{1}{2}gt^2$

$\\ \sf\longmapsto h=0t+\dfrac{1}{2}10t^2$

$\\ \sf\longmapsto 500=5t^2$

$\\ \sf\longmapsto t^2=100$

$\\ \sf\longmapsto t=10s$

Now

$\\ \sf\longmapsto h=cT$

$\\ \sf\longmapsto T=\dfrac{h}{c}$

$\\ \sf\longmapsto T=\dfrac{500}{340}$

$\\ \sf\longmapsto T=1.47\approx 1.5s$

Total time:-

$\\ \sf\longmapsto T_{net}=t+T=10+1.5=11.5s$

8 0

2 years ago

NEED HELP ASAP!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

jasenka [17]

Answer:

She pulled the scarf from her neck and wiped her face.

4 0

3 years ago

Which of the following supports the theory of continental drift?

Y_Kistochka [10]

I think it's The fossil record. The same animal fossil is in Africa and South America. The animal could have not swim across so its the fossil record

7 0

3 years ago

The glass in a window is 35 inches wide and 20 inches tall, and standard atmospheric pressure is 14.7 pounds per square inch. Wh

iogann1982 [59]

Answer:103 pounds

Explanation:

Given

width of window $b=35 in.$

height of window $h=20 in.$

standard atmospheric pressure $P_{outside}=14.7 psi$

Also $P_{inside}=1.01P_{outside}$

Thus Net Force on the window will be the algebraic sum of Force due to outside and inside Pressure .

$F_{net}=(P_{inside}-P_{outside})\cdot A$

$F_{net}=P_{outside}(1.01-1)\times 35\times 20$

$F_{net}=14.7\times 0.01\times 35\times 20$

$F_{net}=102.9\ pounds\approx 103\ pounds$

8 0

3 years ago