Tần số ngưỡng của một kim loại có giá trị X. Nếu tần số của chùm sáng chiếu đến bề mặt katot tăng từ 2X đến 4X thì cường độ dòng
1 answer:
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r₁ = distance of point A from charge q₁ = 0.13 m
r₂ = distance of point A from charge q₂ = 0.24 m
r₃ = distance of point A from charge q₃ = 0.13 m
Electric field by charge q₁ at A is given as
E₁ = k q₁ /r₁² = (9 x 10⁹) (2.30 x 10⁻¹²)/(0.13)² = 1.225 N/C towards right
Electric field by charge q₂ at A is given as
E₂ = k q₂ /r₂² = (9 x 10⁹) (4.50 x 10⁻¹²)/(0.24)² = 0.703 N/C towards left
Since the electric field in left direction is smaller, hence the electric field by the third charge must be in left direction
Electric field at A will be zero when
E₁ = E₂ + E₃
1.225 = 0.703 + E₃
E₃ = 0.522 N/C
Electric field by charge "q₃" is given as
E₃ = k q₃ /r₃²
0.522 = (9 x 10⁹) q₃/(0.13)²
q₃ = 0.980 x 10⁻¹² C = 0.980 pC
Answer:
The time taken to stop the box equals 1.33 seconds.
Explanation:
Since frictional force always acts opposite to the motion of the box we can find the acceleration that the force produces using newton's second law of motion as shown below:
Given mass of box = 5.0 kg
Frictional force = 30 N
thus
Now to find the time that the box requires to stop can be calculated by first equation of kinematics
The box will stop when it's final velocity becomes zero
Here acceleration is taken as negative since it opposes the motion of the box since frictional force always opposes motion.