True or False:

A spherical conductor is inside a uniform electric field with a magnitude of 1000 N/C to the right. What is the magnitude and di

Lyrx [107]

Answer:

1.-E=1000N/C to the LEFT

2.-The electric field inside a conductor in electrostatic state is always zero (conductor proprieties).

3.-The voltmeter read 0V as differential voltage between two points from the conductor

Explanation:

1.The electric field inside the conductor must be zero (conductor proprieties). Then the charges create a electric field equal an opposite to the external electric field. In other words E=1000N/C to the LEFT

2. The electric field inside a conductor in electrostatic state is always zero. As shown in the figure the electric field induced by the charges in the sphere surface cancelled the EXTERN electric field.

3.If the Electric field inside the conductor is zero, that means that the Voltage in the hole conductor is constant (conductor proprieties). In other words the the voltmeter read 0v as differential voltage between two points from the conductor.

3 0

3 years ago

Can we use a clinical thermometer to measure the temperature of a candle flame

miv72 [106K]

In theory, yes. The 2 problems are the materials used for clinical thermometers, & the temperature capacity of the clinical thermometer. If anything, change the material & extend the measurement threshold. At that point, it wouldn´t be used for clinical garbage anymore.

3 0

4 years ago

lanet R47A is a spherical planet where the gravitational acceleration on the surface is 3.45 m/s2. A satellite orbitsPlanet R47A

qaws [65]

$2.6×10^6\:\text{m}$

Explanation:

The acceleration due to gravity g is defined as

$g = G\dfrac{M}{R^2}$

and solving for R, we find that

$R = \sqrt{\dfrac{GM}{g}}\:\:\:\:\:\:\:(1)$

We need the mass M of the planet first and we can do that by noting that the centripetal acceleration $F_c$ experienced by the satellite is equal to the gravitational force $F_G$ or

$F_c = F_G \Rightarrow m\dfrac{v^2}{r} = G\dfrac{mM}{r^2}\:\:\:\:\:(2)$

The orbital velocity <em>v</em> is the velocity of the satellite around the planet defined as

$v = \dfrac{2\pi r}{T}$

where <em>r</em><em> </em>is the radius of the satellite's orbit in meters and <em>T</em> is the period or the time it takes for the satellite to circle the planet in seconds. We can then rewrite Eqn(2) as

$\dfrac{4\pi^2 r}{T^2} = G\dfrac{M}{r^2}$