Answer:
The magnification is -6.05.
Explanation:
Given that,
Focal length = 34 cm
Distance of the image =2.4 m = 240 cm
We need to calculate the distance of the object

Where, u = distance of the object
v = distance of the image
f = focal length
Put the value into the formula



The magnification is



Hence, The magnification is -6.05.
Answer:
I = I₀ + M(L/2)²
Explanation:
Given that the moment of inertia of a thin uniform rod of mass M and length L about an Axis perpendicular to the rod through its Centre is I₀.
The parallel axis theorem for moment of inertia states that the moment of inertia of a body about an axis passing through the centre of mass is equal to the sum of the moment of inertia of the body about an axis passing through the centre of mass and the product of mass and the square of the distance between the two axes.
The moment of inertia of the body about an axis passing through the centre of mass is given to be I₀
The distance between the two axes is L/2 (total length of the rod divided by 2
From the parallel axis theorem we have
I = I₀ + M(L/2)²
Given Information:
Wavelength of the red laser = λr = 632.8 nm
Distance between bright fringes due to red laser = yr = 5 mm
Distance between bright fringes due to laser pointer = yp = 5.14 mm
Required Information:
Wavelength of the laser pointer = λp = ?
Answer:
Wavelength of the laser pointer = λp = ?
Explanation:
The wavelength of the monochromatic light can be found using young's double slits formula,
y = Dλ/d
y/λ = D/d
Where
λ is the wavelength
y is the distance between bright fringes.
d is the double slit separation distance
D is the distance from the slits to the screen
For the red laser,
yr/λr = D/d
For the laser pointer,
yp/λp = D/d
Equating both equations yields,
yr/λr = yp/λp
Re-arrange for λp
λp = yp*λr/yr
λp = (5*632.8)/5.14
λp = 615.56 nm
Therefore, the wavelength of the small laser pointer is 615.56 nm.
Answer:
α= 1.3 10-5 ºC⁻¹
Explanation:
La dilatación termica de los cuerpos esta dada por la relación
ΔL = L₀ α ( T -T₀)
en este caso nos piden el coeficiente de dilatación térmica
α =DL/L₀ DT
calculemos
α = ( 100,13 -100)/[100 (100 – 0)]
α = 1,3 10-5 ºC⁻¹
Traduction
The thermal expansion of bodies is given by the relationship
ΔL = L₀ α (T -T₀)
in this case they ask us for the coefficient of thermal expansion
α = ΔL / L₀ ΔT
let's calculate
α = (100,13 -100) / [100 (100 - 0)]
α= 1.3 10-5 ºC⁻¹
Answer:
<em>we</em><em> </em><em>have</em><em> </em><em>to</em><em> </em><em>use</em><em> </em><em>formula</em><em> </em><em>of</em><em> </em><em>volume</em><em> </em><em>to</em><em> </em><em>find</em><em> </em><em>volume</em><em> </em><em>of</em><em> </em><em>a</em><em> </em><em>cuboid</em><em>.</em><em> </em><em>(</em><em> </em><em>i.e</em><em> </em><em>v</em><em> </em><em>=</em><em> </em><em>l</em><em> </em><em>×</em><em>b</em><em> </em><em>×</em><em>h</em><em>)</em>
Explanation:
here, let your length of cuboid be x cm, breadth be y cm and height be z cm .
now, formula to find volume of cuboid = length ×
breadth × height.
so, v( volume)= l (length)× b (breadth)× h (height)
or, v= x cm × y cm × z cm
therefore, volume is xyz cm^3..... answer.
<em><u>hope</u></em><em><u> </u></em><em><u>it helps</u></em><em><u>.</u></em><em><u>.</u></em>