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3 years ago

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In a cold winters day, if you left a cup of water sitting outside, it could freeze. Heat is transferred out of the water. Descri

be the behavior of the water molecules and how temperature is affected. Enter your answer in the space provided.

1 answer:

Agata [3.3K]3 years ago

4 0

If you were to sit a hot cup of water out side it would frezze faster

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The center of the Hubble space telescope is 6940 km from Earth’s center. If the gravitational force between Earth and the telesc

Law Incorporation [45]

The gravitational force between two objects is given by:
$F=G \frac{m_1 m_2}{r^2}$
where
G is the gravitational constant
m1 and m2 are the masses of the two objects
r is the separation between the two objects

The distance of the telescope from the Earth's center is $r=6940 km=6.94 \cdot 10^6 m$ , the gravitational force is $F=9.21 \cdot 10^4 N$ and the mass of the Earth is $m_1=5.98 \cdot 10^{24} kg$ , therefore we can rearrange the previous equation to find m2, the mass of the telescope:
$m_2 = \frac{Fr^2}{Gm_1}= \frac{(9.21 \cdot 10^4 N)(6.94\cdot 10^6)^2}{(6.67\cdot 10^{-11})(5.98\cdot 10^{24})} =11121 kg$

6 0

3 years ago

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You and your lab partner, Mel, are engrossed in a chemistry activity using beans to compute the average atomic mass of an elemen

Vikentia [17]

Answer:

C:

Explanation:

either C or A but A seems unlikely after multiple attempts. Although the question doesn't make it clear whether the balance is electric either way it could be wrong in someway and seems to be the most likely.

5 0

3 years ago

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A capacitor is formed from two concentric spherical conducting shells separated by vacuum. The inner sphere has radius 11.0 cm ,

viktelen [127]

Part A)
First of all, let's convert the radii of the inner and the outer sphere:
$r_A = 11.0 cm = 0.110 m$
$r_B = 16.5 cm=0.165 m$
The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is
$C=4 \pi \epsilon _0 \frac{r_A r_B}{r_B- r_A}=4\pi(8.85 \cdot 10^{-12}C^2m^{-2}N^{-1}) \frac{(0.110m)(0.165m)}{0.165m-0.110m}=$
$=3.67\cdot 10^{-11}F$

Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:
$Q=CV=(3.67\cdot 10^{-11}F)(100 V)=3.67\cdot 10^{-9}C$

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:
$E\cdot (4 \pi r^2) = \frac{Q}{\epsilon _0}$
from which
$E(r) = \frac{Q}{4 \pi \epsilon_0 r^2}$
In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get
$E(0.111m)=2680 N/C$

And so, the energy density at r=0.111 m is
$U= \frac{1}{2} \epsilon _0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(2680 N/C)^2=3.17 \cdot 10^{-5}J/m^3$

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor: $Q=3.67 \cdot 10^{-9}C$ . We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so
$E(0.164 m)= \frac{Q}{4 \pi \epsilon_0 r^2}=1228 N/C$

And therefore, the energy density at this distance from the center is
$U= \frac{1}{2}\epsilon_0 E^2 = \frac{1}{2} (8.85\cdot 10^{-12}C^2m^{-2}N^{-1})(1228 N/C)^2=6.68 \cdot 10^{-6}J/m^3$

8 0

3 years ago

g n diffraction, the formula for minima is given by a times s i n (theta )equals m lambda, where a is the width of the slit, the

ki77a [65]

Answer:

θ = 22.2

Explanation:

This is a diffraction exercise

a sin θ = m λ

The extension of the third zero is requested (m = 3)

They indicate the wavelength λ = 630 nm = 630 10⁻⁹ m and the width of the slit a = 5 10⁻⁶ m

sin θ = m λ / a

sin θ = 3 630 10⁻⁹ / 5 10⁻⁶

sin θ = 3.78 10⁻¹ = 0.378

θ = sin⁻¹ 0.378

to better see the result let's find the angle in radians

θ = 0.3876 rad

let's reduce to degrees

θ = 0.3876 rad (180º /π rad)

θ = 22.2º

4 0

3 years ago

While an airplane is in flight, four forces act on it. Thrust is caused by the airplane's propellers pushing through the surroun

IceJOKER [234]

The answer would be weight

4 0

3 years ago

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