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Gennadij [26K]
3 years ago
11

Why would a positive current flow from the positive end of a wire towards its negative end?

Physics
1 answer:
irinina [24]3 years ago
5 0

because positive and negative attract

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Two conducting spheres are each given a charge Q. The radius of the larger sphere is three times greater than that of the smalle
padilas [110]

Answer:

Explanation:

If E₀ is the electric field outside the smaller sphere and r is the radius of larger sphere.

E₀ = kQ/r²

The radius of the larger sphere is 3r and the charge on both sphere is same then the electric field outside the larger sphere is given as

E = kQ/(3r)² = kQ/9r² = 1/9 (kQ/r²)= 1/9 x E₀

hence the correct option is e.

5 0
3 years ago
If the radio waves transmitted by a radio station have a frequency of 76.5 mhz, what is the wavelength of the waves, in meters?
____ [38]
Given: Velocity of light    c = 3.00 x 10⁸ m/s

Frequency f = 7.65 x 10⁷/s

Required: Wavelength λ = ?

Formula:  λ = c/f

λ = 3.00 x 10⁸ m/s/7.65 x 10⁷/s

λ = 3.92 m
3 0
3 years ago
What happens to an object after being transported to three planets in the solar system?
alukav5142 [94]
The weight changes but the mass will stay the same.
6 0
3 years ago
A pulsar is a rapidly rotating neutron star that emits a radio beam the way a lighthouse emits a light beam. We receive a radio
DanielleElmas [232]

Answer:

\alpha =-2.2669642\times^{-10}rad/s^2

Explanation:

Angular acceleration is defined by \alpha =\frac{\Delta \omega}{\Delta t}=\frac{\omega_f-\omega_i}{\Delta t}

Angular velocity is related to the period by \omega=\frac{2\pi}{T}

Putting all together:

\alpha =\frac{\frac{2\pi}{T_f}-\frac{2\pi}{T_i}}{\Delta t}=\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})

Taking our initial (i) point now and our final (f) point one year later, we would have:

\Delta t=1\ year=(365)(24)(60)(60)s=31536000
s

T_i=0.0786s

T_f=0.0786s+7.03\times10^{-6}s

So for our values we have:

\alpha =\frac{2\pi}{\Delta t}(\frac{1}{T_f}-\frac{1}{T_i})=\frac{2\pi}{31536000s}(\frac{1}{0.0786s+7.03\times10^{-6}s}-\frac{1}{0.0786s})=-2.2669642\times^{-10}rad/s^2

Where the minus sign indicates it is decelerating.

8 0
3 years ago
Ml(d^2θ/dt^2) =-mgθ
Nata [24]

The equation of motion of a pendulum is:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\dfrac{g}{\ell}\sin\theta,

where \ell it its length and g is the gravitational acceleration. Notice that the mass is absent from the equation! This is quite hard to solve, but for <em>small</em> angles (\theta \ll 1), we can use:

\sin\theta \simeq \theta.

Additionally, let us define:

\omega^2\equiv\dfrac{g}{\ell}.

We can now write:

\dfrac{\textrm{d}^2\theta}{\textrm{d}t^2} = -\omega^2\theta.

The solution to this differential equation is:

\theta(t) = A\sin(\omega t + \phi),

where A and \phi are constants to be determined using the initial conditions. Notice that they will not have any influence on the period, since it is given simply by:

T = \dfrac{2\pi}{\omega} = 2\pi\sqrt{\dfrac{g}{\ell}}.

This justifies that the period depends only on the pendulum's length.

4 0
3 years ago
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