What potential difference is required across a 8Ω resistor to cause 2.0 A to flow through it?

I need the answer for both of the questions please

Lady_Fox [76]

But even more pain on pain and then pain and pain ya feel me and even more pain okay and yes more pain

5 0

3 years ago

What happens to light waves from a star as the star moves away from Earth?

AURORKA [14]

<h2>Answer: Light waves have a redshift due to the Doppler effect </h2>

The astronomer Edwin Powell Hubble observed several celestial bodies, and when obtaining the spectra of distant galaxies he observed the spectral lines were displaced towards the red (red shift), whereas the nearby galaxies showed a spectrum displaced to the blue.

From there, Hubble deduced that the farther the galaxy is, the more redshifted it is in its spectrum. <u>The same happens with the stars and this phenomenom is known as the Doppler effect. </u>

This phenomenon refers to the change in a wave perceived frequency (or wavelength=color) when the emitter of the waves, and the receiver (or observer in the case of light) move relative to each other. For example, as a star moves away from the Earth, its espectrum turns towards the red.

8 0

3 years ago

A gun is fired on a day when the speed of sound is 335 m/s and an echo is heard 0.75 seconds later. How far away is the object t

cricket20 [7]

Answer:

v= 335 m/s

2∆t= 0.75 s

∆x= v.∆t → ∆x= 335×½×0.75 = 125.625 m

8 0

2 years ago

Three point charges have equal magnitudes, two being positive and one negative. These charges are fixed to the corners of an equ

Irina-Kira [14]

Answer:

Magnitude of the net force on q₁-

Fn₁=1403 N

Magnitude of the net force on q₂+

Fn₂= 810 N

Magnitude of the net force on q₃+

Fn₃= 810 N

Explanation:

Look at the attached graphic:

The charges of the same sign exert forces of repulsion and the charges of opposite sign exert forces of attraction.

Each of the charges experiences 2 forces and these forces are equal and we calculate them with Coulomb's law:

F= (k*q*q)/(d)²

F= (9*10⁹*3*10⁻⁶*3*10⁻⁶)(0.01)² =810N

Magnitude of the net force on q₁-

Fn₁x= 0

Fn₁y= 2*F*sin60 = 2*810*sin60° = 1403 N

Fn₁=1403 N

Magnitude of the net force on q₃+

Fn₃x= 810- 810 cos 60° = 405 N

Fn₃y= 810*sin 60° = 701.5 N

$Fn_{3} = \sqrt{405^{2}+701.5^{2} }$

Fn₃ = 810 N

Magnitude of the net force on q₂+

Fn₂ = Fn₃ = 810 N

6 0

3 years ago

A wheel is rotating freely at angular speed 420 rev/min on a shaft whose rotational inertia is negligible. A second wheel, initi

Mashcka [7]

Answer:60 rev/min

Explanation:

Given

angular speed of first shaft $\omega _1=420\ rev/min$

Moment of inertia of second shaft is seven times times the rotational speed of the first i.e. If I is the moment of inertia of first wheel so moment of inertia of second is 7 I

As there is no external torque therefore angular momentum is conserved

$L_1=L_2$

$I_1\omega _1=I_2\omega _2$

$I\times (420)=7 I\times (\omega _2)$

$\omega _2=\frac{420}{7}$

$\omega _2=60\ rev/min$

8 0

3 years ago