The particle with sharp ends have the slowest rate of deposition
Answer: Option C
<u>Explanation:</u>
As per aerosol physics, deposition is a process where aerosol particles accumulate or settle on solid surfaces. Thereby, it reduces the concentration of particles in the air. Deposition velocity (rate of deposition) defines from F = vc, where v is deposition rate, F denotes flux density and c refers concentration.
Deposition velocity is slowest for particles of intermediate-sized particles because the frictional force offers resistance to the flow. Density is directly proportional to the deposition rate so clearly shows that high-density particles settle faster. Due to friction, round and large-sized particles deposit faster than oval/flattened sediments.
Answer:
The position is 8.18cm from the mirror.
Nature is b=virtual
Size is 1.82cm
Explanation:
Note that for a convex mirror, the image distance and the focal length are negative;
Given
Object height H0 = 4cm
object distance u = 18cm
Radius of curvature R = 30cm
Since f = R/2
f = 30/2
f = -15cm
Recall that:
Since the image distance is negative, this shows that the image is a virtual image.
To get the size:
When both particles, the electron and the proton move at the same speed, they may have differences with their de Broglie wavelength, the particle that would have a longer wavelength would be the proton since the wavelength is in direct proportionality with the mass of the particle.
Answer:
b) Nothing will happen, the sea saw will still be balanced.
Explanation:
b) Nothing will happen, the sea saw will still be balanced.
Reason:-
When two kids are balanced, the sum of torques on the seesaw will be zero.
if each kid, reduces their distances by half, then the torque of each kid will be half and the sum of torque of each on the seesaw will be zero.
Therefore the seesaw is balanced
Answer:
3.86×10⁶ Newton/coulombs
Explaination:
Applying,
E = F/q....................... Equation 1
Where E = Electric Field, F = Force, q = charge.
From the question,
Given: F = 5.4×10⁻¹ N, q = -1.4×10⁻⁷ coulombs
Substitute these values into equation 1
E = 5.4×10⁻¹/ -1.4×10⁻⁷
E = -3.86×10⁶ Newtons/coulombs
Hence the magnitude of the electric field created by the
negative test charge is 3.86×10⁶ Newton/coulombs
