Answer:
vB = - 0.176 m/s (↓-)
Explanation:
Given
(AB) = 0.75 m
(AB)' = 0.2 m/s
vA = 0.6 m/s
θ = 35°
vB = ?
We use the formulas
Sin θ = Sin 35° = (OA)/(AB) ⇒ (OA) = Sin 35°*(AB)
⇒ (OA) = Sin 35°*(0.75 m) = 0.43 m
Cos θ = Cos 35° = (OB)/(AB) ⇒ (OB) = Cos 35°*(AB)
⇒ (OB) = Cos 35°*(0.75 m) = 0.614 m
We apply Pythagoras' theorem as follows
(AB)² = (OA)² + (OB)²
We derive the equation
2*(AB)*(AB)' = 2*(OA)*vA + 2*(OB)*vB
⇒ (AB)*(AB)' = (OA)*vA + (OB)*vB
⇒ vB = ((AB)*(AB)' - (OA)*vA) / (OB)
then we have
⇒ vB = ((0.75 m)*(0.2 m/s) - (0.43 m)*(0.6 m/s) / (0.614 m)
⇒ vB = - 0.176 m/s (↓-)
The pic can show the question.
Answer:
the answer how you analyzs the problwm
The answer is true because if the effect is neglected, the saturation id region is considered true
Answer: a) 1.05kW b) 3.78MJ c) 5.3 bars
Explanation :
A)
Conversions give 900 kcal as 900000 x 4.2 J/cal {4.2 J/cal is the standard factor}
= 3780kJ
And 1 hour = 3600s
Therefore, Power in watts = 3780/3600 = 1.05kW = 1050W
B)
At 15km/hour a 15km run takes 1 hour.
1 hour is 3600s and the runner burns 1050 joule per second.
Energy used in 1 hour = 3600 x 1050 J/s
= 3780000 J or 3.78MJ
C)
1 mile = 1.61km so 13.1 mile is 13.1 x 1.61 = 21.1km
15km needs 3.78 MJ of energy therefore 21.1km needs 3.78 x 21.1/15 = 5.32MJ =5320 kJ
Finally,
1 Milky Way = 240000 calories = 4.2 x 240000 J = 1008000J or 1008kJ
This means that the runner needs 5320/1008 = 5.3 bars
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I had a couple of answers for this, but when I checked nothing
was right, so im not sure.
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