Answer:
Explanation:
Given
Sphere A has a net charge of +5 Q
Sphere B has a net charge of -3 Q
Sphere C has a net charge of 7 Q
When sphere B is touched to sphere A then charge is redistributed among the sphere such that charge on each sphere is
Now sphere A and B has charge Q
When sphere A is touched to sphere C
Net charge after removal
Now sphere A and C has 4 Q charge
Now sphere C and B is touched
Net charge on C and B is given by
At the end of process charge on sphere A is 4Q
Charge on B and C is 2.5Q
I believe the correct answer is the bottom one. Hope this helped!
-TTL
Answer:
specifically 30º west of the north
Explanation:
This exercise can be solved using the composition of vectors, let's call the speed of the river that goes east (v₁ = 8 mi / h) to the speed of the boat (v₂ = 16 mi / h) and the desired speed from south to north (v₃ =?). Look at the attachment to be clear about the vectors
Let's use the Pythagorean triangle
v₃² = v₁² + v₂²
Let's calculate v₂
v₂² = v₃² - v₁²
v₂ = √ (16² - 8²)
v₂ = 13.86 mi / h
To find the angle we use trigonometry
sin θ = v1 / v3
θ = sin⁻¹ (v1 / v3)
θ = sin⁻¹ (8/16)
θ = 30º
The boat must be pointed at 30º with respect to the south-north direction, specifically 30º west of the north
Answer:
X = 5.48 mm
Explanation:
for single slit
By Rayleigh criterian
where d = slit width =0.5 mm
wavelength
therefore maximum of value of X can be calculated from above
X = 5.48 mm
Answer:
0.68 m
Explanation:
We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s
V=frequency*wavelength
Then wavelength is given by 350/500=0.68 m
Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed