Answer:
= 1.7 cm
Explanation:
The magnification of the compound microscope is given by the product of the magnification of each lens
M = M₀ 
M = - L/f₀ 25/
Where f₀ and are the focal lengths of the lens and eyepiece, respectively, all values in centimeters
In this exercise they give us the magnification (M = 400X), the focal length of the lens (f₀ = 0.6 cm), the distance of the tube (L = 16 cm), let's look for the focal length of the eyepiece ()
= - L / f₀ 25 / M
Let's calculate
= - 16 / 0.6 25 / (-400)
= 1.67 cm
The minus sign in the magnification is because the image is inverted.
= 1.7 cm
<h3><u>Answer;</u></h3>
Radius = 0.0818 m
Angular velocity = 2.775 × 10^7 rad/sec
<h3><u>Explanation;</u></h3>
The mass of proton m=1.6748 × 10^-27 kg;
Charge of electron e= 1.602 × 10^-19 C;
kinetic energy E= 2.7 MeV
= 2.7 × 10^6 × 1.602 × 10^-19 J;
= 4.32 × 10^-13 Joules
But; K.E =0.5m*v^2,
Hence v=√(2K.E/m)
Velocity = 2.27 × 10^7 m/s
Angular velocity, ω = v/r
Therefore; V = ωr
Hence; V = √(2K.E/m) = ωr
r= √(2E/m)/w = √E*√(2*m)/(eB)
= √E * √(2×1.6748×10^-27)/(1.602×10^-19 ×2.9)
but E = 4.32 × 10^-13 Joules
r = 0.0818 m
Angular speed
Angular velocity, ω = v/r , where r is the radius and v is the velocity
Therefore;
Angular velocity = 2.27 × 10^7 / 0.0818 m
= 2.775 × 10^7 rad /sec
Answer:
vf=94.4 m/s
Explanation:
acceleration is the final velocity minus initial velocity divided by time
a = (vf-vi)/t
Given:
a= 14.2 m/s^2
vi= 0 (at rest)
t = 6.6
Solve for vf
a = (vf-vi)/t
a*t=vf-vi
(14.2)*(6.6)=vf - 0
vf=94.4 m/s
Answer:
The magnitude of the force that the 6.3 kg block exerts on the 4.3 kg block is approximately 41.9 N
Explanation:
Forces on block 4.3 kg are:
63N to the right and R21 (contact force from the 6.3 kg block) to the left
Net force on 4.3 kg block is: 63 N - R21
Forces on the 6.3 kg block are:
R12 to the right (contact force from the 4.3 kg block) and 11 N to the left.
So net force on the 6.3 kg block is: R12 - 11 N
According to the action-reaction principle the contact forces R21 and R12 must be equal in magnitude (let's call them simply "R").
Then, since the blocks are moving with the SAME acceleration, we equal their accelerations:
a1 = (63 N - R)/4.3 = (R - 11 N)/6.3 = a2
solve for R by cross multiplication
6.3 (63 - R) = 4.3 (R - 11)
396.9 - 6.3 R = 4.3 R - 47.3
369.9 + 47.3 = 10.6 R
444.2 = 10.6 R
R = 444.2 / 10.6
R = 41.90 N
