Oh, what a lovely question! Imagine a small ladybug resting on a leaf, its delicate body measuring about 1 cm in length. Nature is full of such beautiful and intricate creations, each one a tiny masterpiece waiting to be appreciated. Just like that little ladybug, we can find wonder and beauty in the smallest of things around us.
An example of an object that is 1 cm in length is a standard paperclip. Paperclips typically measure around 1 cm in length, making them a common everyday object that fits this measurement. Other examples could include a small button, a LEGO brick stud, or a standard sewing pin.
1 cm
pencil eraser
It takes 10 mm to cover the same length as one cm, so the cm is larger.
It is a length of 1.9 centimetres.
A Dime!
The length of a centimeter is approximately the width of a standard paperclip or the tip of a typical pencil.
The image distance (61 cm) is positive since the image is on the same side of the lens as the object. Using the lens formula (1/f = 1/d_o + 1/d_i), where d_o is the object distance (12 cm) and d_i is the image distance, the focal length (f) of the lens is approximately 15 cm.
The mirror formula for concave mirrors is 1/f = 1/d_o + 1/d_i, where f = focal length, d_o = object distance, and d_i = image distance. Given f = 10 cm, d_i = 30 cm, we can solve for d_o: 1/10 = 1/d_o + 1/30. Solving for d_o gives d_o = 15 cm. The corresponding object is located 15 cm away from the mirror.
616 cubed centimeters.
Using the lens formula (1/f = 1/do + 1/di) and the magnification formula (m = -di/do) where m = -4, you can solve for the focal length (f). Given the object distance (do = -15 cm), you can calculate the focal length to be 10 cm.
The volume of the object is calculated by multiplying its length, width, and height: 20 cm * 5 cm * 1 cm = 100 cm³. To find the density, divide the mass (500 g) by the volume (100 cm³): 500 g / 100 cm³ = 5 g/cm³. The density of the object is 5 g/cm³.
An ordinary fingernail (the pink part) is approx 1 cm in length.
Using the lens formula (1/f = 1/do + 1/di), where f is the focal length, do is the object distance, and di is the image distance, we can solve for f. Once we have the focal length, we can use the magnification equation (magnification = hi/ho = -di/do) to find the height of the image where hi is the height of the image and ho is the height of the object.
By unit of length and distance and conversion ,we can say that 1 cm=10 mm 56 mm=5.6 cm
1/o + 1/i = 1/ff = (o x i)/(o + i)f = 11.1 cm (rounded)
length x width x height x specific weight example : 10 cm x 5 cm x 2 cm x 1 kg/cm3 = 100 kg
If the object is kept in place while adjusting the lens, the real image will also move as the lens is adjusted. If the image moves closer to the lens, the focal length of the lens has decreased. If the image moves farther away from the lens, the focal length has increased.