Force on a proton when they are moving inside the magnetic field is given by

so here while protons are going upwards we can say its direction of velocity is towards +Y and magnetic field is towards us which is along +Z direction so they will have force due to this motion.
As per above equation of force we can have direction of force which is along +X direction.
So here both proton will bend towards +X direction and start moving in curved path.
Answer:
Tension in the string will increase
Explanation:
As we know that tension in the string at any angle with the vertical is given as

now we have

also we know that
angular speed of the stone is directly depending on the time period of the motion
so it is given as

since the frequency of the revolution is increased from n = 1 rev/s to 2 rev/s
so the angular speed would be doubled
So here we can say that
tension in the string will increase when we will increase the frequency of revolution.
Answer:
The gauge pressure in Pascals inside a honey droplet is 416 Pa
Explanation:
Given;
diameter of the honey droplet, D = 0.1 cm
radius of the honey droplet, R = 0.05 cm = 0.0005 m
surface tension of honey, γ = 0.052 N/m
Apply Laplace's law for a spherical membrane with two surfaces
Gauge pressure = P₁ - P₀ = 2 (2γ / r)
Where;
P₀ is the atmospheric pressure
Gauge pressure = 4γ / r
Gauge pressure = 4 (0.052) / (0.0005)
Gauge pressure = 416 Pa
Therefore, the gauge pressure in Pascals inside a honey droplet is 416 Pa
Answer:
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Answer:

Explanation:
Given:
Solute Diffusion rate = 4.0 × 10⁻¹¹ kg/s
Area of cross-section = 0.50 cm²
Length of channel =0.25 cm
Now for the new channel
Area of cross-section = 0.30 cm²
Length of channel =0.10 cm
let the Solute Diffusion rate of new channel = s
now equating the diffusion rate per unit volume for both the channels

thus,
