To find the surface area of a single cube we first nees to take the cube root of 8cm3 which is 2.
Now we know that the length of each side is 2 and we can find the area of one side by doing 2x2 which is 4.
To find the total surface area of one cube we do 4 times 6 side giving us a total of 24cm2.
To find the total surface area of the 8 individual cubes, we multiply 24cm2 by 8 to give us a total of 192cm2.
Now to find the total surface area of the one large cube, we know that each side of one of the small cubes is 4cm2 and the large cube is set up so that there are two levels of four cubes right on top of each other. So, the total area of each side of the large cube is 4cm2 times 4 which gives us 16cm2.
Then we multiply 16cm2 by 6 sides to give us a total surface area of 96cm2.
The ratio of the surface area of the single large cube comapred to the total surface area of the single cubes is 96:192
We can further simplify this ratio:
96:192
48:96
24:48
12:24
6:12
3:6
1:2
1 Amp = 1 Coulomb/sec
1 Coulomb/sec = 6.25*10^18 electrons/sec
Therefore,
5.0 A = 5 C/s = 5*6.25*10^18 = 3.125*10^19 e/s
In 10 second, number of electrons are calculated as;
Number of electrons through the device = 3.125*10^19*10 = 3.125*10^20 electrons
Answer:
The object will travel at the speed of 16 m/s.
Explanation:
Given
To determine
How fast is the object traveling?
<u>Important Tip:</u>
The product of the mass and velocity of an object — momentum.
Using the formula
where
Thus, in order to determine the speed of the object, all we need to do is to substitute p = 64 and m = 4 in the formula
switch the equation
divide both sides by 4
simplify
m/s
Therefore, the object will travel at the speed of 16 m/s.
Answer:
beam of light converges to a point A. A lens is placed in the path of the convergent beam 12 cm from P.
To find the point at which the beam converge if the lens is (a) a convex lens of focal length 20 cm, (b) a concave lens of focal length 16 cm
Solution:
As per the given criteria,
the the object is virtual and the image is real (as the lens is placed in the path of the convergent beam)
(a) lens is a convex lens with
focal length, f=20cm
object distance, u=12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
20
1
+
12
1
⟹
v
1
=
60
3+5
⟹v=7.5cm
Hence the image formed is real, at 7.5cm from the lens on its right side.
(b) lens is a concave lens with
focal length, f=−16cm
object distance, 12cm
applying the lens formula, we get
f
1
=
v
1
−
u
1
⟹
v
1
=
f
1
+
u
1
⟹
v
1
=
−16
1
+
12
1
⟹
v
1
=
48
−3+4
⟹v=48m
Hence the image formed is real, at 48 cm from the lens on the right side.
Type of energy was used is kinetic energy.
