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2 years ago

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I. The substance in model A has the highest density.

1 answer:

Lynna [10]2 years ago

6 0

Answer:

If you dont have the answer already its i and iii i believe

Explanation:

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A 250 - ω resistor is connected in series with a 4.80 - μf capacitor. the voltage across the capacitor is vc=(7.60v)⋅sin[(120rad

galben [10]

<span>553 ohms The Capacitive reactance of a capacitor is dependent upon the frequency. The lower the frequency, the higher the reactance, the higher the frequency, the lower the reactance. The equation is Xc = 1/(2*pi*f*C) where Xc = Reactance in ohms pi = 3.1415926535..... f = frequency in hertz. C = capacitance in farads. I'm assuming that the voltage and resistor mentioned in the question are for later parts that are not mentioned in this question. Reason is that they have no effect on the reactance, but would have an effect if a question about current draw is made in a later part. With that said, let's calculate the reactance. The 120 rad/s frequency is better known as 60 Hz. Substitute known values into the formula. Xc = 1/(2*pi* 60 * 0.00000480) Xc = 1/0.001809557 Xc = 552.6213302 Rounding to 3 significant figures gives 553 ohms.</span>

6 0

3 years ago

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A 234.0 g piece of lead is heated to 86.0oC and then dropped into a calorimeter containing 611.0 g of water that initally is at

Vaselesa [24]

Answer: $24.70 ^{\circ}C$

Explanation:

Given

mass of lead piece $m_l=234 gm\approx 0.234 kg$

mass of water in calorimeter $m_w=611 gm\approx 0.611 kg$

Initial temperature of water $T_w=24^{\circ}C$

Initial temperature of lead piece $T_l=24^{\circ}C$

we know heat capacity of lead and water are $125.604 J/kg-k$ and $4.184 kJ/kg-k$ respectively

Let us take $T ^{\circ}C$ be the final temperature of the system

Conserving energy

heat lost by lead=heat gained by water

$m_lc_l(T_l-T)=m_wc_w(T-T_w)$

$0.234\times 125.604(86-T)=0.611\times 4.184\times 1000(T-24)$

$86-T=\frac{0.611\times 4.184\times 1000}{29.391}(T-24)$

$86-T=86.97T-2087.49$

$T=\frac{2173.491}{87.97}=24.70^{\circ}C$

3 0

2 years ago

A uniform, spherical, 1900.0 kg shell has a radius of 5.00 m. Find the gravitational force this shell exerts on a 1.80 kg point

Mandarinka [93]

Answer:

$F=9.09\times 10^{-9} N$

Explanation:

We are given that

Mass of spherical shell, $m_1$ =1900 kg

Mass= $m_2=1.80 kg$

Radius of shell=r=5 m

Distance between two masses=r=5.01 m

Because distance measure from center .

Gravitational force

$F=G\frac{m_1m_2}{r^2}$

$G=6.67\times 10^{-11} Nm^2/kg^2$

Using the formula

$F=6.67\times 10^{-11}\times \frac{1900\times 1.80}{(5.01)^2}$

$F=9.09\times 10^{-9} N$

Hence,the gravitational force = $F=9.09\times 10^{-9} N$

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2 years ago

As distance increases, force decrease and vise versa, as distance decreases then, force increases. This is a ________________ re

Sonbull [250]

It is indirectly proportional or inverse so it would be C

8 0

2 years ago

Which statement is TRUE about white light?

Olegator [25]

B: <span>It contains all of the wavelengths of the visible light spectrum.
Hope this helps mate =)</span>

6 0

2 years ago

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