Answer:
first lens v = 48 cm
second lens v = -15.6 cm
magnification = 1.67
final image is virtual
and final image is upright
Explanation:
given data
distance = 16 cm
focal length f1 = 12 cm
focal length f2 = 10.0 cm
to find out
location of the final image and magnification and Type of image
solution
we apply here lens formula that is
1/f = 1/v + 1/u .....................1
put here all value and find v for 1st lens
1/12 = 1/v + 1/16
v = 48 cm
and find v for 2nd lens
here u = 20- 48 = -28
- 1/10 = 1/v - 1/28
v = -15.6 cm
and
magnification = first lens (v/u) × second lens ( v/u)
magnification = (-15.6/-28) × ( 48/16)
magnification = 1.67
so here final image is virtual
and final image is upright
Answer:
his type of shock is called inelastic
Explanation:
This exercise is for vehicle crashes, which corresponds to exercise is momentum conservation.
We must begin by defining a system formed by the two cars so that the forces during the crash have been intense and the moment is preserved.
Looking for the moments
initial. Before the crash
p₀ = m₁ v₁₀
final. After the crash
p_f = (m₁ + m₂) v
the conservation of the moment is written
p₀ = p_f
m₁ v₁₀ = (m₁ + m₂) v
This type of shock is called inelastic and has the characteristics that the kinetic energy is not conserved.
Answer:
t = Δa / v
Explanation:
To know which option is not true, we shall fine a relationship between acceleration (a), velocity (v), time (t) and radius (r). This is illustrated below:
Acceleration can simply be defined as the rate of change of velocity with time. Mathematically, it is expressed as shown below:
Acceleration = change in velocity / time
a = Δv / t ..... (1)
But
Δv = v₂ – v₁
Substitute the value of Δv into equation (1)
a = Δv / t
a = v₂ – v₁ / t ....... (2)
From equation (1), make Δv the subject of the equation.
a = Δv / t
Cross multiply
Δv = at .... (3)
From equation (1), make t the subject of the equation.
a = Δv / t
Cross multiply
at = Δv
Divide both side by a
t = Δv /a ...... (4)
From circular motion, centripetal's force is given by:
F = mv²/r
F = ma꜀
Therefore,
ma꜀ = mv²/r
Cancel out m
a꜀ = v²/r
SUMMARY:
a = Δv / t
a = v₂ – v₁ / t
Δv = at
t = Δv /a
a꜀ = v²/r
Considering the options given in question above, t = Δa / v is not a true statement.
Answer:![\theta =79.47 North\ of\ west](https://tex.z-dn.net/?f=%5Ctheta%20%3D79.47%20North%5C%20of%5C%20west)
Explanation:
Given
Mass of airplane =21500 kg
Force due to jet engines=35100 N
Force from wind is 14900 N at angle of 75 south of west
Resolving forces
![F_x=14900cos75=-3856.403 N](https://tex.z-dn.net/?f=F_x%3D14900cos75%3D-3856.403%20N)
![F_y=35100-14900sin75](https://tex.z-dn.net/?f=F_y%3D35100-14900sin75)
![F_y=35100-14392.29=20707.70 N](https://tex.z-dn.net/?f=F_y%3D35100-14392.29%3D20707.70%20N)
Therefore acceleration in x direction
![a_x=\frac{F_x}{m}=\frac{-3856.403}{21500}=-0.179 m/s^2](https://tex.z-dn.net/?f=a_x%3D%5Cfrac%7BF_x%7D%7Bm%7D%3D%5Cfrac%7B-3856.403%7D%7B21500%7D%3D-0.179%20m%2Fs%5E2)
![a_y=\frac{F_y}{m}=\frac{20707.70}{21500}=0.963 m/s^2](https://tex.z-dn.net/?f=a_y%3D%5Cfrac%7BF_y%7D%7Bm%7D%3D%5Cfrac%7B20707.70%7D%7B21500%7D%3D0.963%20m%2Fs%5E2)
Direction of acceleration
![tan\theta =\frac{a_y}{a_x}=\frac{0.963}{0.179}=-5.38](https://tex.z-dn.net/?f=tan%5Ctheta%20%3D%5Cfrac%7Ba_y%7D%7Ba_x%7D%3D%5Cfrac%7B0.963%7D%7B0.179%7D%3D-5.38)
![\theta =79.47\ North\ of\ west](https://tex.z-dn.net/?f=%5Ctheta%20%3D79.47%5C%20North%5C%20of%5C%20west)