32? I could be wrong but I’m going with that answer choice
Answer:
r = 0.05 m = 5 cm
Explanation:
Applying ampere's law to the wire, we get:
where,
r = distance of point P from wire = ?
μ₀ = permeability of free space = 4π x 10⁻⁷ N/A²
I = current = 2 A
B = Magnetic Field = 8 μT = 8 x 10⁻⁶ T
Therefore,
<u>r = 0.05 m = 5 cm</u>
Nice couch lol
and aluminum I think
Answer:
4.45×10¯¹¹ N
Explanation:
From the question given above, the following data were obtained:
Mass of ball (M₁) = 4 Kg
Mass of bowling pin (M₂) = 1.5 Kg
Gravitational constant (G) = 6.67×10¯¹¹ Nm²/Kg²
Distance apart (r) = 3 m
Force of attraction (F) =?
The force of attraction between the ball and the bowling pin can be obtained as follow:
F = GM₁M₂ / r²
F = 6.67×10¯¹¹ × 4 × 1.5 / 3²
F = 4.002×10¯¹⁰ / 9
F = 4.45×10¯¹¹ N
Therefore, the force of attraction between the ball and the bowling pin is 4.45×10¯¹¹ N
Answer:
Explanation:
Threshold frequency = 4.17 x 10¹⁴ Hz .
minimum energy required = hν where h is plank's constant and ν is frequency .
E = 6.6 x 10⁻³⁴ x 4.17 x 10¹⁴
= 27.52 x 10⁻²⁰ J .
wavelength of radiation falling = 245 x 10⁻⁹ m
Energy of this radiation = hc / λ
c is velocity of light and λ is wavelength of radiation .
= 6.6 x 10⁻³⁴ x 3 x 10⁸ / 245 x 10⁻⁹
= .08081 x 10⁻¹⁷ J
= 80.81 x 10⁻²⁰ J
kinetic energy of electrons ejected = energy of falling radiation - threshold energy
= 80.81 x 10⁻²⁰ - 27.52 x 10⁻²⁰
= 53.29 x 10⁻²⁰ J .
