Responda:
1) E = 6 × 10 ^ 6NC ^ -1 2) Q = 6 × 10 ^ -5
Explicação:
Dado o seguinte:
Carga (q) = 3uC = 3 × 10 ^ -6C
Força elétrica (Fe) = 18N
Intensidade do campo elétrico (E) =?
1)
Lembre-se:
Força elétrica (Fe) = carga (q) * Intensidade do campo elétrico (E)
Fe = qE; E = Fe / q
E = 18N / (3 × 10 ^ -6C)
E = 6N / 10 ^ -6C
E = 6 × 10 ^ 6NC ^ -1
2)
Lembre-se:
E = kQ / r ^ 2
E = intensidade do campo elétrico
Q = carga de origem
r = distância de espera = 30cm = 30/100 = 0,3m
K = 9,0 × 10 ^ 9
6 × 10 ^ 6 = (9,0 × 10 ^ 9 * Q) / 0,3 ^ 2
9,0 × 10 ^ 9 * Q = 6 × 10 ^ 6 * 0,09
Q = 0,54 × 10 ^ 6 / 9,0 × 10 ^ 9
Q = 0,06 × 10 ^ (6-9)
Q = 0,06 × 10 ^ -3
Q = 6 × 10 ^ -5 = 60 × 10 ^ -6 = 60μC
Equation: Mass x Velocity = Momentum
Answer: 93 x 13 = 1,209
Answer:
Explanation:
We need 2 different equations for this problem: first the velocity of sound equation, then the frequency of the sound equation.
The velocity of sound is found in:
v = 331.5 + .606T
We need to find that first in order to fill it into the frequency equation which is
where v is the velocity we will find the part a, f is frequency and lambda is the wavelength. Starting with the velocity of the sound:
v = 331.5 + .606(25) and
v = 331.5 + 15 and rounding correctly using the rules for sig fig when adding:
v = 347 m/s
Filling that into the frequency equation:
and
so
Answer: 
Explanation:
We are told both planets describe a circular orbit around the star S. So, let's approach this problem begining with the angular velocity 
of the planet P1 with a period 
:
(1)
Where:
is the velocity of planet P1
is the radius of the orbit of planet P1
Finding :
(2)
(3)
(4)
On the other hand, we know the gravitational force 
between the star S with mass 
and the planet P1 with mass 
is:
(5)
Where 
is the Gravitational Constant and its value is 
In addition, the centripetal force 
exerted on the planet is:
(6)
Assuming this system is in equilibrium:
(7)
Substituting (5) and (6) in (7):
(8)
Finding :
(9)
(10)
Finally:
(11) This is the mass of the star S
Answer:
c. The incident light must have at least as much energy as the electron work function
Explanation:
In photoelectric effect, electrons are emitted from a metal surface when a light ray or photon strikes it. An electron either absorbs one whole photon or it absorbs none. After absorbing a photon, an electron either leaves the surface of metal or dissipate its energy within the metal in such a short time interval that it has almost no chance to absorb a second photon. An increase in intensity of light source simply increase the number of photons and thus, the number of electrons, but the energy of electron remains same. However, increase in frequency of light increases the energy of photons and hence, the
energy of electrons too.
Therefore, the energy of photon decides whether the electron shall be emitted or not. The minimum energy required to eject an electron from the metal surface, i.e. to overcome the binding force of the nucleus is called ‘Work Function’
Hence, the correct option is:
<u>c. The incident light must have at least as much energy as the electron work function</u>
