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Yakvenalex [24]
3 years ago
10

What is .00245 in scientific notation?

Physics
1 answer:
Sloan [31]3 years ago
4 0

Answer:

2.45 x 10^-2

Explanation:

All numbers in scientific notation or standard form are written in the form

m × 10n, where m is a number between 1 and 10 ( 1 ≤ |m| < 10 ) and the exponent n is a positive or negative integer.

To convert 0.0245 into scientific notation, follow these steps:

Step 1) Move the decimal 2 times to right in the number so that the resulting number, m = 2.45, is greater than or equal to 1 but less than 10

Step 2) Since we moved the decimal to the right the exponent n is negative

n = -2

Step 3) Write in the scientific notation form m × 10n

= 2.45 × 10-2

Therefore,

2.45 × 10-2 is the scientific notation form of 0.0245 number and 2.45e-2 is the scientific e-notation form for 0.0245

Here are some examples of decimal to scientific notation calculator

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You see a marble sign. Its letters are too worn to read. What caused this?
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A wall, acted upon by a force of 20 N, does not move. The work done on the wall in this process is
jenyasd209 [6]

Answer:

0 (Zero)

Explanation:

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Work done on an object is the force applied on that object in the direction of Displacement of the object .

means that ,

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When the wall is acted upon by the force of 20 N , Wall does not move ,

Thus , displacement of the wall is zero.

So, W = 20×0 = 0.

Thus, Work done on the wall is zero.

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3 years ago
A toy rocket is launched vertically from ground level (y = 0.00 m), at time t = 0.00 s. The rocket engine provides constant upwa
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The toy rocket is launched vertically from ground level, at time t = 0.00 s. The rocket engine provides constant upward acceleration during the burn phase. At the instant of engine burnout, the rocket has risen to 72 m and acquired a velocity of 30 m/s. The rocket continues to rise in unpowered flight, reaches maximum height, and falls back to the ground with negligible air resistance.

The total energy of the rocket, which is a sum of its kinetic energy and potential energy, is constant.

At a height of 72 m with the rocket moving at 30 m/s, the total energy is m*9.8*72 + (1/2)*m*30^2 where m is the mass of the rocket.

At ground level, the total energy is 0*m*9.8 + (1/2)*m*v^2.

Equating the two gives: m*9.8*72 + (1/2)*m*30^2 = 0*m*9.8 + (1/2)*m*v^2

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=> v^2 = 11556/5

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3 years ago
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