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Marrrta [24]
3 years ago
15

In the two-stage cooling method, what is the maximum amount of time allowed to cool food from 135 degrees F or more to 70 degree

s F or less?
Physics
1 answer:
Tresset [83]3 years ago
5 0
The problem ask to find the maximum amount of time that is allowed to cool food from 135 degree F or more to 70 degree F or less and the allowed time would be that the food would be cool of is around to 4 to 6 hours. I hope you are satisfied with my answer and feel free to ask for more of you have questions and further clarifications 
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Object A is moving due east, while object B is moving due north. They collide and stick together in a completely inelastic colli
serious [3.7K]

Answer:

pf = 198.8 kg*m/s

θ = 46.8º N of E.

Explanation:

  • Since total momentum is conserved, and momentum is a vector, the components of the momentum along two axes perpendicular each other must be conserved too.
  • If we call the positive x- axis to the W-E direction, and the positive y-axis to the S-N direction, we can write the following equation for the initial momentum along the x-axis:

       p_{ox} = p_{oAx} + p_{oBx}  (1)

  • We can do exactly the same for the initial momentum along the y-axis:

       p_{oy} = p_{oAy} + p_{oBy}  (2)

  • The final momentum along the x-axis, since the collision is inelastic and both objects stick together after the collision, can be written as follows:

       p_{fx} =  (m_{A} + m_{B} ) * v_{fx}  (3)

  • We can repeat the process for the y-axis, as follows:

       p_{fy} =  (m_{A} + m_{B} ) * v_{fy}  (4)

  • Since (1) is equal to (3), replacing for the givens, and since p₀Bₓ = 0, we can solve for vfₓ as follows:

       v_{fx} = \frac{p_{oAx}}{(m_{A}+ m_{B)}} = \frac{m_{A}*v_{oAx} }{(m_{A}+ m_{B)}} =\frac{17.0kg*8.00m/s}{46.0kg} =  2.96 m/s (5)

  • In the same way, we can find the component of the final momentum along the y-axis, as follows:

       v_{fy} = \frac{p_{oBy}}{(m_{A}+ m_{B)}} = \frac{m_{B}*v_{oBy} }{(m_{A}+ m_{B)}} =\frac{29.0kg*5.00m/s}{46.0kg} =  3.15 m/s (6)

  • With the values of vfx and vfy, we can find the magnitude of the final speed of the two-object system, applying the Pythagorean Theorem, as follows:

      v_{f} = \sqrt{v_{fx} ^{2} + v_{fy} ^{2}} = \sqrt{(2.96m/s)^{2} + (3.15m/s)^{2}} = 4.32 m/s (7)

  • The magnitude of the final total momentum is just the product of the combined mass of both objects times the magnitude of the final speed:

       p_{f} = (m_{A} + m_{B})* v_{f}  = 46 kg * 4.32 m/s = 198.8 kg*m/s (8)

  • Finally, the angle that the final momentum vector makes with the positive x-axis, is the same that the final velocity vector makes with it.
  • We can find this angle applying the definition of tangent of an angle, as follows:

       tg \theta = \frac{v_{fy}}{v_{fx}} = \frac{3.15 m/s}{2.96m/s} = 1.06 (9)

       ⇒ θ = tg⁻¹ (1.06) = 46.8º N of E

8 0
3 years ago
A man goes 150 m due east and then 200 m due north how far is he from starting point
grin007 [14]

Answer:

250 m

Explanation:

Since North and East are 90 degrees from each other, we can treat this as a right-angled triangle, with the distance in each direction being the sides and the distance from the starting point being the hypotenuse.

Hence, sqrt(150^2+200^2) = 250 m.

Hope this helped!

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Answer:

trur

Explanation:

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Claim 2: Molecules speed up when they get energy from other molecules and slow down when they give energy to other molecules.
Energy can’t be destroyed (stated in claim 1) so claim 2 is more than likely to be correct
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Explanation:

Period of a mass on a spring is:

T = 2π√(m/k)

T is inversely proportional with the square root of k.  So as the spring constant increases, the period decreases.

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