A) A concave mirror forming a larger, virtual image
Explanation:
The figure is missing; see attachment.
There are two types of mirror:
- Concave (converging) mirrors: a concave mirror is a mirror that reflects the light inward
- Convex (diverging) mirrors: a convex mirror is a mirror that reflects the light outward
The image formed by a mirror can also be of two types:
- Real image: it is formed on the same side of the object, with respect to the mirror
- Virtual image: it is formed on the opposite side of the object, with respect to the mirror
In the figure of this problem (see attachment), we see that:
- The mirror reflects the light from the object inward --> so it is a concave mirror
- The image is formed on the other side of the mirror --> it is a virtual image
So the correct option is
A) A concave mirror forming a larger, virtual image
Learn more about mirrors:
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Answer:
The chemical equation for photosynthesis is 6CO2+6H2O→C6H12O6+6O2.
Explanation:
Answer : The value of the constant for a second order reaction is, 
Explanation :
The expression used for second order kinetics is:
![kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}](https://tex.z-dn.net/?f=kt%3D%5Cfrac%7B1%7D%7B%5BA_t%5D%7D-%5Cfrac%7B1%7D%7B%5BA_o%5D%7D)
where,
k = rate constant = ?
t = time = 17s
= final concentration = 0.0981 M
= initial concentration = 0.657 M
Now put all the given values in the above expression, we get:


Therefore, the value of the constant for a second order reaction is, 
Answer:
The box will be moving at 0.45m/s. The solution to this problem requires the knowledge and application of newtons second law of motion and the knowledge of linear motion. The vertical component of the force Fp acts vertically upwards against the directio of motion. This causes a constant upward force of 23sin45° to act on the box. Fhe frictional force of 13N also acts vertically upwards and so two forces act upwards against rhe force of gravity resulting un a net force of 0.7N acting kn the box. This corresponds to an acceleration of 0.225m/s². So in w.0s after i start to push v = 0.45m/s.
Explanation:
Answer:
Period refers to the time for something to happen and is measured in seconds/cycle