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3 years ago

5

A spherical weather balloon is filled with hydrogen until its radius is 4.40 m. Its total mass including the instruments it carr

ies is 19.0 kg. (a) Find the buoyant force acting on the balloon, assuming the density of air is 1.29 kg/m3. kN (b) What is the net force acting on the balloon and its instruments after the balloon is released from the ground? kN (c) Why does the radius of the balloon tend to increase as it rises to higher altitude?

1 answer:

ipn [44]3 years ago

5 0

Answer:

4515.49484 N

4329.10484 N

Explanation:

r = Radius of balloon = 4.4 m

m = Mass of balloon with instruments = 19 kg

g = Acceleration due to gravity = 9.81 m/s²

Volume of balloon

$v=\frac{4}{3}\pi r^3\\\Rightarrow v=\frac{4}{3}\pi 4.4^3\\\Rightarrow v=356.8179\ m^3$

The Buoyant force = Weight of the air displaced

$F=\rho vg\\\Rightarrow F=1.29\times 356.8179\times 9.81\\\Rightarrow F=4515.49484\ N$

The buoyant force acting on the balloon is 4515.49484 N

Net force on the balloon

$F_n=F-W\\\Rightarrow F_n=4515.49484-19\times 9.81\\\Rightarrow F_n=4329.10484\ N$

The net force on the balloon is given by 4329.10484 N

As the balloon goes up the pressure outside reduces as the density of air decreases while the air pressure inside the balloon is high hence, the radius of the balloon tend to increase as it rises to higher altitude.

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Voltage=V=120V
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2 years ago

Two 51 g blocks are held 30 cm above a table. As shown in the figure, one of them is just touching a 30-long spring. The blocks

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The concept of this question can be well understood by listing out the parameters given.

The mass of the block = 51 g = 51 × 10⁻³ kg
The distance of the block from the table = 30 cm
Length of the spring = 30 cm

The purpose is to determine the spring constant.

Let us assume that the two blocks are Block A and Block B.

At point A on block A, the initial velocity on the block is zero

i.e. u = 0

We want to determine the time it requires for Block A to reach the table. The can be achieved by using the second equation of motion which can be expressed by using the formula.

$\mathsf{S = ut + \dfrac{1}{2}gt^2}$

From the above formula,

The distance (S) = 30 cm; we need to convert the unit to meter (m).

Since 1 cm = 0.01 m
Then, 30cm = 0.3 m

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∴

inputting the values into the equation above, we have;

$\mathsf{0.3 = (0)t + \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 = \dfrac{1}{2}*(9.80)*(t^2)}$

$\mathsf{0.3 =4.9*(t^2)}$

By dividing both sides by 4.9, we have:

$\mathsf{t^2 = \dfrac{0.3}{4.9}}$

$\mathsf{t^2 = 0.0612}$

$\mathsf{t = \sqrt{0.0612}}$

$\mathsf{t =0.247 \ seconds}$

However, block B comes to an instantaneous rest on point C. This is achieved by the dropping of the block on the spring. During this process, the spring is compressed and it bounces back to oscillate in that manner. The required time needed to get to this point C is half the period, this will eventually lead to the bouncing back of the block with another half of the period, thereby completing a movement of one period.

By applying the equation of the time period of a simple harmonic motion.

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

where the relation between time (t) and period (T) is:

$\mathsf{t = \dfrac{T}{2}}$

T = 2t

T = 2(0.247)

T = 0.494 seconds

$\mathsf{T = 2 \pi \sqrt{\dfrac{m}{k}}}$

By making the spring constant k the subject of the formula:

$\mathsf{\dfrac{T}{2 \pi } = \sqrt{ \dfrac{m}{k}}}$

$\Big(\dfrac{T}{2 \pi }\Big)^2 = { \dfrac{m}{k}$

$\dfrac{T^2}{(2 \pi)^2 }= { \dfrac{m}{k}$

$\mathsf{ T^2 *k = 2 \pi^2*m} \\ \\ \mathsf{ k = \dfrac{2 \pi^2*m}{T^2}}$

$\mathsf{ k =\Big( \dfrac{(2 \pi)^2*(51 \times 10^{-3})}{(0.494)^2} \Big) N/m}$

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Therefore, we conclude that the spring constant as a result of instantaneous rest caused by the compression of the spring is 8.25 N/m.

Learn more about simple harmonic motion here:

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F = mgh / d

W = mg

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B) calculate the force when the stopping distance = 0.345 m

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Fd = mgh hence F = mgh / d

F(net) = W + F = mg ( 1 + $\frac{h}{d} )$

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