Answer:
Explanation:
Let r be the rate of the slower walker in mph
[r + (r + 1.7)](2) = 13
(2r + 1.7)(2) = 13
4r + 3.4 = 13
4r = 9.6
r = 2.4 mph
r + 1.7 = 4.1 mph
The height reached by the two carts after collision is determined as 5.34 m.
<h3>
Initial velocity of Cart A</h3>
Apply the principle of conservation of mechanical energy.
K.E = P.E
v = √2gh
v = √(2 x 9.8 x 12)
v = 15.34 m/s
<h3>Final velocity of the two carts after the collision</h3>
Apply the principle of conservation of linear momentum for inelastic collision.
m₁u₁ + m₂u₂ = v(m₁ + m₂)
8(15.34) + 4(0) = v(8 + 4)
122.72 = 12v
v = 10.23 m/s
<h3>Height reached by both carts</h3>
Apply the principle of conservation of mechanical energy.
P.E = K.E
mgh = ¹/₂mv²
h = v²/(2g)
h = (10.23²) / (2 x 9.8)
h = 5.34 m
Learn more about linear momentum here: brainly.com/question/7538238
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Answer:
W = 1080.914 J
Explanation:
f(x) = 1100xe⁻ˣ
Work done by a variable force moving through a particular distance
W = ∫ f(x) dx (with the integral evaluated between the interval that the force moves through)
W = ∫⁶₀ 1100xe⁻ˣ dx
W = 1100 ∫⁶₀ xe⁻ˣ dx
But the integral can only be evaluated using integration by parts.
∫ xe⁻ˣ dx
∫ vdu = uv - ∫udv
v = x
(dv/dx) = 1
dv = dx
du = e⁻ˣ dx
∫ du = ∫ e⁻ˣ dx
u = -e⁻ˣ
∫ vdu = uv - ∫udv
∫ xe⁻ˣ dx = (-e⁻ˣ)(x) - ∫ (-e⁻ˣ)(dx)
= -xe⁻ˣ - e⁻ˣ = -e⁻ˣ (x + 1)
∫ xe⁻ˣ dx = -e⁻ˣ (x + 1) + C (where c = constant of integration)
W = 1100 ∫⁶₀ xe⁻ˣ dx
W = 1100 [-e⁻ˣ (x + 1)]⁶₀
W = 1100 [-e⁻⁶ (6 + 1)] - [-e⁰ (0 + 1)]
W = 1100 [-0.0173512652 + 1]
W = 1100 × (0.9826487348)
W = 1080.914 J
Hope this Helps!!!
Answer:
3.5x^2 +25x +250-y=0 derivate this eqn you get 7x +25-1 =0 7x =-24 divided the eqn and you get it then inter the value what you get in first eqn
Answer: Peak wavelength
{lambda max}
= 9.7*EXP{-7}meter
Which is approximately,
1 micro-meter.
Explanation: lambda{max} which is peak wavelength is inversely proportional to temperature {T}.This is given by the wiens displacement law.
Lambda max
=max displacement{Xmax} / T
For the first case at T = 6000K
Lambda max = 483 nano-meter
=483*EXP{-9}meter.
So let's solve for max displacement {Xmax}.
Xmax = T*lambda max
= 6000*483*EXP{-9}
=2.898*EXP{-3}kelvin-meter
Xmax would be constant during Temperature change.
Therefore lambda max at 3000K would be,
Lambda max
= {2.898*EXP{-3} K-m} / 3000K
= 9.7*EXP{-7} meter
Which is approximately,
1*EXP{-6} meter= 1 micro-meter
NOTE: EXP used here means 10^.
