Answer:
223 degree
Explanation:
We are given that
Magnitude of resultant vector= 8 units
Resultant vector makes an angle with positive -x in counter clockwise direction
We have to find the magnitude and angle of the equilibrium vector.
We know that equilibrium vector is equal in magnitude and in opposite direction to the given vector.
Therefore, magnitude of equilibrium vector=8 units
x-component of a vector=
Where v=Magnitude of vector
Using the formula
x-component of resultant vector=
y-component of resultant vector=
x-component of equilibrium vector=
y-component of equilibrium vector=
Because equilibrium vector lies in III quadrant
The angle 
lies in III quadrant
In III quadrant ,angle =
Angle of equilibrium vector measured from positive x in counter clock wise direction=180+43=223 degree
Anything that's not supported and doesn't hit anything, and
doesn't have any air resistance, gains 9.8 m/s of downward
speed every second, on account of gravity. If it happens to
be moving up, then it loses 9.8 m/s of its upward speed every
second, on account of gravity.
(64.2 m/s) - [ (9.8 m/s² ) x (1.5 sec) ]
= (64.2 m/s) - [ 14.7 m/s ]
= 49.5 m/s . (upward)
Answer:
Yes
Explanation:
p = momentum of photon
E = energy of photon
c = velocity of light
Units of p = kg m /s
Units of E = kg m^2 / s^2
Units of E / p = {kg m^2 / s^2} / {kg m /s} = m/s
It is the unit of speed, so by the division of energy to the momentum, we get the speed. yes it is correct.
Answer:
Explanation:
Initial velocity u = V₀ in upward direction so it will be negative
u = - V₀
Displacement s = H . It is downwards so it will be positive
Acceleration = g ( positive as it is also downwards )
Using the formula
v² = u² + 2 g s
v² = (- V₀ )² + 2 g H
= V₀² + 2 g H .
v = √ ( V₀² + 2 g H )
Answer:
Explanation:
-The only relevant force is the electrostatic force
-The formula for the electrostatic force is:
E is the electric field and q is the magnitude of the charge.
#Since the electric field is the same in both cases, and the charge of the protons and electrons have the same magnitude, you can state that the magnitude of the electric forces acting in both proton and electron are the same.
-Applying Newton's 2nd Law:
#equate the two forces:
#The equations for velocity in uniform acceleration:
#For the proton:
#For the electron:
The mass values of the proton and electron are:
The speed of the ion is therefore calculated as:
Hence, the ion's speed at the negative plate is
