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4 years ago

Since protons (p*) with a positive charge and neutrons (nº) without a charge are located in the

Physics

1 answer:

konstantin123 [22]4 years ago

3 0

Answer:

A nucleus contains protons and neutrons . Protons are positively charged and neutrons are neutral . So charge of nucleus depends upon protons . Hence , neucleus is positively charged

Explanation:

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An object is thrown straight up with a velocity, in ft/s, given by v(t)= -32t + 83, where t is in seconds, from a height of 46 f

____ [38]

<h2>a) Initial velocity = 83 ft/s</h2><h2>b) Object's maximum speed = 99.4 ft/s</h2><h2>c) Object's maximum displacement = 153.64 ft</h2><h2>d) Maximum displacement occur at t = 2.59 seconds.</h2><h2>e) The displacement is zero when t = 5.70 seconds</h2><h2>f) Object's maximum height = 153.64 ft</h2>

Explanation:

We have velocity

v(t)= -32t + 83

Integrating

s(t) = -16t²+83t+C

At t = 0 displacement is 46 feet

46 = -16 x 0²+83 x 0+C

C = 46 feet

So displacement is

s(t) = -16t²+83t+46

a) Initial velocity is

v(0)= -32 x 0 + 83 = 83 ft/s

Initial velocity = 83 ft/s

b) Maximum velocity is when the object reaches ground, that is s(t) = 0 ft

Substituting

0 = -16t²+83t+46

t = 5.70 seconds

Substituting in velocity equation

v(t)= -32 x 5.70 + 83 = -99.4 ft/s

Object's maximum speed = 99.4 ft/s

c) Maximum displacement is when the velocity is zero

That is

-32t + 83 = 0

t = 2.59 s

Substituting in displacement equation

s(2.59) = -16 x 2.59²+83 x 2.59+46 = 153.64 ft

Object's maximum displacement = 153.64 ft

d) Maximum displacement occur at t = 2.59 seconds.

e) Refer part b

The displacement is zero when t = 5.70 seconds

f) Same as option d

Object's maximum height = 153.64 ft

5 0

4 years ago

A satellite in geostationary orbit is used to transmit data via electromagnetic radiation. The satellite is at a height of 35,00

Rus_ich [418]

Answer:

$6.99535\times 10^{-6}\ V/m$

Explanation:

P = Power Output = 1000 W

r = Radius = 35000000 m

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

c = Speed of light = $3\times 10^8\ m/s$

Intensity of Electric radiation

$I=\frac{P}{A}\\\Rightarrow I=\frac{P}{4\pi r^2}\\\Rightarrow I=\frac{1000}{4\pi\times 35000000^2}\ W/m^2$

Intensity of Electric radiation

$I=\frac{1}{2}c\epsilon_0E_0\\\Rightarrow E_0=\sqrt{\frac{2I}{c\epsilon_0}}\\\Rightarrow E_0=\sqrt{\frac{2\times \frac{1000}{4\pi\times 35000000^2}}{3\times 10^8\times 8.85\times 10^{-12}}}\\\Rightarrow E_0=6.99535\times 10^{-6}\ V/m$

The amplitude of the electric field vector is $6.99535\times 10^{-6}\ V/m$

7 0

4 years ago

During most of its lifetime, s star maintains an equilibrium size in which the inward force of gravity on each atom is balanced

frez [133]

Answer:

$r_f=137493m$

$v=8638940m/s$

Explanation:

During this process the mass $M=2\times10^{30}Kg$ will be considered constant. We start from a radius $r_i=7\times10^8m$ and a period $T_i=30\ days=(30)(24)(60)(60)s=2592000s$ . The final period is $T_f=0.1s$ .

Angular momentum <em>L</em> is conserved in this process. We can use the formula $L=I\omega$ , where I is the momentum of inertia (which for a solid sphere is $I=\frac{2mr^2}{5}$ ) and $\omega=\frac{2\pi }{T}$ is the angular velocity, so we can write the star's angular momentum as: