Answer:
1.
109.6 cm , - 1.74 , real
2.
1.5
Explanation:
1.
d₀ = object distance = 63 cm
f = focal length of the lens = 40 cm
d = image distance = ?
using the lens equation
![\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7B1%7D%7Bd_%7Bo%7D%7D%20%2B%20%5Cfrac%7B1%7D%7Bd%7D)
![\frac{1}{40} = \frac{1}{63} + \frac{1}{d}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B40%7D%20%3D%20%5Cfrac%7B1%7D%7B63%7D%20%2B%20%5Cfrac%7B1%7D%7Bd%7D)
d = 109.6 cm
magnification is given as
![m = \frac{-d}{d_{o}}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B-d%7D%7Bd_%7Bo%7D%7D)
![m = \frac{-109.6}{63}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B-109.6%7D%7B63%7D)
m = - 1.74
The image is real
2
d₀ = object distance = a
d = image distance = - (a + 5)
f = focal length of lens = 30 cm
using the lens equation
![\frac{1}{f} = \frac{1}{d_{o}} + \frac{1}{d}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bf%7D%20%3D%20%5Cfrac%7B1%7D%7Bd_%7Bo%7D%7D%20%2B%20%5Cfrac%7B1%7D%7Bd%7D)
![\frac{1}{30} = \frac{1}{a} + \frac{1}{- (a + 5)}](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B30%7D%20%3D%20%5Cfrac%7B1%7D%7Ba%7D%20%2B%20%5Cfrac%7B1%7D%7B-%20%28a%20%2B%205%29%7D)
a = 10
magnification is given as
![m = \frac{-d}{d_{o}}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B-d%7D%7Bd_%7Bo%7D%7D)
![m = \frac{- (- (a +5))}{a}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B-%20%28-%20%28a%20%2B5%29%29%7D%7Ba%7D)
![m = \frac{(5 + 10)}{10}](https://tex.z-dn.net/?f=m%20%3D%20%5Cfrac%7B%285%20%2B%2010%29%7D%7B10%7D)
m = 1.5
Answer:
The value of the spring constant of this spring is 1000 N/m
Explanation:
Given;
equilibrium length of the spring, L = 10.0 cm
new length of the spring, L₀ = 14 cm
applied force on the spring, F = 40 N
extension of the spring due to applied force, e = L₀ - L = 14 cm - 10 cm = 4 cm
From Hook's law
Force applied to a spring is directly proportional to the extension produced, provided the elastic limit is not exceeded.
F ∝ e
F = ke
where;
k is the spring constant
k = F / e
k = 40 / 0.04
k = 1000 N/m
Therefore, the value of the spring constant of this spring is 1000 N/m
Answer:
magnitude of the magnetic field 0.692 T
Explanation:
given data
rectangular dimensions = 2.80 cm by 3.20 cm
angle of 30.0°
produce a flux Ф = 3.10 ×
Wb
solution
we take here rectangular side a and b as a = 2.80 cm and b = 3.20 cm
and here angle between magnitude field and area will be ∅ = 90 - 30
∅ = 60°
and flux is express as
flux Ф =
.................1
and Ф = BA cos∅ ............2
so B =
and we know
A = ab
so
B =
..............3
put here value
B =
solve we get
B = 0.692 T
Answer:
A maximum
Explanation:
When displacement is maximum, velocity is Zero and vice versa
When displacement is maximum, acceleration is maximum and when it is zero, acc. Is zero