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Nadya [2.5K]
3 years ago
5

Stan and Silvia Waterson go on a vacation and leave their 17 year old daughter, Jessie, in charge of the house and her two young

er siblings. While the elder Watersons are away, a severe storm knocks down power lines and the electricity is off for a week. Since all the food in the refrigerator and freezer spoiled, Jessie talks the owner of a small grocery store into letting her have groceries on credit. When the Waterson parents come back home, the store demands payment from them for the groceries. On what basis must the parents pay for the groceries
Physics
1 answer:
Marina CMI [18]3 years ago
5 0

Answer:

Generally, minors are not allowed to engage in contracts, except when minors seek basic necessities. Basic necessities include food, shelter and clothes (normal clothes, not designer clothes). If a minor engages in any contract regarding basic necessities, it is considered a valid contract. If a minor engages in contracts to purchase anything else not considered a basic necessity, the contracts are voidable.

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Explanation:

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A box slides down a 31° ramp with an acceleration of 0.99 m/s2. Determine the coefficient of kinetic friction between the box an
tino4ka555 [31]

Answer:\mu=0.48

Explanation:

Given

inclination \theta =31^{\circ}

Acceleration of object=0.99 m/s^2

Now using FBD

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Which of the following is an example of a physical change?
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3 years ago
A sealed tank containing seawater to a height of 12.8 m also contains air above the water at a gauge pressure of 2.90 atm . Wate
exis [7]

Answer:

The velocity of water at the bottom, v_{b} = 28.63 m/s

Given:

Height of water in the tank, h = 12.8 m

Gauge pressure of water, P_{gauge} = 2.90 atm

Solution:

Now,

Atmospheric pressue, P_{atm} = 1 atm = 1.01\tiems 10^{5} Pa

At the top, the absolute pressure, P_{t} = P_{gauge} + P_{atm} = 2.90 + 1 = 3.90 atm = 3.94\times 10^{5} Pa

Now, the pressure at the bottom will be equal to the atmopheric pressure, P_{b} = 1 atm = 1.01\times 10^{5} Pa

The velocity at the top, v_{top} = 0 m/s, l;et the bottom velocity, be v_{b}.

Now, by Bernoulli's eqn:

P_{t} + \frac{1}{2}\rho v_{t}^{2} + \rho g h_{t} = P_{b} + \frac{1}{2}\rho v_{b}^{2} + \rho g h_{b}

where

h_{t} -  h_{b} = 12.8 m

Density of sea water, \rho = 1030 kg/m^{3}

\sqrt{\frac{2(P_{t} - P_{b} + \rho g(h_{t} - h_{b}))}{\rho}} =  v_{b}

\sqrt{\frac{2(3.94\times 10^{5} - 1.01\times 10^{5} + 1030\times 9.8\times 12.8}{1030}} =  v_{b}

v_{b} = 28.63 m/s

5 0
3 years ago
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