Jak wpłynie na czas oczekiwania na pierwsze ogrzanie wody (pytanie 1) zwiększenie mocy

A force of 8800 N is exerted on a piston that has an area of 0.010 m^2. What is the force exerted by a second piston that has an

professor190 [17]

Explanation:

It is given that,

Force on piston, F₁ = 8800 N

Area, $A_1=0.01\ m^2$

Area, $A_2=0.04\ m^2$

Let F₂ is the force exerted on the second piston. Using Pascal's law as :

Pressure at piston 1 = Pressure at piston 2

$\dfrac{F_1}{A_1}=\dfrac{F_2}{A_2}$

$F_2=\dfrac{F_1}{A_1}\times A_2$

$F_2=\dfrac{8800}{0.01}\times 0.04$

$F_2=35200\ N$

So, the force exerted by a second piston is 35200 N. Hence, this is the required solution.

8 0

3 years ago

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IN WHAT WAY HAS TECHNOLOGY USED ABOARD THE INTERNATIONAL SPACE STATION BENEFITTED HUMANS BACK ON EARTH?

drek231 [11]

Answer

The ways are;

• By commercializing low earth orbit

• By supporting water purification efforts worldwide

• By bringing space station ultrasound equipment on earth

• By improving eye surgery with space hardware

Explanation

There is a revolutionary commercial pathways that is currently opening access to space.For example, NASA is purchasing commercial cargo resupply to space station which enable U.S business to develop a competition in selling services to others while freeing NASA resources for deep space exploration. Using technology developed in space, the Water Security Corporation has deployed systems using NASA water-processing technology to ensure drinkable water availability for human survival. Medical care in remote regions is now facilitated by use of small ultrasound units, tele-medicine and remote guided methods similar to those used in space. The Eye Tracking Device experiment has enable researchers to have an insight into corrective laser surgeries.

6 0

3 years ago

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What is the oscillation period of your eardrum when you are listening to the A4 note on a piano (frequency 440 Hz)?

Contact [7]

The period is the inverse of the frequence:

$T=\dfrac{1}{f}=\dfrac{1}{440}=0.0022\overline{72}$

4 0

3 years ago

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You charge an initially uncharged 65.7-mf capacitor through a 39.1-Ï resistor by means of a 9.00-v battery having negligible int

uysha [10]

In a RC-circuit, with the capacitor initially uncharged, when we connect the battery to the circuit the charge on the capacitor starts to increase following the law:
$Q(t) = Q_0 (1-e^{-t/\tau})$
where t is the time, $Q_0 = CV$ is the maximum charge on the capacitor at voltage V, and $\tau = RC$ is the time constant of the circuit.
Using this law, we can answer all the three questions of the problem.

1) Using $R=39.1 \Omega$ and $C= 65.7 mF=65.7\cdot 10^{-3}F$ , the time constant of the circuit is:
$\tau = RC=(39.1 \Omega)(65.7 \cdot 10^{-3}F)=2.57 s$

2) To find the charge on the capacitor at time $t=1.95 \tau$ , we must find before the maximum charge on the capacitor, which is
$Q_0 = CV=(65.7 \cdot 10^{-3}F)(9 V)=0.59 C$
And then, the charge at time $t=1.95 \tau$ is equal to
$Q(1.95 \tau) = Q_0 (1-e^{-t/\tau})=(0.59 C)(1-e^{-1.95})=0.51 C$

3) After a long time (let's say much larger than the time constant of the circuit), the capacitor will be fully charged, this means its charge will be $Q_0 = 0.59 C$ . We can see this also from the previous formule, by using $t=\infty$ :
$Q(t) = Q_0 (1-e^{-\infty})=Q_0(1-0) = 0.59 C$

4 0

3 years ago

abody starts from rest and accelerate uniformly at 8m/s/s.calculate the distance travelled by the body in 10s?

vagabundo [1.1K]

Explanation:

using the formula: S=ut+½gt², where u=0, S=?, g=8m/s², t=10seconds.

S=ut+½gt² ("ut" term will cancel because u=0).

=> S= ½gt²

=>S = ½×8×10²

=>S = 4×100

=>S = 400m .

Therefore, the distance traveled by the body in 10s is 400m.

hope this helps you.

6 0

2 years ago