Its larger and if u where wondering to positive ions are smaller
Answer:
O The particles of the medium move more slowly and there are fewer chances to transfer energy.
Explanation:
Various media are made up of particles. These particles are in constant motion according to the kinetic theory of matter. Recall that temperature has been defined as the average kinetic energy of the particles in a medium. Hence, for any given medium, the velocity of particle motion increases or decreases linearly with temperature.
The speed of particles in any medium increases or decreases as the temperature of the medium increases or decreases as emphasised above. Hence, at low temperature, the velocity of waves set up by the motion of particles in a medium decreases and transfer the wave energy to neighbouring particles occurs more slowly than at high temperatures.
Answer:
Both of them reach the lake at the same time.
Explanation:
We have equation of motion s = ut + 0.5at²
Vertical motion of James : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²

Vertical motion of John : -
Initial velocity, u = 0 m/s
Acceleration, a = g
Displacement, s = h
Substituting,
s = ut + 0.5 at²
h = 0 x t + 0.5 x g x t²

So both times are same.
Both of them reach the lake at the same time.
Mass (m)=55kg
acceleration (a)=9.81 m/s^2, this is the acceleration due to gravity.
initial velocity=0m/s. The skydiver doesn’t start with any speed because she is on the plane or helicopter.
final velocity=16m/s This is the velocity (speed) the skydiver reaches
The equation we use is KE=.5mv^2
Kinetic energy=.5 mass x velocity^2
KE=.5(55kg)(16m/s)^2
KE=.5(55kg)(256m/s)
KE=.5(14080J)
J=Joules
KE=7040J
Kinetic energy is 7040 Joules (J)
Hope this helps
Answer:
V = 192 kV
Explanation:
Given that,
Charge, 
Distance, r = 0.3 m
We need to find the electric potential at a distance of 0.3 m from a point charge. The formula for electric potential is given by :

So, the required electric potential is 192 kV.