An object distance is
presented as s = 5f and we know that the mirror equation relates the image
distance to the object distance and the focal length.
The mirror equation is
1/f = 1/s + 1/s’ where the variable f stands for
the focal length of the mirror. Variable (s)
represents the distance between the mirror surface and the object and the
variable <span>(s’) represents the distance between the mirror surface and
the image. </span>
In addition, a concave mirror
will have a positive focal length (f) and a convex mirror will have a negative
focal length (f).
Now, we then have 1/f = 1/5f
+ 1/s’ which is s’ = 5f/4
Then we get the magnification
ratio that expresses the size or amount of magnification or reduction of the
object or image and to get the magnification, we use this equation: M= s’/s
M= 5f/4x5f
s’ = 1/4s
Therefore, the image height
is one fourth of the object height
vib. motion motion of wire of guitar
circular motion revolution of earth around sun
1ml 1cm3
1m3 100cm3
volume of liquid measuring cylinder
Answer:
Time = 80.91 seconds
Explanation:
Given the following data;
Velocity = 5.50 m/s.
Distance = 445 meters
To find the time;
Velocity can be defined as the rate of change in displacement (distance) with time. Velocity is a vector quantity and as such it has both magnitude and direction.
Mathematically, velocity is given by the equation;

Substituting into the formula, we have;
5.5 = 445/time
Time = 445/5.5
Time = 80.91 seconds
Answer:
The heat loss per unit length is 
Explanation:
From the question we are told that
The outer diameter of the pipe is 
The thickness is
The temperature of water is
The outside air temperature is 
The water side heat transfer coefficient is 
The heat transfer coefficient is 
The heat lost per unit length is mathematically represented as
![\frac{Q}{L} = \frac{2 \pi (T - Ta)}{ \frac{ln [\frac{d}{D} ]}{z_1} + \frac{ln [\frac{d}{D} ]}{z_2}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%5Cpi%20%28T%20-%20Ta%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_1%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7Bd%7D%7BD%7D%20%5D%7D%7Bz_2%7D%7D)
Substituting values
![\frac{Q}{L} = \frac{2 * 3.142 (363 - 263)}{ \frac{ln [\frac{0.104}{0.002} ]}{300} + \frac{ln [\frac{0.104}{0.002} ]}{20}}](https://tex.z-dn.net/?f=%5Cfrac%7BQ%7D%7BL%7D%20%20%20%3D%20%5Cfrac%7B2%20%2A%203.142%20%28363%20-%20263%29%7D%7B%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B300%7D%20%20%2B%20%20%5Cfrac%7Bln%20%5B%5Cfrac%7B0.104%7D%7B0.002%7D%20%5D%7D%7B20%7D%7D)


Answer:
y = 77.74 10⁻⁵ m
Explanation:
For this exercise we can use Newton's second law
F = m a
a = F / m
a = 4.9 10⁻¹⁶ / 9.1 10⁻³¹
a = 0.538 10¹⁵ m / s
This is the vertical acceleration of the electron.
Now let's use kinematics to find the time it takes to move the
x= 29 mm = 29 10⁻³ m
On the x axis
v = x / t
t = x / v
t = 29 10⁻³ / 1.7 10⁷
t = 17 10⁻¹⁰ s
Now we can look for vertical distance at this time.
y =
t + ½ a t²
y = 0 + ½ 0.538 10¹⁵ (17 10⁻¹⁰)²
y = 77.74 10⁻⁵ m