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valina [46]
3 years ago
8

Ahmed kept a balloon from his birthday for several days until it began to deflate and slowly sink to the ground. He wondered how

the balloon was losing its ability to float. He decided to use his Physical Science textbook to get more answers.
What chapter should Ahmed read to learn more about how things function or behave?

Laws
Theories
Hypotheses
Observations
Physics
2 answers:
Nikolay [14]3 years ago
8 0
Trust me, the answer is laws
harina [27]3 years ago
5 0
Most likely theories because there are many theories on why balloons lose their air
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<em>All the three light bulbs will glow</em>

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Sholpan [36]

Answer:

SO4

Explanation:

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3 0
3 years ago
In an RLC series circuit that includes a source of alternating current operating at fixed frequency and voltage, the resistance
maw [93]

Answer:

Capacitive Reactance is 4 times of resistance

Solution:

As per the question:

R = X_{L} = j\omega L = 2\pi fL

where

R = resistance

X_{L} = Inductive Reactance

f = fixed frequency

Now,

For a parallel plate capacitor, capacitance, C:

C = \frac{\epsilon_{o}A}{x}

where

x = separation between the parallel plates

Thus

C ∝ \frac{1}{x}

Now, if the distance reduces to one-third:

Capacitance becomes 3 times of the initial capacitace, i.e., x' = 3x, then C' = 3C and hence Current, I becomes 3I.

Also,

Z = \sqrt{R^{2} + (X_{L} - X_{C})^{2}}

Also,

Z ∝ I

Therefore,

\frac{Z}{I} = \frac{Z'}{I'}

\frac{\sqrt{R^{2} + (R - X_{C})^{2}}}{3I} = \frac{\sqrt{R^{2} + (R - \frac{X_{C}}{3})^{2}}}{I}

{R^{2} + (R - X_{C})^{2}} = 9({R^{2} + (R - \frac{X_{C}}{3})^{2}})

{R^{2} + R^{2} + X_{C}^{2} - 2RX_{C} = 9({R^{2} + R^{2} + \frac{X_{C}^{2}}{9} - 2RX_{C})

Solving the above eqn:

X_{C} = 4R

6 0
3 years ago
a projectile is shot horizontally from the edge of a cliff, 230m above the ground. the projectile lands 300m from base of the cl
bekas [8.4K]

Answer:

The time taken by the projectile to hit the ground is 6.85 sec.

Explanation:

Given that,

Vertical height of cliff = 230 m

Distance = 300 m

Suppose, determine the time taken by the projectile to hit the ground.

We need to calculate the time

Using second equation of motion

s=ut+\dfrac{1}{2}gt^2

Where, s = vertical height of cliff

u = initial vertical velocity

g = acceleration due to gravity

Put the value in the equation

230=0+\dfrac{1}{2}\times9.8\times t^2

t=\sqrt{\dfrac{230\times2}{9.8}}

t=6.85 sec

Hence, The time taken by the projectile to hit the ground is 6.85 sec.

7 0
3 years ago
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