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svp [43]
2 years ago
15

the pressure inside a latex balloon is nearly the same as the pressure outside. if you let a helium balloon go, and if, as it ri

ses, it stays at a constant temperature, the volume of the balloon will
Physics
1 answer:
olga_2 [115]2 years ago
3 0

Answer:       If you let a helium balloon go, and if, as it rises, it stays at a constant temperature, the volume of the balloon will. The magnitude of the buoyant force is equal to the weight of the amount of fluid that has the same total volume as the object.

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7. A particle is forced to move in a straight line path. It returns to the starting point after 10 s. The total distance covered
lutik1710 [3]

d. All three statements are true.

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1 year ago
Darren filled ocean water, fresh water, bottled water, and tap water into four different containers. He then dropped identical g
elena-14-01-66 [18.8K]
I would say container 1
6 0
3 years ago
Read 2 more answers
An L-R-C series circuit, R = 160 Ω , L = 0.790 H , and C = 1.30×10−2 μF . The source has a voltage amplitude of 140 V and a freq
wolverine [178]

Answer: a) 1 b) 61 W c) 61 W

Explanation:

a) The  Power Factor (also known as cos φ), is defined by the difference in phase between current and voltage, in a RLC series circuit, and is expressed as follows:

cos φ = R / Z = R / \sqrt{(R)^{2} + (Xl -Xc)^{2} }

In resonance, XL =XC, so the circuit behaves like it were purely resistive, so Z=R.

Replacing in the expression for power factor, we have:

cos φ = R/Z = R/R = 1

This means that in resonance, current and voltage are in phase each other.

b) The average power delivered by the source, in resonance, is simply the power dissipated at the resistance R, as follows:

Pavg = V² rms / R = V₀² / √2 / R = 61 W

c) If the circuit remains in resonance, the average power , which does not depends on frequency provided this condition remains, keeps the same, i.e. , 61 W.

7 0
3 years ago
Can someone please help me with this
SashulF [63]
There’s no picture or question
3 0
3 years ago
Playing near a road construction site, a child falls over a barrier and down onto a dirt slope that is angled downward at 33° to
Debora [2.8K]

Answer:

\mu_{k} \approx 0.719

Explanation:

The equations of equilibrium for the child are: (x' in the direction parallel to slope, y' in the direction perpendicular to slope)

\Sigma F_{x'} = m\cdot g \cdot \sin \theta - \mu_{k}\cdot N = m\cdot a\\\Sigma F_{y'} = N - m\cdot g\cdot \cos \theta = 0

After some algebraic manipulation, an expression for the coefficient of kinetic friction is obtained:

m\cdot g\cdot \sin \theta - \mu_{k}\cdot m \cdot g \cos \theta = m \cdot a

g \cdot (\sin \theta - \mu_{k}\cdot \cos \theta) = a

\mu_{k}\cdot \cos \theta = \sin \theta - \frac{a}{g}

\mu_{k} = \frac{1}{\cos \theta}\cdot (\sin \theta - \frac{a}{g} )

\mu_{k} = \frac{1}{\cos 33^{\textdegree}}\cdot \left(\sin 33^{\textdegree}-\frac{(-0.57\,\frac{m}{s^{2}}) }{9.807\,\frac{m}{s^{2}} } \right)

\mu_{k} \approx 0.719

5 0
3 years ago
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