Ayúdenme tmb NECESITO LA SOLUCIÓN DE EL PROBLEMA
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Physics, 22.06.2019 08:40
An isolated conducting spherical shell carries a positive charge. part a which statement (or statements) about the electric field and the electric potential inside and outside the spherical shell is correct? which statement (or statements) about the electric field and the electric potential inside and outside the spherical shell is correct? electric potential inside the shell is constant and outside the shell is changing as 1/r2 both the electric potential and the electric field does change with r inside and outside the spherical shell electric potential inside and outside the shell is constant, but not zero electric potential inside the shell is constant and outside the shell is equal to zero electric field inside and outside the shell is constant (does not change with the position r), but is not equal to zero electric field inside and outside the shell is changing as 1/r (where r is the distance from the center of the sphere) electric field inside is equal to zero and outside the shell is constant, but not zero electric potential inside the shell is constant and outside the shell is changing as 1/r electric field inside and outside the shell is changing as 1/r2 electric field inside is equal to zero and outside the shell is changing as 1/r2 electric field inside and outside the shell is zero electric field inside is constant and outside the shell is changing as 1/r
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Physics, 22.06.2019 17:00
If you put a helium-filled balloon in the refrigerator, what will happen?
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Physics, 22.06.2019 18:30
Ablock of mass m slides on a horizontal frictionless table with an initial speed v0 . it then compresses a spring of force constant k and is brought to rest. the acceleration of gravity is 9.8 m/s2. how much is the spring compressed x from its natural length? 1) x = v0*sqrt(k/(mg)) 2) x=v0*sqrt(m/k) 3) x=v0*((mk)/g) 4) x=v0*sqrt(k/m) 5) x=v0*(m/kg) 6) x=v0*sqrt((mg)/k) 7) x=(v0)^2/(2g) 8) x=v0*(k/(mg)) 9) x=(v0)^2/(2m) 10) x=v0*((mg)/k)
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Physics, 23.06.2019 04:00
Can anyone plz answer any of these. even if its one it will
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