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sergiy2304 [10]
3 years ago
9

You are holding a shopping basket at the grocery store with two 0.55-kg cartons of cereal at the left end of the basket. the bas

ket is 0.78 m long. where should you place a 1.8-kg half gallon of milk, relative to the left end of the basket, so that the center of mass of your groceries is at the center of the basket?

Physics
1 answer:
stepan [7]3 years ago
5 0
Refer to the diagram shown below.

The basket is represented by a weightless rigid beam of length 0.78 m.
The x-coordinate is measured from the left end of the basket.

The mass at x=0 is 2*0.55 = 1.1 kg.
The weight acting at x = 0 is W₁ = 1.1*9.8 = 10.78 N

The mass near the right end is 1.8 kg.
Its weight is W₂ = 1.8*9.8 = 17.64 N

The fulcrum is in the middle of the basket, therefore its location is
x = 0.78/2 = 0.39 m.

For equilibrium, the sum of moments about the fulcrum is zero.
Therefore
(10.78 N)*(0.39 m) - (17.64 N)*(x-0.39 m) = 0
4.2042 - 17.64x + 6.8796 = 0
-17.64x = -11.0838
x = 0.6283 m

Answer:  0.63 m from the left end.

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Hi

The answer to this question is B. Reaction

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2 years ago
A(n) 0.2 kg object is swung in a vertical circular path on a string 0.1 m long. The acceleration of gravity is 9.8 m/s2 . If a c
Leya [2.2K]

Answer:

T=83.37N

Explanation:

Since the object is under a circular motion, according to Newton's second law, when the object is at the top of the circle we have:

\sum F_y: T-mg=F_c

Where F_c is the centripetal force and is given by:

F_c=ma_c=m\frac{v^2}{r}

Replacing and solving for T:

T=m\frac{v^2}{r}+mg\\T=0.2kg\frac{(6.38\frac{m}{s})^2}{0.1m}+0.2kg(9.8\frac{m}{s^2})\\T=83.37N

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3 years ago
A 2,000 kg railroad car is coasting on a track at a constant velocity of 20 m/s. As the car coasts under a loading ramp, a 500 k
BlackZzzverrR [31]

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the car with the hay should slow to 16m/s if the bale of hay is dropped into it.

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3 years ago
In the nuclear fission process mass is converted into energy. Determine the total mass that must be converted to energy in one y
brilliants [131]

Answer:

1) Mass that needs to be converted at 100% efficiency is 0.3504 kg

2) Mass that needs to be converted at 30% efficiency is 1.168 kg

Explanation:

By the principle of mass energy equivalence we have

E=mc^{2}

where,

'E' is the energy produced

'm' is the mass consumed

'c' is the velocity of light in free space

Now the energy produced by the reactor in 1 year equals

Energy=Power\times time\\\\\therefore Energy=1\times 10^{9}\times 365\times 24\times 3600\\\\Energy=31.536\times 10^{15}Joules

Thus the mass that is covertred at 100% efficiency is

mass=\frac{Energy}{c^{2}}\\\\mass=\frac{31.536\times 10^{15}}{(3\times 10^{8})^{2}}\\\\mass=\frac{31.536\times 10^{15}}{9\times 10^{16}}\\\\\therefore mass=0.3504kg

Part 2)

At 30% efficiency the mass converted equals

mass|_{30}=\frac{mass|_{100}}{0.3}\\\\mass|_{30}=\frac{0.3504}{0.3}\\\\mass|_{30}=1.168kg

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3 years ago
Can any ideal gas power cycle have a thermal efficiency greater than 55 percent when using thermal energy reservoirs at 627∘C at
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Answer:

Since the maximum thermal efficiency is higher than 55 percent, there can be a power cycle with these reservoir temperature with an efficiency higher than 55 percent.

Explanation:

The maximum thermal efficiency is determined from the given temperature

nth Carnot = 1- TL/TH

Where TL= 17+273= 290k

TH= 627*273= 900K.

nth Carnot = 1- 290/900 = 0.68

0.68*100 = 68 percent

8 0
3 years ago
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