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Natasha_Volkova [10]
3 years ago
10

The metric system is now known as the international system of units. a. T b. F

Physics
1 answer:
inn [45]3 years ago
6 0
<span>The metric system is the oldest name for the international system of units. The answer is <u>a. True. </u>SI unit or the international systems of units are based on seven basic units; the meter, kilogram, second, ampere, Kelvin, candela and mole. All of these basic units are divided into multiples by a power of ten. For example in meters, 1 meter is equal to: 1000 millimeter, 100 centimeter, 10 decimeter, 0.1 decameter, 0.01 hectometer, 0.001 kilometer and so on and so forth.</span>
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A pendulum consisting of a 0.5 kg mass tied to a 0.3 m string is set into oscillation at the same moment that a stone is dropped
lara31 [8.8K]

Answer:

2.72 cycles

Explanation:

First of all, let's find the time that the stone takes to reaches the ground. The stone moves by uniform accelerated motion with constant acceleration g=9.8 m/s^2, and it covers a distance of S=44.1 m, so the time taken is

S=\frac{1}{2}at^2\\t=\sqrt{\frac{2S}{a}}=\sqrt{\frac{2(44.1m)}{9.8 m/s^2}}=3 s

The period of the pendulum instead is given by:

T=2 \pi \sqrt{\frac{L}{g}}=2 \pi \sqrt{\frac{0.3 m}{9.8 m/s^2}}=1.10 s

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3 years ago
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The membrane that surrounds a certain type of living cell has a surface area of 5.1 x 10-9 m2 and a thickness of 1.4 x 10-8 m.
ziro4ka [17]

a) The charge on the outer surface is 1.2\cdot 10^{-12} C

b) The number of ions is 7.5\cdot 10^6

Explanation:

a)

The membrane behaves as a parallel plate capacitor, whose capacity is given by the equation

C=\frac{k\epsilon_0 A}{d}

where

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d=1.4\cdot 10^{-8} m is the distance between the two plates

Substituting,

C=\frac{(4.3)(8.85\cdot 10^{-12})(5.1\cdot 10^{-9})}{1.4\cdot 10^{-8}}=1.4\cdot 10^{-11} F

The capacity of the membrane is related to the potential difference between the two surfaces by

C=\frac{Q}{\Delta V}

where here we have

Q = excess charge on one surface

\Delta V = 85.5 mV = 0.0855 V is the potential difference between the two surfaces

Solving for Q, we find

Q=C\Delta V=(1.4\cdot 10^{-11})(0.0855)=1.2\cdot 10^{-12} C

b)

We said that the net charge on the outer surface is

Q=1.2\cdot 10^{-12} C

The charge of one K+ ions is equal to the electron charge

+e=1.6\cdot 10^{-19} C

Therefore, the number of ions on the outer surface can be found by dividing the total charge by the charge of a single ion:

N=\frac{Q}{e}=\frac{1.2\cdot 10^{-12}}{1.6\cdot 10^{-19}}=7.5\cdot 10^6

Learn more about capacity:

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