Answer
given,
mass = 100 kg
acceleration = 10 m/s²
A mass 20 kg slides over 100 kg block
acceleration = 3 m/s²
horizontal friction exerted by the 100 kg block on 20 kg
using newton's second law
F - f = 0
F = f
f = ma
f = 20 × 3
f = 60 N
now net force acting on the 100 kg block
F_net = m a
F_net = 100 x 10
F_net = 1000 N
after 20 kg block falls the acceleration of the bock
F = 1000 +60
F = 1060 N
acceleartion on the block
a = 10.60 m/s²
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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.
Learn more here: brainly.com/question/21413915
Answer:
The flux is calculated as φ=BAcosθ. The flux is thereforemaximum when the magnetic field vector is perpendicular to theplane of the loop. We may also deduce that the flux is zero whenthere is no component of the magnetic field that is perpendicularto the loop.
when angle is zero then flux is maximium because when angle zerocos is maximium
I believe the answer is A
