How did you take a picture?
Answer:
The first two statements are false
The third statement is true
Explanation:
<u>The dot product assures that the integrand is always nonnegative.</u>
The dot product may be negative, this could ocurr when the vectors are directed oposite each other, for example take the unitary vector i and -i its doct product will give -1.
Another way to consider this is to take the definition of the dot product in terms of teh angle between the vetcors:
![\vec{A} \cdot \vec{B} = |\vec{A}| \times |\vec{B}| cos(\theta)](https://tex.z-dn.net/?f=%5Cvec%7BA%7D%20%5Ccdot%20%5Cvec%7BB%7D%20%3D%20%7C%5Cvec%7BA%7D%7C%20%5Ctimes%20%7C%5Cvec%7BB%7D%7C%20cos%28%5Ctheta%29)
When θ>π :
cos(θ)<0
<u>The dot product indicates that only the component of the force perpendicular to the path contributes to the integral</u>
In fact the dot product is a projection of the vectors, the perpendicular component may be obtained using the cross product
<u>The dot product indicates that only the component of the force parallel to the path contributes to the integral.</u>
This one is true, since the dot product gives the projection of one vector to another, that is, the parallel component of the vector among the other one
Answer:
82.8 J
Explanation:
The work done to raise the crate is ...
PE = Mgh = (3 kg)(9.8 m/s^2)(2 m) = 58.8 J
The kinetic energy added to send the box flying is ...
KE = (1/2)Mv^2 = (1/2)(3 kg)(4 m/s)^2 = 24 J
So, the total work involved in this activity is ...
58.8 J +24 J = 82.8 J
Answer:
a. f₀ = 6.355 Hz ; b. Δf = 6.35 Hz
Explanation:
Given Data:
Length of wire = l =347 m ;
Tension in wire = T = 65.2 * 10⁶N ;
Linear density = μ = 3.35 kg/m ;
Solution:
a)
Fundamental mode = f₀ = (1/2l)*(sqr.root(T/μ))
By putting the values, we get
f₀ = (1/2(347))*![\sqrt[]{\frac{65.2 * 10^6}{3.35} }](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B65.2%20%2A%2010%5E6%7D%7B3.35%7D%20%7D)
f₀ = 6.355 Hz
b)
To find the frequency difference between successive modes we need to find frequency of second harmonic first
f₁ = (2/2l)*(sqr.root(T/μ))
f₁ = (2/2(347))*![\sqrt[]{\frac{65.2 * 10^6}{3.35} }](https://tex.z-dn.net/?f=%5Csqrt%5B%5D%7B%5Cfrac%7B65.2%20%2A%2010%5E6%7D%7B3.35%7D%20%7D)
f₁ = 12.71 Hz
Difference is:
Δf = f₁ - f₀ = 12.71 - 6.355
= 6.35 Hz