Answer:
Net force = - 1500 N
Explanation:
We calculate the net force acting using Newton's second Law:
The relative velocity of the athlete relative to the ground is 5.2 m/s
The given parameters;
constant velocity of the athlete, V = 5.2 m/s
let the velocity of the ground = Vg = 0
The relative velocity concept helps us to determine the velocity of a moving object relative to a stationary observer.
The athlete is the moving object in this question while the ground is stationary.
The relative velocity of the athlete relative to the ground is calculated as follows;
Thus, the relative velocity of the athlete relative to the ground is 5.2 m/s
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Answer:
Explanation:
<u>Charge of an Electron</u>
Since Robert Millikan determined the charge of a single electron is
Every possible charged particle must have a charge that is an exact multiple of that elemental charge. For example, if a particle has 5 electrons in excess, thus its charge is
Let's test the possible charges listed in the question:
. We have just found it's a possible charge of a particle
. Since 3.2 is an exact multiple of 1.6, this is also a possible charge of the oil droplets
this is not a possible charge for an oil droplet since it's smaller than the charge of the electron, the smallest unit of charge
cannot be a possible charge for an oil droplet because they are not exact multiples of 1.6
Finally, the charge is four times the charge of the electron, so it is a possible value for the charge of an oil droplet
Summarizing, the following are the possible values for the charge of an oil droplet:
Answer:
6.54 x 10^-10 N/kg
Explanation:
mass of the shell, M = 135 kg
radius of shell, R = 1.49 m
distance from the centre of the shell, r = 3.71 m
The gravitational filed is given by
E = 6.54 x 10^-10 N/kg
Thus, the gravitational field intensity at a distance 3.71 m is 6.54 x 10^-10 N/kg.
Answer:
a) F_{e} - F_{m} = 0, b) v = 666.67 m / s
Explanation:
For the proton to move y-axis the sum of the electric and magnetic force must be zero, therefore
= 0
a) ∑ F = 0
F_{e} - F_{m} = 0
b) we write the forces
q E = q v B
v = E / B
Let's calculate
v = 300 / 0.45
v = 666.67 m / s