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Viefleur [7K]
2 years ago
10

What is called the process of finding exact quantity of a substance?​

Physics
1 answer:
Tema [17]2 years ago
7 0

Answer:

Measurement is called the process of finding exact quantity of a substances.

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A swimmer takes 45 seconds to complete a 100-meter lap. Calculate the swimmer’s average speed.
g100num [7]
2.22 m/s. Average speed is Total distance over time it took. 100m/45s = 2.22m/s
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Give a example of frequency which an elephant can hear and a tiger cannot hear include a unit in ansawer
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write a good scientific question that can be answered by Guadalupe based on what is described above. Label the independent and d
kakasveta [241]
To do that, we would need to know what is "described above".
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In Fig.25-46,how much charge is stored on the parallel-plate capacitors by the 12.0 V battery? One is filled with air,and the ot
Fittoniya [83]

Answer:

Q=7.9\times 10^{-10}\ C

Explanation:

Given that

V= 12 V

K=3

d= 2 mm

Area=5.00 $ 10#3 m2

Assume that

$ = Multiple sign

# = Negative sign

A=5\times 10^{-3}\ m^2

We Capacitance given as

For air

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{\varepsilon _oA}{d}

C_1=\dfrac{8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}

C_1=2.2\times 10^{-11}\ F

C_2=\dfrac{K\varepsilon _oA}{d}

C_2=\dfrac{3\times 8.85\times 10^{-12}\times 5\times 10^{-3}}{2\times 10^{-3}}\ F

C_2=6.6\times 10^{-11}\ F

Net capacitance

C=C₁+C₂

C=8.8\times 10^{-11}\ F

We know that charge Q given as

Q= C V

Q=12\times 6.6\times 10^{-11}\ C

Q=7.9\times 10^{-10}\ C

6 0
3 years ago
an astronaut weighs 8.00x10^2 on the surface of earth. what is the weight of the astronaut 6.37x10^6 meters above the surface of
Marta_Voda [28]
To solve this problem, we use the Law of Universal Gravitation:

F = Gm1m2/d^2

where m1 and m2 are two objects. In this case, earth and man. d is the distance between the objects. Lastly, G is the gravitational constant. Since the mass of the earth and man are constant, this is lumped up with G into k. The equation would be:

F = k/d^2
k = Fd^2
F_{1} d_{1} ^{2} =F_{2} d_{2} ^{2}

The radius of earth, d1, is equal to 6.371E+6 m. Thus, d2 = 2d1

(8E+2)(d1)^2 = F2(2d1)^2

(8E+2)(d1)^2 = 4F2(d1)^2

(8E+2)=4F2
F2 = 200 Newtons
4 0
3 years ago
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