Answer:
Its approx location is (5.18,1.9)
Explanation:
Using F( 5,2) = ( xy-1, y²-11)
= ( 5*2-¹, 2²-11)
= (9,-5)
= so at point t=1.02
(5,2)+(1.02-1)*(9,-5)
(5,2)+( 0.02)*(9,-5)
(5+0.18, 2-0.1)
= ( 5.18, 1.9)
Answer:
The mass flow rate is 2.37*10^-4kg/s
The exit velocity is 34.3m/s
The total flow of energy is 0.583 KJ/KgThe rate at which energy leave the cooker is 0.638KW
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Answer:
<em>c. ABBA counterbalancing
</em>
Explanation:
The student should not use the method because it is a progressive error management technique for each subject by introducing all <em>treatment circumstances twice, first in one sequence, then in the other (AB, BA) by subject counterbalancing.</em>
If participants experience conditions more than once, they experience the conditions first in one order, then the opposite order.
Answer:
a. A = 0.735 m
b. T = 0.73 s
c. ΔE = 120 J decrease
d. The missing energy has turned into interned energy in the completely inelastic collision
Explanation:
a.
4 kg * 10 m /s + 6 kg * 0 m/s = 10 kg* vmax
vmax = 4.0 m/s
¹/₂ * m * v²max = ¹/₂ * k * A²
m * v² = k * A² ⇒ 10 kg * 4 m/s = 100 N/m * A²
A = √1.6 m ² = 1.26 m
At = 2.0 m - 1.26 m = 0.735 m
b.
T = 2π * √m / k ⇒ T = 2π * √4.0 kg / 100 N/m = 1.26 s
T = 2π *√ 10 / 100 *s² = 1.99 s
T = 1.99 s -1.26 s = 0.73 s
c.
E = ¹/₂ * m * v²max =
E₁ = ¹/₂ * 4.0 kg * 10² m/s = 200 J
E₂ = ¹/₂ * 10 * 4² = 80 J
200 J - 80 J = 120 J decrease
d.
The missing energy has turned into interned energy in the completely inelastic collision
Answer: I think Its the Height is 11.76 Meters (38.582677 Feet) between the bridge and the ground
Explanation: Supposing that where not counting air resistance in the equation, the equation 
states that 1/2 multiplied by earths gravitational acceleration multiplied by the amount of time to reach the bottom: 2.4 seconds equals 11.76 meters of height between the bridge and the ground.
