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4 years ago

Will mark as BRAINLIEST.....

Physics

1 answer:

Jlenok [28]4 years ago

5 0

Answer: initially the packet was ascending up with the balloon.

Taking upward as positive direction;

initial velocity, u = 4.9 m/s

final velocity = v m/s

initial height, h₁ = 245 m

final height, h₂ = 0

a = -9.8 m/s²

time taken = t seconds

s = ut + 0.5at²

⇒ (h₂-h₁) = ut + 0.5at²

⇒ 0-245 = 4.9t + 0.5×(-9.8)×t²

⇒ -245 = 4.9t - 4.9t²

⇒ 4.9t² -4.9t -245 =0

Solving it, we get t = 7.59s

v = u + at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/s

So velocity is 69.48 m/s downward

Explanation:

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Anna007 [38]

The average speed is determined by the following formula:

average speed = [sum of (speed * time for which that speed was traveled)] / total time

average speed = [(83 * 26 + 41 * 52 + 60 * 45 + 0 * 15) / 60] / [(26 + 52 + 45 + 15) / 60]
*note: The division by 60 is to convert minutes to hours. We see that the 60 cancels from the top and bottom of the division

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The total distance traveled is equivalent to the numerator of the fraction we used in the first part. This is:
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4 years ago

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dolphi86 [110]

Increase; newton’s second law of motion

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3 years ago

Josh is much bigger than taylor. after the push, which of the two is moving faster?

egoroff_w [7]

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3 years ago

Suppose a two-level system is in a bath with temperature 247 K. The energy difference between the two states is 1.1 × 10-21 J. W

Alik [6]

Answer:

The probability of higher energy state is 0.4200.

Explanation:

Given that,

Temperature = 247 K

Energy difference between two states $E_{2}-E_{1}=1.1\times10^{-21}\ J$

We need to calculate the probability of higher energy state

Probability of $E_{1}= e^{-\beta E_{1}}$

Probability of $E_{2}= e^{-\beta E_{2}}$

The total probability is

$e^{-\beta E_{1}}+e^{-\beta E_{1}}=1$

Here, E₁ = lower energy state

E₂ = higher energy state

Put the value of E₁ in to the formula

$e^{-\beta(E_{2}-1.1\times10^{-21})}+e^{-\beta E_{1}}=1$

$e^{-\beta E_{2}}(e^{\beta 1.1\times10^{-21}}+1)=1$

$e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{KT}}}$

Here, $\beta=\dfrac{1}{KT}$

Put the value into the formula

$e^{\beta E_{2}}=\dfrac{1}{1+e^{\dfrac{1.1\times10^{-21}}{1.380\times10^{-23}\times247}}}$

$e^{\beta E_{2}}=0.4200$

Hence, The probability of higher energy state is 0.4200.

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3 years ago

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