We do not feel the gravitational forces from objects other than the Earth because they are weak.
The addition of vectors involve both magnitude and direction. In this case, we make use of a triangle to visualize the problem. The length of two sides were given while the measure of the angle between the two sides can be derived. We then assign variables for each of the given quantities.
Let:
b = length of one side = 8 m
c = length of one side = 6 m
A = angle between b and c = 90°-25° = 75°
We then use the cosine law to find the length of the unknown side. The cosine law results to the formula: a^2 = b^2 + c^2 -2*b*c*cos(A). Substituting the values, we then have: a = sqrt[(8)^2 + (6)^2 -2(8)(6)cos(75°)]. Finally, we have a = 8.6691 m.
Next, we make use of the sine law to get the angle, B, which is opposite to the side B. The sine law results to the formula: sin(A)/a = sin(B)/b and consequently, sin(75)/8.6691 = sin(B)/8. We then get B = 63.0464°. However, the direction of the resultant vector is given by the angle Θ which is Θ = 90° - 63.0464° = 26.9536°.
In summary, the resultant vector has a magnitude of 8.6691 m and it makes an angle equal to 26.9536° with the x-axis.
When I went through with the math, the answer I came upon was:
<span>6.67 X 10^14 </span>
<span>Here is how I did it: First of all we need to know the equation. </span>
<span>c=nu X lamda </span>
<span>(speed of light) = (frequency)(wavelength) </span>
<span>(3.0 X 10^8 m/s) = (frequency)(450nm) </span>
<span>We want the answer in meters so we need to convert 450nm to meters. </span>
<span>450nm= 4.5 X 10^ -7 m </span>
<span>(3.0 X 10^8 m/s) = (frequency)(4.5 X 10^ -7 m) </span>
<span>Divide the speed of light by the wavelength. </span>
<span>(3.0 X 10^8m/s) / (4.5 X 10^ -7m) =6.67 X 10^ 14 per second or s- </span>
<span>Answer: 6.67 X 10^14 s- hope this helps</span>
Answer:
Yes is large enough
Explanation:
We need to apply the second Newton's Law to find the solution.
We know that,
And we know as well that
Replacing the aceleration value in the equation force we have,
Substituting our values we have,
The weight of the person is then,
<em>We can conclude that force on the ball is large to lift the ball</em>
The statement about pointwise convergence follows because C is a complete metric space. If fn → f uniformly on S, then |fn(z) − fm(z)| ≤ |fn(z) − f(z)| + |f(z) − fm(z)|, hence {fn} is uniformly Cauchy. Conversely, if {fn} is uniformly Cauchy, it is pointwise Cauchy and therefore converges pointwise to a limit function f. If |fn(z)−fm(z)| ≤ ε for all n,m ≥ N and all z ∈ S, let m → ∞ to show that |fn(z)−f(z)|≤εforn≥N andallz∈S. Thusfn →f uniformlyonS.
2. This is immediate from (2.2.7).
3. We have f′(x) = (2/x3)e−1/x2 for x ̸= 0, and f′(0) = limh→0(1/h)e−1/h2 = 0. Since f(n)(x) is of the form pn(1/x)e−1/x2 for x ̸= 0, where pn is a polynomial, an induction argument shows that f(n)(0) = 0 for all n. If g is analytic on D(0,r) and g = f on (−r,r), then by (2.2.16), g(z) =
