How much must you raise the

deff fn [24]

Answer:

the magnetic field can be used to make electricity

Explanation:

Moving a magnet around a coil of wire, or moving a coil of wire around a magnet, pushes the electrons in the wire and creates an electrical current. Electricity generators essentially convert kinetic energy (the energy of motion) into electrical energy

8 0

3 years ago

I just need the answer to number 1 (with work please) <br> (Newton’s Second Law)

Valentin [98]

Answer:

wish I could help

Explanation:

I been rereading this and I can't solve it lemme go ask people in ma house real quick

8 0

3 years ago

In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 1.40×1016 kg and a ra

Arturiano [62]

A) 8.11 m/s

For a satellite orbiting around an asteroid, the centripetal force is provided by the gravitational attraction between the satellite and the asteroid:

$m\frac{v^2}{(R+h)}=\frac{GMm}{(R+h)^2}$

where

m is the satellite's mass

v is the speed

R is the radius of the asteroide

h is the altitude of the satellite

G is the gravitational constant

M is the mass of the asteroid

Solving the equation for v, we find

$v=\sqrt{\frac{GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula,

$v=\sqrt{\frac{(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=8.11 m/s$

B) 11.47 m/s

The escape speed of an object from the surface of a planet/asteroid is given by

$v=\sqrt{\frac{2GM}{R+h}}$

where:

$G=6.67\cdot 10^{-11} m^3 kg^{-1}s^{-2}$

$M=1.40\cdot 10^{16}kg$

$R=8.20 km=8200 m$

$h=6.00 km = 6000 m$

Substituting into the formula, we find:

$v=\sqrt{\frac{2(6.67\cdot 10^{-11})(1.40\cdot 10^{16}kg)}{8200 m+6000 m}}=11.47 m/s$

5 0

3 years ago

WGVU-AM is a radio station that serves the Grand Rapids, Michigan area. The main broadcast frequency is 1480kHz. At a certain di

Yuki888 [10]

Answer:

a

$\lambda = 202.7 \ m$

b

$w = 9.3 *10^{6} \ rad/s$

c

$k = 0.031 m^{-1}$

d

$E_{max} = 9.0 *10^{-3} \ V/m$

Explanation:

From the question we are told that

The frequency of the radio station is $f= 1480 \ kHz = 1480 *10^{3}\ Hz$

The magnitude of the magnetic field is $B = 3.0* 10^{-11} \ T$

Generally the wavelength is mathematically represented as

$\lambda = \frac{c}{f}$

Here c is the speed of light with value $c = 3.0 *10^{8} \ m/s$

So

$\lambda = \frac{3.0 *10^{8}}{ 1480 *10^{3}}$

=> $\lambda = 202.7 \ m$

Generally the angular frequency is mathematically represented as

$w = 2 \pi * f$

=> $w = 2 * 3.142 * 1480 *10^{3}$

=> $w = 9.3 *10^{6} \ rad/s$

Generally the wave number is mathematically represented as

=> $k = \frac{2 \pi }{\lambda}$

=> $k = \frac{2 * 3.142 }{ 202.7}$

=> $k = 0.031 m^{-1}$

Generally the amplitude of the electric field at this distance from the transmitter is mathematically represented as

$E_{max} = c * B$

=> $E_{max} = 3.0 *10^{8} * 3.0* 10^{-11}$

=> $E_{max} = 9.0 *10^{-3} \ V/m$

3 0

3 years ago

Two solid steel shafts (G = 77.2 GPa) are connected to a coupling disk B and to fixed supports at A and C. Take T = 1.3 kN·m. De

My name is Ann [436]

Answer:

τ (bc. max) =25.37 MPa

Explanation:

From the question, T = 1.3Kn.m;

(G = 77.2 GPa) and from the image of this solid shaft system i attached;

d(ab) = 50mm; d(bc) = 38mm; L(ab) =0.2m and L(bc) = 0.25m

So ΣT = 0 → Ta + Tc = 1.3Kn.m

So the system is statically indeterminate.

Let's check at the equation that makes it compatible ;

ψ = 0 and ψ(c/b) + ψ(b/a) + ψ(a) = 0

[{T(bc)} / {J(bc)G}] + [{T(ab)} / {J(ab)G}] + 0 =

ΣT = 0 and T(bc) = T(c)

ΣT = 0 and T(ab) = T(c) - 1.3 Kn.m

Now,

[(T(c) x 250mm)/{(π/2)(19^4)}] + [{((T(c) - 1300000 Nmm) x 200mm}/{(π/2)(25^4)}] = 0

So, T(c) = 273374 Nmm = 273.374Nm

T(a) = 1300Nm - 273.374Nm = 1026. 63Nm

From the beginning we saw that;

T(bc) = T(c) = 273.374Nm

Now let's find the maximum shear stress in shaft BC;

τ (bc. max) = {τ (bc) x r(bc)} / J(bc)

τ (bc. max) = (273374Nmm x 19mm)/ {(π/2)(19^4)} = 5194106 / 204707

= 25.37 MPa

3 0

3 years ago