Here a cat is running at constant speed which is given as 10 km/h for 5s
So here the average speed is defined as total distance moved in total time interval
so here it is given by
since
here speed of cat is constant so it will remain the same
And hence the average speed and instantaneous speed at any instant for this duration will remain the same
so here answer would be
<em>average speed = 10 km/h</em>
<em>instantaneous speed = 10 km/h</em>
Place the object in between the two jaws such that they touch opposite ends of the object making sure the object is held firmly but don’t press too tight. If you need to measure an internal diameter, then insert the upper jaws in to the cavity and open them till they touch the sides. Tighten the locking screw to hold the jaws in position.
Note the position of the vernier scale zero on the main scale. The main scale reading is the division just before where the zero mark of the vernier scale is aligned. So is the zero mark aligns just after the fifth division between 3 and 4 the main scale reading is then 3.5.
The next step is to take the vernier scale reading. To do this find the mark on the vernier scale which lines up perfectly with a mark on the main scale. The vernier reading can then be found by multiplying the least value of the vernier scale with the number of divisions till that mark. For example if the least value is 0.01 mm and the 7thmark of the vernier scale is lined up perfectly then the vernier scale reading is 7 x 0.01 = 0.07.
The final step is to add the main scale and vernier readings to get the final measurement. For example 3.5 + 0.07 = 3.57 mm.
E S *
The "E" represents Earth, "S" represent Sun, and the "*" represents the nearest star(which is Proxima Centauri).
The main thing to worry about here is units, so ill label everything out.
D'e,s'(Distance between earth and sun) = .<span>00001581 light years
D'e,*'(Distance between earth and Proxima) = </span><span>4.243 light years
Now this is where it gets fun, we need to put all the light years into centimeters.(theres alot)
In one light year, there are </span>9.461 * 10^17 centimeters.(the * in this case means multiplication) or 946,100,000,000,000,000 centimeters.
To convert we multiply the light years we found by the big number.
D'e,s'(Distance between earth and sun) = 1.496 * 10^13 centimeters<span>
D'e,*'(Distance between earth and Proxima) = </span><span>4.014 * 10^18 centimeters
</span>
Now we scale things down, we treat 1.496 * 10^13 centimeters as a SINGLE centimeter, because that's the distance between the earth and the sun. So all we have to do is divide (4.014 * 10^18 ) by (<span>1.496 * 10^13 ).
Why? because that how proportions work.
As a result, you get a mere 268335.7 centimeters.
To put that into perspective, that's only about 1.7 miles
A lot of my numbers came from google, so they are estimations and are not perfect, but its hard to be on really large scales.</span>
Answer:
'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
Explanation:
The question is incomplete, find the complete question in the comment section.
Concave mirrors is an example of a curved mirror. The outer surface of a concave mirror is always coated. On the concave mirror, we have what is called the central axis or principal axis which is a line cutting through the center of the mirror. The points located on this axis are the Pole, the principal focus and the centre of curvature. <em>The focus point is close to the curved mirror than the centre of curvature.</em>
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During the formation of images, one of the incident rays (rays striking the plane surface) coming from the object and parallel to the principal axis, converges at the focus point after reflection because all incident rays striking the surface are meant to reflect out. <em>All incident light striking the surface all converges at a point on the central axis known as the focus.</em>
Based on the explanation above, it can be concluded that 'Incident rays that are parallel to the central axis are sent through a point on the near side of the mirror'.
