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bearhunter [10]
4 years ago
8

When is a zero not significant?

Physics
2 answers:
maria [59]4 years ago
8 0

Answer:

Not between significant digits.

Explanation:

A zero not significant when it's not between significant digits.

tigry1 [53]4 years ago
5 0

Answer:

Zero are not significant when indicating the decimal point position

Explanation:

The number zero has a special meaning with respect to its meaning, when it is placed to the right as a decimal, its meanings so precision measurements were made, for example 10 cm and 10.0 cm, the first one has an accuracy of one centimeter and the second has an accuracy of 0.1 cm.

In the case of being placed to the left of all numbers, the meaning is to indicate where the decimal point goes, but it does not give information about the accuracy of the measurement, in this second case the zeros are not significant, for example: 0.1 0.0001 in these two cases are not significant

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When running your engine, you cause debris, rocks and propeller blast to be directed towards people or other aircraft. Is this c
beks73 [17]

Answer:

Yes, it is reckless. This is because it is the responsibility of the pilot to make sure that the direction of the propeller blast is away from people or other aircraft and in a safe direction.

Explanation:

Yes, it is reckless to let the propeller blast face people and other aircraft. This is because it is the responsibility of the pilot to make sure that the direction of the propeller blast is away from people or other aircraft and in a safe direction. People and other aircraft can be injured by the debris and the rocks that are scattered by the engine of the aircraft.

3 0
3 years ago
An electron is accelerated within a particle accelerator using a 100 MV electric potential. The 100 MeV electron moves along an
Delicious77 [7]

Answer:

The length of the tube is 3.92 m.

Explanation:

Given that,

Electric potential = 100 MV

Length = 4 m

Energy = 100 MeV

We need to calculate the value of \gamma

Using formula of relativistic energy

E=m_{0}c^2(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

Put the value into the formula

1.6\times10^{-15}= 9.1\times`10^{-31}\times9\times10^{16}(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)=\dfrac{1.6\times10^{-15}}{9.1\times10^{-31}\times9\times10^{16}}

Here, \gamma-1=(\dfrac{1}{\sqrt{1-\dfrac{v^2}{c^2}}}-1)

\gamma-1=0.01953

\gamma=0.01953+1

\gamma=1.01953

We need to calculate the length

Using formula of length

L'=\dfrac{L}{\gamma}

Put the value into the formula

L'=\dfrac{4}{1.01953}

L'=3.92\ m

Hence, The length of the tube is 3.92 m.

8 0
3 years ago
If a short-wave radio station broadcasts on a frequency of 9.065 megahertz (MHz), what is the wavelength of
Yakvenalex [24]

Answer:

If the radio wave is on an FM station, these are in Megahertz. A megahertz is one ... Typical radio wave frequencies are about 88~108 MHz .

Explanation:

To calculate the wavelength of a radio wave, you will be using the equation: Speed of a wave = wavelength X frequency.

Since radio waves are electromagnetic waves and travel at 2.997 X

10

8

meters/second, then you will need to know the frequency of the radio wave.

If the radio wave is on an FM station, these are in Megahertz. A megahertz is one million hertz. If the radio wave is from an AM radio station, these are in kilohertz (there are one thousand hertz in a kilohertz). Hertz are waves/second. Hertz is usually the label for the frequency of electromagnetic waves.

To conclude, to determine the wavelength of a radio wave, you take the speed and divide it by the frequency.

Typical radio wave frequencies are about

88

~

108

MHz

. The wavelength is thus typically about

3.41

×

10

9

~

2.78

×

10

9

nm

.

7 0
2 years ago
How many animals lost each hour
joja [24]

Answer:

562

Explanation:

6 0
3 years ago
A wheel of radius R, mass M, and moment of inertia I is mounted on a frictionless, horizontal axles. A light cord wrapped around
Alex_Xolod [135]

Answer:

\alpha =\frac{m*g*R}{I-m*R^2}

a = \frac{m*g*R^2}{I-m*R^2}

T=\frac{I*m*g}{I-m*R^2}

Explanation:

By analyzing the torque on the wheel we get:

T*R=I*\alpha    Solving for T:   T=I/R*\alpha

On the object:

T-m*g = -m*a    Replacing our previous value for T:

I/R*\alpha-m*g = -m*a

The relation between angular and linear acceleration is:

a=\alpha*R

So,

I/R*\alpha-m*g = -m*\alpha*R

Solving for α:

\alpha =\frac{R*m*g}{I+m*R^2}

The linear acceleration will be:

a =\frac{R^2*m*g}{I+m*R^2}

And finally, the tension will be:

T =\frac{I*m*g}{I+m*R^2}

These are the values of all the variables: α, a, T

8 0
3 years ago
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