1) Size of the image: 2 cm
In order to calculate the size of the image, we can use the following proportion:
where
p = 80 m is the distance of the tree from the pinhole
q = 20 cm = 0.2 m is the distance of the image from the pinhole
= 8 m is the heigth of the object
is the height of the image
By re-arranging the proportion, we find
2) Magnification: 0.0025
The magnification of a camera is given by the ratio between the size of the image and the size of the real object:
so, in this problem we have
<h2>
Answer:</h2>
(a) 10N
<h2>
Explanation:</h2>
The sketch of the two cases has been attached to this response.
<em>Case 1: The box is pushed by a horizontal force F making it to move with constant velocity.</em>
In this case, a frictional force 
is opposing the movement of the box. As shown in the diagram, it can be deduced from Newton's law of motion that;
∑F = ma -------------------(i)
Where;
∑F = effective force acting on the object (box)
m = mass of the object
a = acceleration of the object
∑F = F -
m = 50kg
a = 0 [At constant velocity, acceleration is zero]
<em>Substitute these values into equation (i) as follows;</em>
F - = m x a
F - = 50 x 0
F - = 0
F = -------------------(ii)
<em>Case 2: The box is pushed by a horizontal force 1.5F making it to move with a constant velocity of 0.1m/s²</em>
In this case, the same frictional force is opposing the movement of the box.
∑F = 1.5F -
m = 50kg
a = 0.1m/s²
<em>Substitute these values into equation (i) as follows;</em>
1.5F - = m x a
1.5F - = 50 x 0.1
1.5F - = 5 ---------------------(iii)
<em>Substitute </em><em> = F from equation (ii) into equation (iii) as follows;</em>
1.5F - F = 5
0.5F = 5
F = 5 / 0.5
F = 10N
Therefore, the value of F is 10N
<em />
<span>1.0 m/s
Momentum = mass x velocity
Total Momentum before any collision = total momentum afterwards
4.0 x 3.0= 12 :g x momentum before (x g because using weight)
Afterwards, if the velocity of the two joined is v then we get:
'momentum x g'=12v
so 12v=12
so v=1m/s</span>
Explanation:
1200 is your answer for this question
