A squirrel standing on a branch shoots an acorn toward a hole in the tree. To determine how the potential energy of the acorn ch
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Answer:
Increasing its charge
Increasing the field strength
Explanation:
For a charged particle moving in a circular path in a uniform magnetic field, the centripetal force is provided by the magnetic force, so we can write:
where
q is the charge
v is the velocity
B is the magnetic field
m is the mass
r is the radius of the orbit
The period of the motion is
Re-arranging for r
And substituting into the previous equation
Solving for T,
So we see that the period is:
- proportional to the charge and the magnetic field
- inversely proportional to the mass and the square of the speed
So the following will increase the period of the particle's motion:
Increasing its charge
Increasing the field strength
Explanation:
Given data:
d = 30 mm = 0.03 m
L = 1m
S = 70 Mpa
Δd = -0.0001d
Axial force = ?
validity of elastic deformation assumption.
Solution:
O'₂ = Δd/d = (-0.0001d)/d = -0.0001
For copper,
v = 0.326 E = 119×10³ Mpa
O'₁ = O'₂/v = (-0.0001)/0.326 = 306×10⁶
∵δ = F.L/E.A and σ = F/A so,
σ = δ.E/L = O'₁ .E = (306×10⁻⁶).(119×10³) = 36.5 MPa
F = σ . A = (36.5 × 10⁻⁶) . (π/4 × (0.03)²) = 25800 KN
S = 70 MPa > σ = 36.5 MPa
∵ elastic deformation assumption is valid.
so the answer is
F = 25800 K N and S > σ