Answer: D. 5cm
Explanation:
Given the following :
Focal length (f) = - 6.0 cm
Height of object = 15.0cm
Distance of object from mirror (u) = 12.0cm
Height of image produced by the mirror =?
Firstly, we calculate the distance of the image from the mirror.
Using the mirror formula
1/f = 1/u + 1/v
1/v = 1/f - 1/u
1/v = 1/-6 - 1/12
1/v = - 1/6 - 1/12
1/v = (- 2 - 1) / 12
1/v = - 3 / 12
v = 12 / - 3
v = - 4
Using the relation :
(Image height / object height) = (- image distance / object distance)
Image height / 15 = - (-4) / 12
Image height / 15 = 4 / 12
Image height = (15 × 4) / 12
Image height = 60 / 12
Image height = 5cm
Answer:
The mass of the block, M =T/(3a +g) Kg
Explanation:
Given,
The upward acceleration of the block a = 3a
The constant force acting on the block, F₀ = Ma = 3Ma
The mass of the block, M = ?
In an Atwood's machine, the upward force of the block is given by the relation
Ma = T - Mg
M x 3a = T - Ma
3Ma + Mg = T
M = T/(3a +g) Kg
Where 'T' is the tension of the string.
Hence, the mass of the block in Atwood's machine is, M = T/(3a +g) Kg
Answer:
Explanation:
The question is one that examine the physical fundamental of mechanics of a cylindrical vessel .
We would use the Euler' equation and some coriolis and centripetal force formula.
The fig below explains it.
Answer:
5.95 A
Explanation:
From the question
R = ρL/A..................... Equation 1
Where R = resistance of the tungsten wire, ρ = Resistivity of the tungsten wire, L = length, A = cross sectional area.
Given: L = 1.5 m, A = 0.8 mm² = 0.8×10⁻⁶ m, ρ = 5.60×10⁻⁸ Ω.m
Substitute these values into equation 1
R = 1.5(5.60×10⁻⁸)/0.8×10⁻⁶
R = 0.084 Ω.
Finally, using Ohm law,
V = IR
Where V = Voltage, I = current
Make I the subject of the equation
I = V/R............... Equation 2
I = 0.5/0.084
I = 5.95 A
It will be
E = mgh.
where h and g are constant thus
m can be written as 4/3πr^3*density
E = 4/3πr^3* density
E? = 4/3π(2R)^3* density
= 4/3π8r^3
thus the e will be 4/3π8r^3* density/4/3πr^3*density nd thus you get 8E ..
