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3 years ago

When is zero considered significant

Physics

2 answers:

Semenov [28]3 years ago

7 0

Answer:

The trailing zero to the right in the decimal place and the zero present between the non zero digit are significant.

Explanation:

The rules which describes about the significant digits are discussed below as:

All the digits which are non zero are significant. Example 178 has three significant digits.

The zero which is present in between two non zero digits are consider significant. Example 507 has 3 significant digits.

The leading zeroes are not consider significant. Example 0.00676 has 3 significant digits only.

Trailing zero to the right in the decimal is consider significant. Example 3.0 has 2 significant digits.

Trailing zero which is present in the whole number without any decimal are not significant. Example 9800 has only 2 significant digit.

vazorg [7]3 years ago

3 0

If a zero is found between significant digits, it is significant. Zeros can be used as (insignificant) place holders to the left of significant digits if the number is a decimal. For example, a mass of 42 g has two significant digits. Expressed in kilograms, the mass of 0.042kg should still have two significant digits.

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A copper rod of length 27.5 m has its temperature increases by 35.9 degrees celsius. how much does its length increase?(unit=m)

gavmur [86]

<h2>The increase in length = 1.87 x 10⁻²</h2>

Explanation:

When copper rod is heated , its length increases

The increase in length can be found by the relation

L = L₀ ( 1 + α ΔT )

here L is the increased length and L₀ is the original length

α is the coefficient of linear expansion and ΔT is the increase in temperature .

The increase in length = L - L₀ = L₀ x α ΔT

Substituting all these value

Increase in length = 27.5 x 1.7 x 10⁻⁵ x 35.9

= 1.87 x 10⁻² m

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3 years ago

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When you throw a ball up in the air, it travels up and then stops instantaneously before falling back down. At the point where i

Gnoma [55]

Answer:

The ball stops instantaneously at the topmost point of the motion.

Explanation:

Assume we have thrown a ball up in the air. For that we have given a force on the ball and it acquires an initial velocity in the upward direction.

The forces that resist the motion of the ball in the upward direction are the force of gravity and air resistance. The ball will instantaneously come to rest when the velocity of the ball reduces to zero.

The two forces acting in the downward direction reduces its speed continuously and it becomes zero at the topmost point.

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3 years ago

A 23 kg body is moving through space in the positive direction of an x axis with a speed of 130 m/s when, due to an internal exp

babymother [125]

Answer:

a) Vx = 1088m/s

b) Vy = -162.93m/s

c) 5246745J

Explanation:

Mass of unbroken body = 23kg

Its velocity along +ve X-axis = 130m/s

Mass of first broken body, m1= 9.4kg

Its velocity along +ve X-axis = 130m/s

Nass of 2nd broken body, m2 = 6.1kg

Its velocity long-lived X - axis = -550m/s

Mass of 3rd broken body = ?

m3 = (23 - 9.4 - 6.1)kg

m3 = 7.5kg

Let velocity along the x-axis = Vx

Let the velocity along the x-axis = Vy

Applying law of conservation of momentum along x-axis

a) m1×0 + m2×(-550) + m3×(Vx) =M × 130

9.4 × 0 + 6.1× (-550) + 7.5(Vx) = 23 ×130

0 + (-5170) + 7.5Vx = 2990

2990 + 5170 = 7.5Vx

8160 = 7.5Vx

Vx = 8160/7.5

Vx = 1088m/s

b) Aplying conservation of momentum along the x-axis

(m1×130) + (m2 × 0) + (m3× Vy) = 0

(9.4 × 130) + (6.1 ×550) + 7.5Vy = 0

1222 + 0 + 7.5Vy = 0

1222 = -7.5Vy

Vy = 1222/(-7.5)

Vy = -262.93m/s

c) The energy released or change in KE is given by:

1/2[(m1v1^2) + (m2v2^2) +(m3Vx^2) ]= MV^2

Change in KE = 1/2[ 9.4× 130^2 + 6.1 × 550^2 + 7.5 × 1088^2 ] - 1/2(23 × 130^2)

Change in KE = 1/2[158860 + 1845250 + 8878080] - 1/2[388700]

Change in KE = 5441095 - 194350

Change in KE = 5246745J

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3 years ago

Explain the Big Bang and how it started

AlexFokin [52]

The big bang is how astronomers explain the way the universe began. It is the idea that the universe began as just a single point, then expanded and stretched to grow as large as it is right now (and it could still be stretching).

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3 years ago

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A 16 g piece of Styrofoam carries a net charge of -8.6 µC and floats above the center of a large horizontal sheet of plastic tha

PilotLPTM [1.2K]

Answer:

the charge per unit area on the plastic sheet is - 3.23 x 10⁻⁷ C/m²

Explanation:

given information:

styrofoam mass, m = 16 g = 0.016 kg

net charge, q = - 8.6 μC

to calculate the charge per unit area on the plastic sheet, we can use the following equation:

$F_{e} = mg$

where

$F_{e} =$ the force between the electric field

m = mass

g = gravitational force

$F_{e} =qE$

where

q = charge

E = electric field

and

E = σ/2ε₀

where

ε₀ = permitivity

thus

$F_{e} =qE$

mg = qσ/2ε₀

σ = (2mg ε₀)/q

= 2 (0.016) (9.8) (8.85 x 10⁻¹²)/( - 8.6 x 10⁻⁶)

= - 3.23 x 10⁻⁷ C/m²

8 0

3 years ago

When is zero considered significant​

When is zero considered significant