Answer:
Explanation:
Let the magnetic field be B = B₁i + B₂j + B₃k
Force = I ( L x B ) , I is current , L is length and B is magnetic field .
In the first case
force = - 2.3 j N
L = 2.5 i
puting the values in the equation above
- 2.3 j = 8 [ 2.5 i x ( B₁i + B₂j + B₃k )]
= - 20 B₃ j + 20 B₂ k
comparing LHS and RHS ,
20B₃ = 2.3
B₃ = .115
B₂ = 0
In the second case
L = 2.5 j
Force = I ( L x B )
2.3i−5.6k = 8 ( 2.5 j x (B₁i + B₂j + B₃k )
= - 20 B₁ k + 20B₃ i
2.3i−5.6k = - 20 B₁ k + 20B₃ i
B₃ = .115
B₁ = .28
So magnetic field B = .28 i + .115 B₃
Part A
x component of B = .28 T
Part B
y component of B = 0
Part C
z component of B = .115 T .
Answer:
(a) 104 N
(b) 52 N
Explanation:
Given Data
Angle of inclination of the ramp: 20°
F makes an angle of 30° with the ramp
The component of F parallel to the ramp is Fx = 90 N.
The component of F perpendicular to the ramp is Fy.
(a)
Let the +x-direction be up the incline and the +y-direction by the perpendicular to the surface of the incline.
Resolve F into its x-component from Pythagorean theorem:
Fx=Fcos30°
Solve for F:
F= Fx/cos30°
Substitute for Fx from given data:
Fx=90 N/cos30°
=104 N
(b) Resolve r into its y-component from Pythagorean theorem:
Fy = Fsin 30°
Substitute for F from part (a):
Fy = (104 N) (sin 30°)
= 52 N
Answer:
Maximum velocity will be 17.651 m /sec
Explanation:
We have given a rope breaks when tension reaches 205 N
Mass m = 0.477 kg
Radius of the circle r = 0.725 m
Tension in the rope is given by




v = 17.651 m /sec
<span>The runner is moving by uniformly accelerated motion, starting from rest (so, his initial velocity is zero). The law of motion of the runner is
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where x(t) is the distance covered after time t, and a is the acceleration of the runner. By re-arranging the formula, we get
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We know the runner has covered a distance of S=12m in t=4.0 s, and if we plug these numbers into the equation, we find the acceleration of the runner:
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