Answer:
To find the distance from the 3.0 μC point charge where the net electric field is zero in the presence of the external uniform electric field, you can use the principle that the electric fields due to the point charge and the external field will cancel each other at that point.
The electric field due to a point charge is given by Coulomb's law:
E_point_charge = k * (|q| / r^2),
where:
E_point_charge is the electric field due to the point charge.
k is Coulomb's constant (approximately 8.99 x 10^9 Nm^2/C^2).
|q| is the magnitude of the point charge (3.0 μC = 3.0 x 10^-6 C).
r is the distance from the point charge.
The external uniform electric field has a magnitude of 1.6 x 10^4 N/C. Let's denote this as E_external.
To find the point where the net electric field is zero, you want the magnitudes of the electric fields due to the point charge and the external field to be equal. So:
E_point_charge = E_external.
Substitute the expressions for both electric fields:
k * (|q| / r^2) = E_external.
Now, plug in the known values:
(8.99 x 10^9 Nm^2/C^2) * (3.0 x 10^-6 C / r^2) = 1.6 x 10^4 N/C.
Now, solve for r:
3.0 x 10^-6 C / r^2 = (1.6 x 10^4 N/C) / (8.99 x 10^9 Nm^2/C^2).
r^2 = (3.0 x 10^-6 C / (1.6 x 10^4 N/C)) * (8.99 x 10^9 Nm^2/C^2).
r^2 = (1.87 x 10^-11 m^2).
Take the square root of both sides to find r:
r ≈ 4.32 x 10^-6 m.
So, the net electric field is zero at a distance of approximately 4.32 x 10^-6 meters from the 3.0 μC point charge in the direction opposite to the external uniform electric field.
Explanation:
Answer:
To estimate the final error on the density of the cube, we can consider the errors introduced by both the measurement of its volume and its weight.
1. Volume Measurement:
- The side length of the cube is given as 10 centimeters, and your ruler can measure to 1 mm accuracy.
- So, the error in measuring the side length is ±0.05 cm (half of the smallest measurement unit).
- To calculate volume, you need to cube the side length: Volume = (10 cm)^3 = 1000 cm^3.
- Using the error propagation rule, the relative error in volume is ±0.05 cm / 10 cm = ±0.005.
2. Weight Measurement:
- The weight is given as 1 kg nominally, which is equivalent to 1000 g.
- Your scale has a precision down to 0.1 g.
- So, the error in measuring the weight is ±0.1 g / 1000 g = ±0.0001 (0.01%) relative error.
Now, to calculate the final error in density, you need to consider both errors in volume and weight:
Density = Weight / Volume
Relative Error in Density = (Relative Error in Weight) + (Relative Error in Volume)
Relative Error in Density = (0.0001) + (0.005) = 0.0051 or 0.51%
So, the final estimated error on the density of the cube is approximately ±0.0051 g/cm^3 or ±0.51%.
The density of the cube is calculated using its mass and volume, with potential errors from the measurements of these quantities leading to a total estimated density error of approximately ±3.01%.
The density of an object is given by the formula density = mass/volume. In this case, the mass of the cube is given as 1 kg (or 1000 g for consistency with the scale's precision), and the volume of the cube can be calculated from the given side length using the formula for the volume of a cube, volume = side³, which equals 1000 cm³.
However, there are measurement errors associated with both the ruler and scale. The ruler can measure to the nearest mm (or 0.1 cm), so the error is ±0.1 cm on each measurement of the cube's sides, leading to a volume error of about ±3%. The scale can measure to the nearest 0.1 g, which gives a mass error of about ±0.01%. The total error in the density, obtained by summing these errors, is therefore approximately ±3.01%.
#SPJ2
Answer:
508Hz
Explanation:
A tuning fork with a frequency of 512 Hz is used to tune a violin. When played together, beats are heard with a frequency of 4 Hz. The string on the violin is tightened and when played again, the beats have a frequency of 2 Hz. The original frequency of the violin was ______.
When two sound waves of different frequency approach your ear, the alternating constructive and destructive interference causes the sound to be alternatively soft and loud - this phenomenon is beat production
frequency is the number of oscillation a wave makes in one seconds.
f1-f2=beats
therefore f1=512Hz
f2=?
beats=4Hz
512Hz-f2=4Hz
f2=512-4
f2=508Hz
the original frequency of the violin is 508Hz
The original frequency of the violin was 508 Hz. This is based on the principle of beats, where the beat frequency is the absolute difference in frequency between the two sources - in this case, the tuning fork and the violin string.
The original frequency of the violin string can be found using the principle of beats. The frequency of the beats is equal to the absolute difference in frequency between the two sources - in this case, the tuning fork and the violin string.
Initially, the beat frequency was heard as 4 Hz. This indicates that the original frequency of the violin was either 512 Hz + 4 Hz = 516 Hz, or 512 Hz - 4 Hz = 508 Hz. However, when the violin string was tightened, the beat frequency decreased to 2 Hz, which means the frequency of the note it was producing increased.
Therefore, the violin must have initially been producing a note with lower frequency (508 Hz), and even after tightening the string, the note it now produces (510 Hz) remains lower than that of the tuning fork.
#SPJ11
Answer:
Explanation:
The experience is magnanimous and wonderful when witnessing the hotspot site. It was a great feeling! I could witness the red hot lava oozing out to the ocean floor from the subsurface like water from a hot-spring, and immediately, a cooling effect occurs. At that current time, the gas was coming up in the form of gas bubbles. I believed there was a mini-volcano located inside the ocean! The scene was fascinating, just like inserting a red-hot metallic-iron ball into the cool water, thereby creating a fizzing of gas and the ball's immediate cooling. It was a remarkable experience of witnessing an underwater volcano with the erupting lava and its immediate cooling to form a solid material.
the Answer is:
12.321
B. 12,000J
C. 5,880J
D. 2,940J