Answer:
= 1.9 cm
Explanation:
The magnification of a microscope is the product of the magnification of the eyepiece by the magnifier with the objective
M = M₀ 
Where M₀ is the magnification of the objective and is the magnification of the eyepiece.
The eyepiece is focused to the near vision point (d = 25 cm)
= 25 /
The objective is focused on the distances of the tube (L)
M₀ = -L / f₀
Substituting
M = - L/f₀ 25/
1) Let's look for the focal length of the eyepiece (faith)
= - L 25 / f₀ M
M = 400X = -400
= - 12 25 /0.40 (-400)
= 1.875 cm
Let's approximate two significant figures
= 1.9 cm
Answer:
TRUE
Explanation:
Balance forces are usually defined as the two distinct force that acts on an object but in opposite directions. These two acting forces are equal in size or magnitude. When this type of force is applied on any object, it signifies that the object is stationary or it is moving at a constant speed and in the same direction.
This force is comprised of two most important properties namely the strength and direction. When any of the two forces is higher then it result in the motion of the object.
Thus, the above given statement is TRUE.
Answer:
Explanation:
Given that,
Mass m = 6.64×10^-27kg
Charge q = 3.2×10^-19C
Potential difference V =2.45×10^6V
Magnetic field B =1.6T
The force in a magnetic field is given as Force = q•(V×B)
Since V and B are perpendicular i.e 90°
Force =q•V•BSin90
F=q•V•B
So we need to find the velocity
Then, K•E is equal to work done by charge I.e K•E=U
K•E =½mV²
K•E =½ ×6.64×10^-27 V²
K•E = 3.32×10^-27 V²
U = q•V
U = 3.2×10^-19 × 2.45×10^6
U =7.84×10^-13
Then, K•E = U
3.32×10^-27V² = 7.84×10^-13
V² = 7.84×10^-13 / 3.32×10^-27
V² = 2.36×10^14
V=√2.36×10^14
V = 1.537×10^7 m/s
So, applying this to force in magnetic field
F=q•V•B
F= 3.2×10^-19 × 1.537×10^7 ×1.6
F = 7.87×10^-12 N
It results change only in it's kinetic energy, it's KE will increase in accord with the work-energy theorem
