D, all notebooks would hit the floor at the same time. The time it takes to hit the floor is independent of their weight, but rather dependent on the acceleration of gravity. Since gravity is constant, they will all hit the floor at the same time.
Answer:
m³/(kg⋅s²)
Explanation:
Hello.
In this case, since the involved formula is:
By writing a dimensional analysis with the proper algebra handling, we obtain:
![N[=]G*\frac{kg*kg}{m^2}\\ \\kg*\frac{m}{s^2}[=]G *\frac{kg*kg}{m^2}\\\\G[=]\frac{kg*m*m^2}{kg^2*s^2}\\ \\G[=]\frac{m^3}{kg*s^2}](https://tex.z-dn.net/?f=N%5B%3D%5DG%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%20%5C%5Ckg%2A%5Cfrac%7Bm%7D%7Bs%5E2%7D%5B%3D%5DG%20%2A%5Cfrac%7Bkg%2Akg%7D%7Bm%5E2%7D%5C%5C%5C%5CG%5B%3D%5D%5Cfrac%7Bkg%2Am%2Am%5E2%7D%7Bkg%5E2%2As%5E2%7D%5C%5C%20%5C%5CG%5B%3D%5D%5Cfrac%7Bm%5E3%7D%7Bkg%2As%5E2%7D)
Thus, answer is:
m³/(kg⋅s²)
Note that the [=] is used to indicate the units of G.
Best regards
Answer:
97.5%
Explanation:
By the empirical rule (68-95-99.7),
- 68% of data are within <em>μ </em>- <em>σ</em> and <em>μ </em>+ <em>σ</em>
- 95% of data are within <em>μ </em>- 2<em>σ</em> and <em>μ </em>+ 2<em>σ</em>
- 99.7% of data are within <em>μ </em>- 3<em>σ</em> and <em>μ </em>+ 2<em>σ</em>
<em>σ </em> and <em>μ</em> are the standard deviation and the mean respectively.
From the question,
<em>μ</em> = 7.2 cm
<em>σ</em> = 0.38 cm
7.96 = 7.2 + (<em>n</em> × 0.38)
<em>n</em> = 2
Hence, 7.96 represents <em>μ </em>+ 2<em>σ</em>.
P(X < <em>μ </em>+ 2<em>σ</em>) = P(X < <em>μ</em>) + P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>)
P(X < <em>μ</em>) is the percentage less than the mean = 50%.
P(<em>μ</em> < X < <em>μ </em>+ 2<em>σ</em>) is half of P(<em>μ </em>- 2<em>σ</em> < X < <em>μ </em>+ 2<em>σ</em>) = 95% ÷ 2 = 47.5%.
Considering this, for apples that are no more than 7.96 cm,
P(X < 7.96) = P(X < 7.2) + P(7.2 < X < 7.96) = 50% + 47.5% = 97.5%
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Answer:
AFter 3.5 s, the wagon is moving at: 
Explanation:
Let's start by finding first the net force on the wagon, and from there the wagon's acceleration (using Newton's 2nd Law):
Net force = 250 N + 178 N = 428 N
Therefore, the acceleration from Newton's 2nd Law is:
So now we apply this acceleration to the kinematic expression for velocity in an object moving under constant acceleration:
