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2 years ago

15

Suppose a 4,000-kg elephant is hoisted 20 m above Earth’s surface. What will the elephant’s gravitational potential energy be? (

Show your work in the space below.)

2 answers:

Len [333]2 years ago

5 0

Answer: 784000 J or 784 KJ

Explanation:

U = m * g * h

U = 4000 * 9.8 * 20

U = 784000 J

Hope this was helpful:D

Iteru [2.4K]2 years ago

3 0

Answer: 784000 J

Explanation:

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The site<br> of proteins is called what

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The current and the potential difference in an inductor are in phase. B. The current lags the potential difference by π/2 in an

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Answer:

The current lags the potential difference by π/2 in an inductor

Explanation:

The potential difference leads to the current by $\frac{\pi}{2}$ . Alternate signals such as current and voltage -in this case- are periodic, this means that this signals are repeated at fixed spaces of time. Thus, In an inductor the current lags the potential difference by $\frac{\pi}{2}$ .

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2 years ago

in a one dimensional collision, a 4kg object and 6 kg onject have initial velocity. calculate the magnitude of impulse

Greeley [361]

in a one dimensional collision, a 4kg object with 5ms^1 and 6 kg object with 2ms^1 have initial velocity, the magnitude of impulse is 12 , 18

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mass 1 = 4kg

mass 2 = 6kg

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velocity 2 = 2ms^1

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1 year 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.

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3 years ago