A free-falling golf ball strikes the ground and exerts a force on it

How do scientists determine the age of the solar system?

Sholpan [36]

By studying meteorites which are the most ancient material in space.

7 0

3 years ago

You measure an electric field of 1.36×106 N/C at a distance of 0.158 m from a point charge. There is no other source of electric

marysya [2.9K]

Answer:

The Electric flux will be $0.42\times10^6\ \rm N.m^2/C$

Explanation:

Given

Strength of the Electric Field at a distance of 0.158 m from the point charge is

$E=1.36\times10^6\ \rm N/C$

We know that the flux of the Electric Field can be calculated by using Gauss Law which is given by

$\int E.dA=\dfrac{q_{in}}{\epsilon_0}\\$

Let consider a sphere of radius 0.158 m as Gaussian Surface at a distance of 0.158 m from the point charge and Let $\phi$ be the flux of the Electric Field coming out\passing through it which is given by

$\phi=\int E.dA=1.36\times10^6 \times4\pi \times 0.158^2\\\\=0.42\times10^6\ \rm N.m^2/C$

It can be observed that same amount of flux which is passing through the Gaussian sphere of radius 0.158 is also passing through the Gaussian sphere of radius 0.142 m at a distance of 0.142 m from its centre.

Also it can be observed that the charge inside the two Gaussian Sphere mentioned have same value so the Flux of electric field through them will also be same.

So the electric flux through the surface of sphere that has given charge at its centre and that has radius 0.142 m is $0.42\times10^6\ \rm N.m^2/C$

8 0

3 years ago

If the sun became cooler but kept the same size it would emit

jek_recluse [69]

If the sun would become cooler having constant size, it would emit less ultraviolet light and less visible light than what it currently gives to earth. Hope this answers the question. Have a nice day. Thank you for posting here.

4 0

3 years ago

What are the characteristics of globular cluster stars?

AlekseyPX

<span>Old age and hundreds of thousands to millions of member stars.</span>

3 0

3 years ago

A beam of alpha particles ( q = +2e, mass = 6.64 x 10-27 kg) is accelerated from rest through a potential difference of 1.8 kV.

Mrrafil [7]

Answer:

The magnetic field required required for the beam not to be deflected is $B = 0.0036T$

Explanation:

From the question we are told that

The charge on the particle is $q = +2e$

The mass of the particle is $m = 6.64 *10^{-27} kg$

The potential difference is $V_a = 1.8 kV = 1.8 *10^{3} V$

The potential difference between the two parallel plate is $V_b = 120 V$

The separation between the plate is $d = 8 mm = \frac{8}{1000} = 8*10^{-3}m$

The Kinetic energy experienced by the beam before entering the region of the parallel plate is equivalent to the potential energy of the beam after the region having a potential difference of 1.8kV

$KE_b = PE_b$