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nikitadnepr [17]

3 years ago

14

A positively-charged object with a mass of 0.129 kg oscillates at the end of a spring, generating ELF (extremely low frequency)

radio waves that have a wavelength of 3.86 × 10^7 m. The frequency of these radio waves is the same as the frequency at which the object oscillates. What is the spring constant of the spring?

1 answer:

9966 [12]3 years ago

7 0

Answer:308 N/m

Explanation:

Given

mass $\left ( m\right )=0.129 kg$

wavelength $\left ( \lambda \right )=3.86\times 10^7$

We know frequency = $\frac{c}{\lambda }=\frac{3\times 10^8}{3.86\tmes 10^7}$

f=7.772 Hz

As the frequency of radio waves is same as the frequency at which object oscillates

$f=\frac{1}{2\pi }\sqrt{\frac{k}{m}}$

$7.772=\frac{1}{2\pi }\sqrt{\frac{k}{0.129}}$

$7.772\times 2\times \pi =\sqrt{\frac{k}{0.129}}$

$k=307.70\approx 308 N/m$

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The nuclei of large atoms, such as uranium, with 9292 protons, can be modeled as spherically symmetric spheres of charge. The ra

Scrat [10]

Answer:

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

Explanation:

Given

Number of protons $= 92$

Radius of nucleus $r_n = 7.4 * 10^{-15} m$

Distance of the electrons $r_1 = 1.0 * 10^ {-10} m$

Part 1

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(7.4* 10^{-15})^2} \\E = 2.42 * 10^{21} N/C$

Part 2

Electric field produced by just outside its surface

$E = \frac{q}{4\pi*E_0* r_n^2 } \\E = \frac{9 * 10^ 9 * 92 * 1.6 * 10^{-19}}{(1* 10^{-15})^2} \\E = 1.3 * 10^{13} N/C$

Part 3

The net electric field inside a uniform shell of negative charge is zero because the electric flux lines cancel out each other

hence, the solution is

Part 1 $E = 2.42 * 10^{21} N/C$

Part 2 $E = 1.3 * 10^{13} N/C$

Part 3 $E = 0$

7 0

2 years ago

Determine the binding energy of an F-19 nucleus. The F-19 nucleus has a mass of 18.99840325 amu. A proton has a mass of 1.00728

Anvisha [2.4K]

Answer:

Energy = 1.38*10^13 J/mol

Explanation:

Total number of proton in F-19 = 9

Total number of neutron in F-19 = 10

Expected Mass of F-19

= 9*1.007 + 10*1.008 = 19.152 u

Actual mass of F-19 = 18.998 u

Energy of one particle of F-19 = 931.5*Δm = 931.5*(19.152-18.998)

= 143.234 MeV

Energy of one mole of F-19 = 143.234*10^6*1.6*10^-19*6.022*10^23

= 1.38*10^13 J/mol

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OlgaM077 [116]

Solution :

The motion in the y direction.

The time taken by the toy rocket to hit the ground,

$S=ut+\frac{1}{2}at^2$

S = distance travelled = 30 m

u = 0 m/s

a = $9.8 \ m/sec^2$

t= time in seconds

Therefore, $30 =\frac{1}{2}9.8 t^2$

t = 2.47 sec

Now motion in the x direction,

u = 12 m/sec

$a=1.6 t \ m/sec^3$

Upon integration 'v' with respect to 't'

$v=\frac{1.6t^2}{2}+12$

Once again integrating with respect to t,

$s=\frac{1.6t^3}{6}+12 t$

$s=\frac{1.6(2.47)^3}{6}+12 (2.47)$

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= 29.65 m

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