The idea that the Earth sits motionless in the Universe at the center of a revolving globe of stars, with the Moon and planets i
1 answer:
You might be interested in
Explanation:
Here is the complete question i guess. The jet plane travels along the vertical parabolic path defined by y = 0.4x². when it is at point A it has speed of 200 m/s, which is increasing at the rate .8 m/s^2. Determine the magnitude of acceleration of the plane when it is at point A.
→ The tangential component of acceleration is rate of increase in the speed of plane so,
→ Now we have to find out the radius of curvature at point A which is 5 Km (from the figure).
dy/dx = d(0.4x²)/dx
= 0.8x
Take the derivative again,
d²y/dx² = d(0.8x)/dx
= 0.8
at x= 5 Km
dy/dx = 0.8(5)
= 4
now insert the values,
→ Now the normal component of acceleration is given by
= (200)²/(87.6×10³)
aₙ = 0.457 m/s²
→ Now the total acceleration is,
a = 0.921 m/s²
Answer : The answer is, 5.97, 24
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 :
5000 is written as
889.9 is written as
In this examples, 5000 and 889.9 are written in the standard notation and and
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 5,970,000,000,000,000,000,000,000 in standard notation.
Now converting this into scientific notation, we get:
As, the decimal point is shifting to left side, thus the power of 10 is positive.
Hence, the answer is,
Now the answer is comparing to
So, m = 5.97 and n = 24
Thus, the answer is, 5.97, 24
Answer:
ℏ
Given:
Principle quantum number, n = 2
Solution:
To calculate the maximum angular momentum, , we have:
(1)
where,
l = azimuthal quantum number or angular momentum quantum number
Also,
n = 1 + l
2 = 1 + l
l = 1
Now,
Using the value of l = 1 in eqn (1), we get:
ℏ