Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
Explanation:
27dB = 2.7 B
So I / I₀ = 10⁻²°⁷ , I₀ is intensity of main sound and I is intensity after reduction.
= 1.99 X 10⁻³
So intensity will reduce by 1.99 X 10⁻³ .
The component of the force in negative z-direction is -0.144 N.
The given parameters;
- <em>current in the wire, I = 2.7 A</em>
- <em>length of the wire, L = (3.2 i + 4.3j) cm</em>
- <em>magnetic filed, B = 1.24 i</em>
The force on the segment of the wire is calculated as follows;
where;
- <em>θ is the angle wire and magnetic field</em>
<em />
The force on the wire segment will be perpendicular in negative z-direction (applying right hand rule), so there won't be any x and y component of the force.
The angle between the wire and the magnetic field is calculated as follows;
The magnitude of the wire length is calculated as follows;
The component of the force in negative z-direction is calculated as;
Thus, the component of the force in negative z-direction is -0.144 N.
Learn more here:brainly.com/question/22719779
<h2>RED!</h2><h3></h3><h3>On the visible spectrum, red has the lowest frequency.</h3><h3>(I'm an amateur astronomer, so I would know.)</h3>
As the distance from a charged particle, "q", increases, the electric potential decreases.
<h3>
Electric potential between particles</h3>
The electric potential between particles is the work done in moving a unit charge from infinity to a certain point against the electrical resistance of the field.
V = Kq/r
where;
- K is Coulomb's constant
- q is the magnitude of the charge
- r is the distance between the charges
Thus, from the formula above, as the distance from a charged particle, "q", increases, the electric potential decreases.
Learn more about electric potential here: brainly.com/question/14306881
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