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Korolek [52]
3 years ago
6

A small bag of sand is released from an ascending hot-air balloon whose upward constant velocity is v0 = 2.45 m/s. knowing that

at the time of the release the balloon was 98.8 m above the ground, determine the time, τ, it takes the bag to reach the ground from the moment of its release.

Physics
1 answer:
lakkis [162]3 years ago
8 0
When you draw an illustration for this problem, you would come up with the same drawing as shown in the picture. As the hot-air balloon travels upwards, there is a slight time when the bag of sand rises up until it reaches the maximum height. Then, it goes back down to the ground. The total time would be t₁ + t₂. The solution is as follows:

H = v₀²/2g = (2.45)²/2(9.81) = 0.306 m
t₁ = H/v₀ = 0.306 m/2.45 m/s = 0.125 s

t₂ = √2(H + 98.8)/g = √2(0.306+ 98.8)/9.81
t₂ = 4.495 s

Total time = 0.125 s + 4.495 s = 4.62 seconds

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During the latter part of your European vacation, you are hanging out at the beach at the gold coast of Spain. As you are laying
adelina 88 [10]

The maximum height attained is 460 m.

<h3>What is the maximum height?</h3>

We know that the final velocity of a body is 0 m/s at the maximum height which is the greatest height that is attained by the body. We now use the formula;

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6 0
1 year ago
A heavy solid disk rotating freely and slowed only by friction applied at its outer edge takes 120 seconds to come to a stop.
alisha [4.7K]

Answer:

The time is 16 min.

Explanation:

Given that,

Time = 120 sec

We need to calculate the moment of inertia

Using formula of moment of inertia

I=\dfrac{1}{2}MR^2

If the disk had twice the radius and twice the mass

The new moment of inertia

I'=\dfrac{1}{2}\times2M\times(2R)^2

I'=8I

We know,

The torque is

\tau=F\times R

We need to calculate the initial rotation acceleration

Using formula of acceleration

\alpha=\dfrac{\tau}{I}

Put the value in to the formula

\alpha=\dfrac{F\times R}{\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{MR}

We need to calculate the new rotation acceleration

Using formula of acceleration

\alpha'=\dfrac{\tau}{I'}

Put the value in to the formula

\alpha=\dfrac{F\times R}{8\times\dfrac{1}{2}MR^2}

\alpha=\dfrac{2F}{8MR}

\alpha=\dfrac{\alpha}{8}

Rotation speed is same.

We need to calculate the time

Using formula angular velocity

\Omega=\omega'

\alpha\time t=\alpha'\times t'

Put the value into the formula

\alpha\times120=\dfrac{\alpha}{8}\times t'

t'=960\ sec

t'=16\ min

Hence, The time is 16 min.

5 0
3 years ago
Suppose that 2 J of work are needed to stretch a spring from its natural length of 30 cm to a length of 42 cm. How far (in cm) b
sergeinik [125]

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched

<u>Explanation:</u>

Work, W = 2 J

Initial distance, x_{1} = 30 cm

Final distance,  = 42 cm

Force, F = 30 N

Stretched length, x = ?

We know,

W = 1/2 kΔx²

Δx = 42-30 cm = 12 cm = 0.12 m

2 J = 1/2 k X (0.12)²

k = 277.77 N/m

According to Hooke's law,

F = kx

30 N = 277.77 X x

x = 0.108 m

x = 10.8  cm

A distance of 10.8 cm beyond its natural length will a force of 30 N keep this spring stretched.

7 0
2 years ago
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