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Alona [7]
2 years ago
9

7. What is the velocity of an object with a distance of 90m south and a time of 5s?

Physics
1 answer:
IrinaK [193]2 years ago
4 0

Answer:

Explanation:

v= s/t

V =90m/5s

V = 8m/s

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A buoy is anchored to the ocean floor a large wave approaches the buoy how will the buoy move as the wave goes by
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8 0
3 years ago
A 2 eV electron encounters a barrier 5.0 eV high and width a. What is the probability b) 0.5 that it will tunnel through the bar
denis23 [38]

Answer:

The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

Explanation:

Given that,

Energy E = 2 eV

Barrier V₀= 5.0 eV

Width = 1.00 nm

We need to calculate the value of \beta

Using formula of \beta

\beta=\sqrt{\dfrac{2m}{\dfrac{h}{2\pi}}(v_{0}-E)}

Put the value into the formula

\beta = \sqrt{\dfrac{2\times9.1\times10^{-31}}{(1.055\times10^{-34})^2}(5.0-2)\times1.6\times10^{-19}}

\beta=8.86\times10^{9}

(a). We need to calculate the tunnel probability for width 0.5 nm

Using formula of tunnel barrier

T=\dfrac{16E(V_{0}-E)}{V_{0}^2}e^{-2\beta a}

Put the value into the formula

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times0.5\times10^{-9}}

T=5.45\times10^{-4}

(b). We need to calculate the tunnel probability for width 1.00 nm

T=\dfrac{16\times 2(5.0-2.0)}{5.0^2}e^{-2\times8.86\times10^{9}\times1.00\times10^{-9}}

T=7.74\times10^{-8}

Hence, The tunnel probability for 0.5 nm and 1.00 nm are  5.45\times10^{-4} and 7.74\times10^{-8} respectively.

6 0
3 years ago
There are 8.8 liters of gas in a piston at a pressure of 1.75 atmospheres. The temperature remains constant, and the gas is comp
saveliy_v [14]

The new pressure P2 is 2.48 atmosphere.

<u>Explanation:</u>

Here, the one of the product of pressure and volume is equal to the products of pressure and volume of other.

By using Boyles's law,

               pressure is inversely proportional to volume,

                                         P1 V1 = P2 V2

where P1, V1 represents the first pressure and volume,

           P2, V2 represents the second pressure and volume

                                              P2 = (P1 V1) / V2

                                                    = (1.75 \times 8.8) / 6.2

                                              P2  = 2.48 atmosphere.  

         

5 0
3 years ago
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