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2 years ago

Precision is how close a series of measurements are to ______

Physics

1 answer:

Leya [2.2K]2 years ago

3 0

Answer:

Accuracy is how close a measured value is to an accepted value. Precision is how close measurements are to one another. To make measurements, you have to evaluate both the accuracy and the precision to get a correct value.

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A 80 ohms resistor, 0.2 H inductance and 0.1 mF capacitor are connected in series across a generator (60 Hz, V rms=120 V). Deter

qaws [65]

Answer:

Impedance = 93.75 ohms

Current = 1.81 A

Explanation:

Resistance = R = 80 ohms

Inductance = L = 0.2 H

Inductive reactance = XL = $X_{L}=$ = ωL = (2πf) L

= 2 (3.14) (60)(0.2) = 75.398 Ohms

Capacitive reactance = 1 / ωC = 1/(2πf)C = 1 / [(2π)(60)(0.1 × 10⁻3)]

= 26.526 Ohms

Impedance = Z = $\sqrt{R^{2} + (X_{L} - X_{C})^{^{2}}} =$

= $\sqrt{8788.511}$ = 93.747 ohms

Voltage = $\sqrt{2}$ × 120 = 169.7056 V

Current = I = V ÷ R = (169.7056) ÷ 93,747 = 1.81 A

8 0

3 years ago

Why don't astronauts use a fountain pen or a ball point pen with liquid ink in a spacecraft

mojhsa [17]

It's probably because there is no gravity which ballpoint pens need for inkflow to the tip.

4 0

3 years ago

asambeis [7]

Answer:

22Volts

Explanation:

The pd at the terminal is known as the emf

Since there are Ten 2.2V cells

Terminal voltage = number of cells * pd of one cell

Terminal voltage = 10 * 2.2

Terminal voltage = 22V

Hence the pd at the battery terminals is 22Volts

4 0

2 years ago

a central concept in quantum mechanics is that both matter and are alternate forms of the same entity and therefore both exhibit

Stels [109]

In quantum mechanics, a central concept is that both matter and energy are alternate forms of the same entity and therefore both exhibit dual characteristics of particles and of waves.

Matter can be defined as anything that has mass and is able to occupy space.

Thus, any physical object or substance that is found on Earth is typically composed of matter.

Similarly, energy is highly affected by the mass of a any physical object or substance just like matter,

Hence, both energy and matter are known to be made up of atoms and as a result of this fact, exhibit dual characteristics of particles and of waves.

A wave can be defined as a disturbance in a medium that progressively transports energy from a source location to another location without the transportation of matter.

In conclusion, this central concept makes it easier for us to better understand the behavior of tiny particles such as electrons.

Find more information: brainly.com/question/17203857

4 0

2 years ago

Read 2 more answers

Compare the wavelengths of an electron (mass = 9.11 × 10−31 kg) and a proton (mass = 1.67 × 10−27 kg), each having (a) a speed o

Ad libitum [116K]

Answer:

Part A:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 6.05x10^{-14}m$ < $\lambda_{electron} = 1.10x10^{-10}m$

Part B:

The proton has a smaller wavelength than the electron.

$\lambda_{proton} = 1.29x10^{-13}m$ < $\lambda_{electron} = 5.525x10^{-12}m$

Explanation:

The wavelength of each particle can be determined by means of the De Broglie equation.

$\lambda = \frac{h}{p}$ (1)

Where h is the Planck's constant and p is the momentum.

$\lambda = \frac{h}{mv}$ (2)

Part A

Case for the electron:

$\lambda = \frac{6.624x10^{-34} J.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

But $J = Kg.m^{2}/s^{2}$

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(6.55x10^{6}m/s)}$

$\lambda = 1.10x10^{-10}m$

Case for the proton:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(6.55x10^{6}m/s)}$

$\lambda = 6.05x10^{-14}m$

Hence, the proton has a smaller wavelength than the electron.

Part B 

For part b, the wavelength of the electron and proton for that energy will be determined.

First, it is necessary to find the velocity associated to that kinetic energy:

$KE = \frac{1}{2}mv^{2}$

$2KE = mv^{2}$

$v^{2} = \frac{2KE}{m}$

$v = \sqrt{\frac{2KE}{m}}$ (3)

Case for the electron:

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{9.11x10^{-31}Kg}}$

but $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{9.11x10^{-31}Kg}}$

$v = 1.316x10^{8}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(9.11x10^{-31}Kg)(1.316x10^{8}m/s)}$

$\lambda = 5.525x10^{-12}m$

Case for the proton :

$v = \sqrt{\frac{2(7.89x10^{-15}J)}{1.67x10^{-27}Kg}}$

But $1J = kg \cdot m^{2}/s^{2}$

$v = \sqrt{\frac{2(7.89x10^{-15}kg \cdot m^{2}/s^{2})}{1.67x10^{-27}Kg}}$

$v = 3.07x10^{6}m/s$

Then, equation 2 can be used:

$\lambda = \frac{6.624x10^{-34}Kg.m^{2}/s^{2}.s}{(1.67x10^{-27}Kg)(3.07x10^{6}m/s)}$

$\lambda = 1.29x10^{-13}m$

Hence, the proton has a smaller wavelength than the electron.

7 0

3 years ago