The force needed to stretch the steel wire by 1% is 25,140 N.
The given parameters include;
- diameter of the steel, d = 4 mm
- the radius of the wire, r = 2mm = 0.002 m
- original length of the wire, L₁
- final length of the wire, L₂ = 1.01 x L₁ (increase of 1% = 101%)
- extension of the wire e = L₂ - L₁ = 1.01L₁ - L₁ = 0.01L₁
- the Youngs modulus of steel, E = 200 Gpa
The area of the steel wire is calculated as follows;

The force needed to stretch the wire is calculated from Youngs modulus of elasticity given as;


Thus, the force needed to stretch the steel wire by 1% is 25,140 N.
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Answer:
6.5e-4 m
Explanation:
We need to solve this question using law of conservation of energy
Energy at the bottom of the incline= energy at the point where the block will stop
Therefore, Energy at the bottom of the incline consists of the potential energy stored in spring and gravitational potential energy=
Energy at the point where the block will stop consists of only gravitational potential energy=
Hence from Energy at the bottom of the incline= energy at the point where the block will stop
⇒
⇒
Also 
where
is the mass of block
is acceleration due to gravity=9.8 m/s
is the difference in height between two positions
⇒
Given m=2100kg
k=22N/cm=2200N/m
x=11cm=0.11 m
∴
⇒
⇒
⇒h=0.0006467m=
Answer:
True
Explanation:
Momentum of an object can be defined as the product of its mass and velocity at which it is travelling. With that in mind, momentum = 3*100=300(kg⋅m/s).
One thing to note is the units mentioned. The SI unit of momentum is kg * m/s as it is the product of mass(kilograms) and velocity(meter per second) and not Newton.
<span>The total
energy stored is the sum of the energy stored in the capacitors. If the
capacitors are series connected
capacitors, then the charging current is the same for both capacitors. This
means that each capacitor stores the same energy and the stored energy is two
times the energy of any of the capacitors.</span>
Answer:
Spring constant, k = 5483.11 N/m
Explanation:
It is given that,
Mass of the organ, m = 2 kg
The natural period of oscillation is, T = 0.12 s
Let k is the spring constant for the spring in the scientist's model. The period of oscillation is given by :



k = 5483.11 N/m
So, the spring constant for the spring in the scientist's model is 5483.11 N/m.