Answer:
v = 6t² + t + 2, s = 2t³ + ½ t² + 2t
59 m/s, 64.5 m
Explanation:
a = 12t + 1
v = ∫ a dt
v = 6t² + t + C
At t = 0, v = 2.
2 = 6(0)² + (0) + C
2 = C
Therefore, v = 6t² + t + 2.
s = ∫ v dt
s = 2t³ + ½ t² + 2t + C
At t = 0, s = 0.
0 = 2(0)³ + ½ (0)² + 2(0) + C
0 = C
Therefore, s = 2t³ + ½ t² + 2t.
At t = 3:
v = 6(3)² + (3) + 2 = 59
s = 2(3)³ + ½ (3)² + 2(3) = 64.5
1 newton is the force needed to accelerate 1 kilogram of mass
at the rate of 1 meter per second² .
1 N = 1 kg-m/s² .
It's a force equal to roughly 3.6 ounces.
A spring that obeys Hooke's law has a spring force constant of 272 N/m. This spring is then stretched by 28.6 cm
Answer:
v=3.66,h-3.66
Explanation:
vertical = 10sin60 - 10sin 30
horizontal =10cos60 + 10cos 30
v = 10×0.8660-10×0.5
h = 10×0.5 + 10 × 0.8660
v=8.660-5.0 = 3.66
h= 5.0-8.660 = -3.66
Answer:
25°C
Explanation:
Using the linear expansivity formula expressed as;
∝ = ΔL/lΔθ
∝ is coefficient of lineat expansion = 1.2 ∙ 10⁻⁵ °C⁻¹
ΔL is the change in length = 6.00036-6
ΔL = 0.00036m
l is the original length = 6m
Δθ is the change in temperature =θ₂-20
Substituting into the formula;
1.2 ∙ 10⁻⁵ °C⁻¹ = 0.00036/6(θ₂-20)
cross multiply
1.2 ∙ 10⁻⁵ * 6 = 0.00036/(θ₂-20)
7.2 ∙ 10⁻⁵= 0.00036/(θ₂-20)
0.00036 = 7.2 ∙ 10⁻⁵(θ₂-20)
0.00036 = 7.2 ∙ 10⁻⁵θ₂-144∙ 10⁻⁵
7.2 ∙ 10⁻⁵θ₂ = 0.00036+0.00144
7.2 ∙ 10⁻⁵θ₂ = 0.0018
θ₂ = 0.0018/0.000072
θ₂ = 25°C
Hence the temperature at which this bar must be acidic for its compression is 6,00036 m is 25°C
