Supposing that the spring is un stretched when θ = 0, and has a toughness of k = 60 N/m.It seems that the spring has a roller support on the left end. This would make the spring force direction always to the left
Sum moments about the pivot to zero.
10.0(9.81)[(2sinθ)/2] + 50 - 60(2sinθ)[2cosθ] = 0 98.1sinθ + 50 - (120)2sinθcosθ = 0 98.1sinθ + 50 - (120)sin(2θ) = 0
by iterative answer we discover that
θ ≈ 0.465 radians
θ ≈ 26.6º
Answer:
6.21 m/s
Explanation:
Using work energy equation then
where d is displacement from initial to final position, v is velocity and subscripts a and b are position A and B respectively, m is mass of collar, g is acceleration due to gravity
Substituting 1 Kg for m, 0.4m for h, 
as 0, 9.81 for g then
Divide distance by the time it takes to travel that distance
the formula for time is divide distance/speed
The magnetic field at center of circular loops of wire is 3.78 x 10¯⁵ T.
We need to know about the magnetic field at the center of circular loops of wire to solve this problem. The magnetic field at the center can be determined as
B = μ₀ . I / 2r
where B is magnetic field, μ₀ is vacuum permeability (4π×10¯⁷ H/m), I is the current and r is radius.
From the question above, we know that:
r = 4 cm = 0.04 m
I = 1.7 A
By substituting the parameter, we get
B = μ₀ . I / 2r
B = 4π×10¯⁷ . 1.7 / (2.0.04)
B = 2.67 x 10¯⁵ T
Due to the perpendicular plane of loops, the total magnetic field at center will be
Btotal = √(2(B²))
Btotal = √(2(2.67 x 10¯⁵²))
Btotal = 3.78 x 10¯⁵ T
Find more on magnetic field at: brainly.com/question/7802337
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Answer:
The Acceleration of the object = 6.4 m/s²
<u>Explanation:</u>
Mass of block (m) = 5 kg
Action force on block, (F₁) = 40 N
<u>To Find:</u>
Acceleration of the object (a) = ?
<u>Required solution:</u>
Frictional force opposing the motion (F₂) = 8 N
Here in this question we have to find Acceleration of the object. So, firstly we have to find Net force of block after that we will find Acceleration of the object on the basis of conditions given above
⇒ Net force = Action force on block - Opposing friction force
⇒ F = F₁ - F₂
⇒ F = 40 - 8
⇒ F = 32 N
Now, we have to two elements that used in formula, Net force and Mass of block.
Net force of the block (F) = 32 N
Mass of block (m) = 5 kg
And we have to find Acceleration of the object.
We can find Acceleration of the object by using the Second law of Newton which says F = ma
Here,
F is the Force in N.
m is the Mass in kg.
a is the Acceleration in m/s².
So let's find Acceleration (a) !
† From second law of Newton
⇛ F = ma
⇛ a = F/m
⇛ a = 32/5
⇛ a = 6.4 m/s²
