Answer:
The electric field intensity is <u>30000 N/C.</u>
Explanation:
Given:
Magnitude of the point charge is, 
Distance of the given point from the point charge is, 
Electric field intensity is directly proportional to the magnitude of point charge and inversely proportional to the square of the distance of the point and the given charge.
Therefore, electric field intensity 'E' at a distance of 'd' from a point charge 'q' is given as:
Plug in 
. Solve for 'E'.
Therefore, the electric field intensity at a point 3 cm from the point charge is 30000 N/C.
Answer:
resistance- decreases current-increases
The answer is
C. periodic fluctuations in the intensity if sound waves.
Answer:
19 N
Explanation:
From the question given above, the following data were obtained:
Pressure (P) = 1.9 kPa
Length (L) = 10 cm
Force (F) =?
Next, we shall convert 1.9 KPa to N/m². This can be obtained as follow:
1 KPa = 1000 N/m²
Therefore,
1.9 KPa = 1.9 KPa × 1000 N/m² / 1 KPa
1.9 KPa = 1900 N/m²
Thus, 1.9 KPa is equivalent to 1900 N/m².
Next, we shall convert 10 cm to m. This can be obtained as follow:
100 cm = 1 m
Therefore,
10 cm = 10 cm × 1 m / 100 cm
10 cm = 0.1 m
Thus, 10 cm is equivalent to 0.1 m
Next, we shall determine the area of the square. This can be obtained as follow:
Length (L) = 0.1 m
Area of square (A) =?
A = L²
A = 0.1²
A = 0.01 m²
Thus, the area of the square is 0.01 m².
Finally, we shall determine the force that must be exerted on the sensor in order for it to turn red. This can be obtained as follow:
Pressure (P) = 1900 N/m²
Area (A) = 0.01 m²
Force (F) =?
P = F/A
1900 = F / 0.01
Cross multiply
F = 1900 × 0.01
F = 19 N
Therefore, a force of 19 N must be exerted on the sensor in order for it to turn red.
