answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
34kurt
2 years ago
10

Which is a correct representation of .000025 in scientific notation?

Physics
2 answers:
lidiya [134]2 years ago
6 0

0.000025 → 2.5 × 10⁻⁵ → 2.5E-5

<h3>Further explanation</h3>

Scientific notation represents the precise way scientists handle exceptionally abundant digits or extremely inadequate numbers in the product of a decimal form of number and powers of ten. Put differently, such numbers can be rewritten as a simple number multiplied by 10 raised to a certain exponent or power. It is a system for expressing extremely broad or exceedingly narrow digits compactly.

Scientific notation should be in the form of  

\boxed{ \ a \times 10^n \ }

where  

\boxed{ \ 1 \leq a \ < 10 \ }

The number 'a' is called 'mantissa' and 'n' the order of magnitude.

From the key question that is being asked, we face the standard form of 0.000025.

\boxed{ \ 0.000025 = \frac{25}{1,000,000} \ }

The coefficient (or mantissa), i.e. 25, is still outside of 1 ≤ a < 10. Both the numerator and denominator are divided by 10.

\boxed{ \ 0.000025 = \frac{2.5}{100,000} \ }

The denominator consists precisely of five zero digits.

Hence, 0,000025 is written in scientific notation as  \boxed{\boxed{ \ 2.5 \times 10^{-5} \ or \ 2.5E - 5 \ }}

The inverse of scientific notation is the standard form. To promptly change scientific notation into standard form, we reverse the process, move the decimal point to the right or left. This expanded form is called the standard form.

<u>A notable example:</u>

\boxed{ \ 3.0 \times 10^{8} \ Hz \ \rightarrow 300,000,000 \ Hz \ or \ 300 \ MHz}

<h3>Learn more</h3>
  1. 0.00069 written in scientific notation brainly.com/question/7263463
  2. Express the pill’s mass in 0.0005 grams using scientific notation or in milligrams brainly.com/question/493592
  3. What 3 digits are in the units period of 4,083,817 brainly.com/question/558692

Keywords: which is, a correct representation, 0.000025, in scientific notation, expanded form, exponent, base, standard form, mantissa, the order of magnitude, power, decimal, very large, small, figures, abundant digits, inadequate

Monica [59]2 years ago
4 0

Answer : The correct answer is, 2.5\times 10^{-5}

Explanation :

Scientific notation : It is the representation of expressing the numbers that are too big or too small and are represented in the decimal form with one digit before the decimal point times 10 raise to the power.

For example :

8000 is written as 8.0\times 10^3

779.9 is written as 7.799\times 10^{-2}

In this examples, 8000 and 779.9 are written in the standard notation and 8.0\times 10^3  and 7.799\times 10^{-2}  are written in the scientific notation.

If the decimal is shifting to right side, the power of 10 is negative and if the decimal is shifting to left side, the power of 10 is positive.

As we are given the 0.000025 in standard notation.

Now converting this into scientific notation, we get:

\Rightarrow 0.000025=2.5\times 10^{-5}

As, the decimal point is shifting to right side, thus the power of 10 is negative.

Hence, the correct answer is, 2.5\times 10^{-5}

You might be interested in
A proton is confined in an infinite square well of width 10 fm. (The nuclear potential that binds protons and neutrons in the nu
kvasek [131]

Answer:

First Question

    E   =   1.065*10^{-12} \  J

Second  Question

   The  wavelength is for an X-ray  

Explanation:

From the question we are told that

     The  width of the wall is  w =  10\ fm =  10*10^{-15 }\ m

     The  first excited state is  n_1  =  2

     The  ground state is   n_0 = 1

Gnerally the  energy (in MeV) of the photon emitted when the proton undergoes a transition is mathematically represented as

          E   =   \frac{h^2 }{ 8 * m  *  l^2 [ n_1^2 - n_0 ^2 ] }

Here  h is the Planck's constant with value  h =  6.62607015 * 10^{-34} J \cdot s

         m is the mass of proton with value m  = 1.67 * 10^{-27} \   kg

So    

          E  =   \frac{( 6.626*10^{-34})^2 }{ 8 * (1.67 *10^{-27})  *  (10 *10^{-15})^2 [ 2^2 - 1 ^2 ] }

=>        E   =   1.065*10^{-12} \  J

Generally the energy of the photon emitted is also mathematically represented as

             E  =  \frac{h * c }{ \lambda }

=>          \lambda  =  \frac{h * c }{E }

=>          \lambda  =  \frac{6.62607015 * 10^{-34} * 3.0 *10^{8} }{ 1.065 *10^{-15 } }

=>         \lambda  =  1.87*10^{-10} \  m

Generally the range of wavelength of X-ray is  10^{-8} \to  1)^{-12}

So this wavelength is for an X-ray.

     

8 0
2 years ago
Each metal is illuminated with 400 nm (3.10 eV) light. Rank the metals on the basis of the maximum kinetic energy of the emitted
34kurt

Answer:

K.E(K) > K.E(Cs) > 0 (others)

Explanation:

Given the Work functions of the metal as

Aluminium (Wo)=4eV

Platinum(Wo) =6.4eV

Cesium (Wo) =2.1eV

Beryllium (Wo) = 5.0eV

Magnesium (Wo) = 3.7eV

Potassium (Wo) = 2.3eV

Using the formula:

K.E = hf - Wo........(1)

Wo = hfo..............(2)

From these the fo can be calculated for all the metals

Where K.E =Kinetic Energy

hf = energy of illumination = 3.10eV

h is Planck constant and has the value 6.6 × 10^-34JS^-1

The frequency f of the illumination is given by

f = 3.10 × 1.6 × 10^-19/6.6 × 10^-34

f = 7.51 × 10¹⁴ Hz..........(*)

Now an electron is only ejected if the threshold frequency of the metal is reached.

The work function has a threshold frequency (fo) for all the metals and this minimum frequency required to required to remove an electron from the surface of a metal.

We need to compare f with fo

If fo >= f there is emission, otherwise there is no emission

So using (2) we calculate for all fo and compare with f

K.E(Al) = 3.10 - 4.0 - 3.10 = -0.9eV, fo = 9.70 × 10¹⁴ Hz (no emission)

K.E(Pt) = 3.10 - 6.40 = -3.30eV, fo = 1.55 × 10^15 Hz, ( no emission)

K.E(Cs) = 3.10 - 2.10 = -1.0eV, fo = 5.09×10¹⁴ Hz, (emission)

K.E(Be) =3.10-5.0 = -1.90eV, fo = 12.12 ×10^15 Hz.,(no emission)

K.E(Mg) = 3.10-3.70 = -0.6eV, fo = 8.97 × 10¹⁴Hz, (no emission)

K.E(K) = 3.10 - 2.30= 0.9eV, fo = 5.58 × 10¹⁴ Hz, (emission)

So the metals whose electron gain Kinetic energy are:

Cesium

Potassium

Others have zero kinetic energy since no electron is emitted.

Hence the rank is:

K.E(K) > K.E(Cs) > 0 (others)

6 0
2 years ago
How much heat Q1 is transferred by 25.0 g of water onto the skin? To compare this to the result in the previous part, continue t
hodyreva [135]

Answer:

The heat transferred  from water to skin  is 6913.5 J.

Explanation:

Given that,

Weight of water = 25.0 g

Suppose that water and steam, initially at 100°C, are cooled down to skin temperature, 34°C, when they come in contact with your skin. Assume that the steam condenses extremely fast. We will further assume a constant specific heat capacity c=4190 J/(kg°K) for both liquid water and steam.

We need to calculate the heat transferred  from water to skin

Using formula for stream

Q=mc\Delta T

Put the value into the formula

Q=25\times10^{-3}\times4190\times(373-307)

Q=6913.5\ J

Hence, The heat transferred  from water to skin  is 6913.5 J.

3 0
2 years ago
A badger is trying to cross the street. Its velocity vvv as a function of time ttt is given in the graph below where rightwards
Mrrafil [7]

Answer: -2.5

Explanation:

1/2(-5)= -2.5

-2.5(1)= -2.5

Got it right in Khan Academy. You’re welcome.

5 0
2 years ago
Determine a formula for the maximum height h that a rocket will reach if launched vertically from the Earth's surface with speed
olga55 [171]

Initially, the energies are:

U_{i}=-\frac{G M_{\varepsilon} m}{r_{e}} \\&#10;=K_{i}=\frac{1}{2} m v_{0}^{2}

At final point, the energies are:

U_{f}=-\frac{G M_{\varepsilon} m}{r_{e}+h} \\&#10;K_{f}=\frac{1}{2} m(0)^{2}=0

Using conservation law of energy,

-\frac{G M_{e} m}{r_{e}}+\frac{1}{2} m v_{0}^{2} &=-\frac{G M_{e} m}{r_{\varepsilon}+h} \\&#10;-\frac{G M_{e}}{r_{e}}+\frac{v_{0}^{2}}{2} &=-\frac{G M_{e}}{r_{e}+h} \\&#10;\frac{-2 G M_{e}+r_{e} v_{0}^{2}}{2 r_{e}} &=-\frac{G M_{e}}{r_{e}+h} \\&#10;\frac{r_{e}+h}{G M_{e}} &=\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}

The equation is further simplified as:

r_{e}+h &=\left(\frac{2 r_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}\right) G M_{e} \\&#10;h &=\frac{2 r_{e} G M_{e}}{2 G M_{e}-r_{e} v_{0}^{2}}-r_{e} \\&#10;&=\frac{2 r_{e} G M_{e}-2 r_{e} G M_{e}+r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}} \\&#10;& h=\frac{r_{e}^{2} v_{0}^{2}}{2 G M_{e}-r_{e} v_{0}^{2}}

7 0
1 year ago
Other questions:
  • What is the total kinetic energy of a 0.15 kg hockey puck sliding at 0.5 m/s and rotating about its center at 8.4 rad/s? The dia
    8·1 answer
  • A satellite completes one revolution of a planet in almost exactly one hour. At the end of one hour, the satellite has traveled
    5·1 answer
  • A physics student stands on the rim of the canyon and drops a rock. The student measures the time for it to reach the bottom to
    6·1 answer
  • A ball hangs on the end of a string that is connected to the ceiling so that it swings like a pendulum. You pull the ball up so
    5·1 answer
  • A 50-kg person stands 1.5 m away from one end of a uniform 6.0-m-long scaffold of mass 70.0 kg.
    15·1 answer
  • Onur drops a basketball from a height of 10\,\text{m}10m10, start text, m, end text on Mars, where the acceleration due to gravi
    8·1 answer
  • If Sienna is using the gymnasium floor to cool off, which position will lower her skin temperature the quickest?
    6·2 answers
  • charge, q1 =5.00μC, is at the origin, a second charge, q2= -3μC, is on the x-axis 0.800m from the origin. find the electric fiel
    12·1 answer
  • 3. A very light bamboo fishing rod 3.0 m long is secured to a boat at the bottom end. It is
    13·1 answer
  • How much does a person weigh if it takes 700 kg*m/s to move them 10 m/s<br><br> NEED ASAP
    14·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!