The image of the object is 8cm to the left of the lens (D)
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What is the image of an object?
The image of an object is said to be the location where light rays from that object intersect with a mirror by reflection.
It is calculated thus:
1÷v = 1÷f - 1÷u
<h3>How to calculate the image of an object</h3>
From the formula
1÷v = 1÷f - 1÷u
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Where </h3>
V = image distance fromthe object
U = object
f = focal length
Substitute the values
1÷v = 1÷8 - 1÷ 4
1÷v = - 1÷8
Make v the subject of formula
v = -8cm
Therefore, the image of the object is 8cm to the left of the lens (D)
Learn more on focal length here:
brainly.com/question/25779311
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Assuming you're working in a 3D cartesian coordinate system, i.e. each point in space has an x, y, and z coordinate, you add up the forces' x/y/z components to find the resultant force.
The non-relativistic formula for low speed v < 0.1c is:
K.E = 0.5mv^2 = 0.5 * 235 * (7)^2 = 5757.5 J
Answer:
4units
Explanation:
To calculate the total distance the beam will travel along this path, you will use the formula for calculating the distance between two coordinates expressed as;
D = √(x2-x1)²+(y2-y1)²
Given the coordinate points
(3,5) and (7,5)
Substitute
D = √(7-3)²+(5-5)²
D = √(7-3)²+0²
D = √4²
D = √16
D = 4
Hence the total distance the beam will travel along this path is 4units
The correct answer is: +5
Explanation:
An object is placed at 0; it means:
Initial position of the object = 0.
Now it moves to 3 units to right, so keeping the standard cartesian coordinate system in mind (in right right x-axis is positive and left x-axis is right), the new position of the object will be +3.
Object now moves 4 units to the left, it means +3 - 4 = -1; object is at the position -1.
Object then moves 6 units to the right, therefore,
Final position of the object = -1 + 6 = +5.
Displacement = Final position - Initial position
Displacement = +5 - 0 = +5
