4. An archer pulls a bowstring back 0.440 m, and the spring constant of the bow is 545 N/m.
1 answer:
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Answer:
From the formula of force:
since AB and k are constants:
x is a constant of proportionality
• when force is 4N, separation distance is 1
therefore, equation becomes
when r is doubled, r becomes 2. find F:
Answer:
8.79*10^6 rad/s
Explanation:
To find the frequency of the circular orbit for an electron you use the following expression, for the radius of the trajectory of an electron, that travels trough a constant magnetic field:
(1)
r: radius of the trajectory
m: mass of the electron = 9.1*10^-31 kg
v: speed of the electron = 1.0*10^6 m/s
q: charge of the electron = 1.6*10^-19 C
B: magnitude of the magnetic field = 5.0*10^-5 T
You use the fact that the angular frequency in a circular motion is given by:
Then, you solve the equation (1) in order to obtain v/r:
Finally, you replace the values of the parameters:
hence, the angular frequency is 8.79*10^6 rad/s
The frequency is:
Answer:
Ф_cube /Ф_sphere = 3 /π
Explanation:
The electrical flow is
Ф = E A
where E is the electric field and A is the surface area
Let's shut down the electric field with Gauss's law
Фi = ∫ E .dA = / ε₀
the Gaussian surface is a sphere so its area is
A = 4 π r²
the charge inside is
q_{int} = Q
we substitute
E 4π r² = Q /ε₀
E = 1 / 4πε₀ Q / r²
To calculate the flow on the two surfaces
* Sphere
Ф = E A
Ф = 1 / 4πε₀ Q / r² (4π r²)
Ф_sphere = Q /ε₀
* Cube
Let's find the side value of the cube inscribed inside the sphere.
In this case the radius of the sphere is half the diagonal of the cube
r = d / 2
We look for the diagonal with the Pythagorean theorem
d² = L² + L² = 2 L²
d = √2 L
we substitute
r = √2 / 2 L
r = L / √2
L = √2 r
now we can calculate the area of the cube that has 6 faces
A = 6 L²
A = 6 (√2 r)²
A = 12 r²
the flow is
Ф = E A
Ф = 1 / 4πε₀ Q/r² (12r²)
Ф_cubo = 3 /πε₀ Q
the relationship of these two flows is
Ф_cube /Ф_sphere = 3 /π