(a) The spring constant is 500 N/m.
(b) The extension of the spring when 25 N force is applied is 0.05 m.
(c) The applied force to cause an extension of 5 mm is 2.5 N.
The given parameters:
- Applied force, F = 10 N
- Extension of the spring, x = 20 mm
The spring constant is calculated as follows;
The extension of the spring when 25 N force is applied is calculated as follows;
The applied force to cause an extension of 5 mm is calculated as follows;
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Answer:
Explanation:
The direction of force will be in upward direction making an angle of θ with the vertical .
Reaction force R = mg - F cosθ
Friction force = μR
= .36 (mg - F cosθ )
Horizontal component of applied force
= F sinθ
For equilibrium
F sinθ = .36 (mg - F cosθ)
F sinθ + .36 F cosθ =.36 mg
F (sinθ + .36 cosθ) = .36 mg
F R( cosδsinθ +sinδ cosθ) = .36 mg ( Rcosδ = 1 . Rsinδ= .36 )
F R sin( θ+δ ) = . 36 mg
F = .36 mg / Rsin( θ+δ )
For minimum F , sin( θ+δ ) should be maximum
sin( θ+ δ ) = sin 90
θ+ δ = 90
Rsinδ / Rcosδ = .36
δ = 20⁰
θ = 70⁰ Ans
Answer:
1.332 N
Explanation:
Net Force = Mass x Acceleration
1.2 x 1.11 = 1.332 N
I'm so sorry if I'm wrong.
Answer:
Height of the rocket be one minute after liftoff is 40.1382 km.
Explanation:
v = velocity of rocket at time t
g = Acceleration due to gravity =
= Constant velocity relative to the rocket = 2,900m/s.
m = Initial mass of the rocket at liftoff = 29000 kg
r = Rate at which fuel is consumed = 170 kg/s
Velocity of the rocket after 1 minute of the liftoff =v
t = 1 minute = 60 seconds'
Substituting all the given values in in the given equation:
Height of the rocket = h
Height of the rocket be one minute after liftoff is 40.1382 km.
Answer:
4. 10.0 m/s²
Explanation:
I) if initial velocity is 'v₀', the final velocity is 'v', the accelaration is 'a', the distance is 'L' and elapsed time if 't', then:
II) using these two equations after substitution v₀=0; v=30 and L=45:
