The rain gets evaporated in to water vapor and is returned to the clouds where they go through condensation and then they poud down as rain or A.K.A, Precipitation.
Answer:
1.
109.6 cm , - 1.74 , real
2.
1.5
Explanation:
1.
d₀ = object distance = 63 cm
f = focal length of the lens = 40 cm
d = image distance = ?
using the lens equation
d = 109.6 cm
magnification is given as
m = - 1.74
The image is real
2
d₀ = object distance = a
d = image distance = - (a + 5)
f = focal length of lens = 30 cm
using the lens equation
a = 10
magnification is given as
m = 1.5
Answer:
None
Explanation:
An scale is the factor by which actual features on ground are enlarged or reduced for representing on a plane. There are different kinds of scales:
- Verbal scale use of words to represent scale information on the map. The distance or linear units are used for depicting this scale on the map. For example: 1 inch = 1 Kilo meter.
- Fractional scale uses the numbers or values for showing the scale instead of words. As the name says, it is represented using a fraction or ratio. Example: 1: 10,000 or 1/10,000
- In large scale more details are shown in a map, however, less area coverage will be shown in a single map as the scale is large and more details are given. Example: 1:500
- Small scale is exactly opposite to the large scale, less details are shown as magnification is not enough, however a large amount of area can be shown in a single map. Example: 1:25,000
- A graphic scale is a bar that has been calibrated to show map distances. On maps that have been reduced or enlarged the original ratio and written scales are incorrect, since the relationship between map distance and real world distance has been altered, graphic scale is enlarged or reduced to the same extent as the map, this makes it the right option.
I hope you find this information useful and interesting! Good luck!
<span>3.78 m
Ignoring resistance, the ball will travel upwards until it's velocity is 0 m/s. So we'll first calculate how many seconds that takes.
7.2 m/s / 9.81 m/s^2 = 0.77945 s
The distance traveled is given by the formula d = 1/2 AT^2, so substitute the known value for A and T, giving
d = 1/2 A T^2
d = 1/2 9.81 m/s^2 (0.77945 s)^2
d = 4.905 m/s^2 0.607542 s^2
d = 2.979995 m
So the volleyball will travel 2.979995 meters straight up from the point upon which it was launched. So we need to add the 0.80 meters initial height.
d = 2.979995 m + 0.8 m = 3.779995 m
Rounding to 2 decimal places gives us 3.78 m</span>
Answer:
90degrees I'm pretty sure
