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avanturin [10]
3 years ago
6

A Person whose weight is 5.20 x 10^2 N is being pulledup

Physics
1 answer:
Gnoma [55]3 years ago
6 0

Answer:

Explanation:

Given

Weight of Person W=5.20\times 10^{2} N

Cave is h=35.1 m deep

Breaking stress T=569 N

Net Force on Person

F_{net}=569-520=49 N

a_{net}=\frac{F_{net}}{\frac{W}{g}}

a_{net}=\frac{49}{\frac{520}{9.8}}

a_{net}=0.923 m/s^2

The shortest time such that the person can be taken out of cave

h=ut+\frac{1}{2}at^2

where

h=distance moved

t=time

a=acceleration

35.1=0+\frac{1}{2}(0.923)(t)^2

t^2=76.05

t=\sqrt{76.05}

t=8.72\ s                    

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The theory of plate tectonics describes how the Earth’s lithosphere is broken into plates and these plates move over the molten
Mumz [18]

Answer:

C

Explanation:

A theory can be changed when new evidence is found. A law doesn't change because it is universally a fact. It doesn't need new evidence to support it.

5 0
3 years ago
The inhabitants of a small island export a cloth made from a plant that grows only on their island. A clothier from New York, be
Inessa [10]

Answer:

Explanation:

purchase price = 400 tepizes / m²

1 tepiz = .625 dollar

purchase price in terms of dollar = 400 x .625 dollar / m²

= 250 dollar / m²

.9144 m = 1 yard

1 m = 1.0936 yard

1m² = 1.196 yard²

price in terms of dollar / yards²

= 250 / 1.196 dollar / yard²

= 209 dollar / yard²

Price of cloth in New York = 120 dollar / yard²

loss = 209 - 120 = 89 dollar / yard²

500 m² = 500 x 1.196 yard²

= 598 yard²

net loss in purchasing 500 m² cloth

= 598 x 89

= 53222 dollar .

7 0
4 years ago
An electron is accelrated by a unifor electric field (1000v/m) pointing vertically upward. Use energy methods to get the magnitu
ExtremeBDS [4]

Explanation:

In the given situation two forces are working. These are:

1) Electric force (acting in the downward direction) = qE

2) weight (acting in the downward direction) = mg

Therefore, work done by all the forces = change in kinetic energy

Hence, qE \times S + mg \times S = 0.5 \times mv^{2}

     1.6 \times 10^{-19} \times 1000 + 9.1 \times 10^{-31} \times 9.8 \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2}

It is known that the weight of electron is far less compared to electric force. Therefore, we can neglect the weight  and the above equation will be as follows.

   (1.6 \times 10^{-19} \times 1000) \times (\frac{0.10}{100}) = 0.5 \times 9.1 \times 10^{-31} \times v^{2
}

         v = sqrt{\frac{1.6 \times 10^{-19}}{(0.5 \times 9.1 \times 10^{-31})}

           = 592999 m/s

Since, the electron is travelling downwards it means that it looses the potential energy.

8 0
3 years ago
Consider a hydrogen atom in the n = 1 state. The atom is placed in a uniform B field of magnitude 2.5 T. Calculate the energy di
dlinn [17]

Answer:

E=29\times 10^{-5}eV

Explanation:

For n-=1 state hydrogen energy level is split into three componets in the presence of external magnetic field. The energies are,

E^{+}=E+\mu B,

E^{-}=E-\mu B,

E^{0}=E

Here, E is the energy in the absence of electric field.

And

E^{+} and E^{-} are the highest and the lowest energies.

The difference of these energies

\Delta E=2\mu B

\mu=9.3\times 10^{-24}J/T is known as Bohr's magneton.

B=2.5 T,

Therefore,

\Delta E=2(9.3\times 10^{-24}J/T)\times 2.5 T\\\Delta E=46.5\times 10^{-24}J

Now,

Delta E=46.5\times 10^{-24}J(\frac{1eV}{1.6\times 10^{-9}J } )\\Delta E=29.05\times 10^{-5}eV\\Delta E\simeq29\times 10^{-5}eV

Therefore, the energy difference between highest and lowest energy levels in presence of magnetic field is E=29\times 10^{-5}eV

6 0
3 years ago
A tourist being chased by an angry bear is running in a straight line toward his car at a speed of 4.2 m/s. The car is a distanc
dem82 [27]

Explanation:

It is known that the relation between speed and distance is as follows.

               velocity = \frac{distance}{time}

As it is given that velocity is 6 m/s and distance traveled by the bear is (d + 29). Therefore, time taken by the bear is calculated as follows.

         t_{bear} = \frac{(d + 29)}{6 m/s} ............. (1)

As the tourist is running in a car at a velocity of 4.2 m/s. Hence, time taken by the tourist is as follows.

              t_{tourist} = \frac{d}{4.2} ............. (2)

Now, equation both equations (1) and (2) equal to each other we will calculate the value of d as follows.

              t_{bear} = t_{tourist}

       \frac{(d + 29)}{6 m/s} = \frac{d}{4.2}

                   4.2d + 121.8 = 6d

                         d = \frac{121.8}{1.8}

                            = 67.66

Thus, we can conclude that the maximum possible value for d is 67.66.

5 0
3 years ago
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