Less, if it’s too big: hard to control and maneuverability for shooting wouldn’t be that good. a smaller wheelchair allows for faster movement and control, along with easier shooting and upper body movement
Technician B is right say that hard water potting i usually jut a Surface problem that can be wahed off.
What do you mean by Hard water?
The amount of dissolved calcium and magnesium in the water determines its hardness. Calcium and magnesium are the main dissolved minerals in hard water. The last time you washed your hands, you might have actually felt the effects of hard water.
What do you mean by acid rain?
Any type of precipitation that contains acidic elements, such as sulfuric or nitric acid, that falls to the ground from the atmosphere in wet or dry forms is referred to as acid rain, also known as acid deposition. Rain, snow, fog, hail, and even acidic dust can fall under this category.
Some plants are sensitive to excessive moisture around their root zone, so it may be necessary to increase drainage when growing plants in pots. Additionally, standing water at the bottom of the pot can cause root rot.
Many university agriculture extension agencies have thoroughly debunked the old garden myth that adding rocks to the bottom of a pot will increase drainage.
Learn more about hard water click here:
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Answer: change the tires
Explanation: you can’t drive on a flat tire
Answer:
The value of critical length = 3.46 mm
The value of volume of fraction of fibers = 0.43
Explanation:
Given data
= 800 M pa
D = 0.017 mm
L = 2.3 mm
= 5500 M pa
= 18 M pa
= 13.5 M pa
(a) Critical fiber length is given by

Put all the values in above equation we get

mm
This is the value of critical length.
(b).Since this critical length is greater than fiber length Than the volume fraction of fibers is given by

Put all the values in above formula we get

= 0.43
This is the value of volume of fraction of fibers.
Answer:
component of acceleration are a = 3.37 m/s² and ar = 22.74 m/s²
magnitude of acceleration is 22.98 m/s²
Explanation:
given data
velocity = 10 m/s
initial time to = 0
distance s = 400 m
time t = 14 s
to find out
components and magnitude of acceleration after the car has travelled 200 m
solution
first we find the radius of circular track that is
we know distance S = 2πR
400 = 2πR
R = 63.66 m
and tangential acceleration is
S = ut + 0.5 ×at²
here u is initial speed and t is time and S is distance
400 = 10 × 14 + 0.5 ×a (14)²
a = 3.37 m/s²
and here tangential acceleration is constant
so velocity at distance 200 m
v² - u² = 2 a S
v² = 10² + 2 ( 3.37) 200
v = 38.05 m/s
so radial acceleration at distance 200 m
ar = 
ar = 
ar = 22.74 m/s²
so magnitude of total acceleration is
A = 
A = 
A = 22.98 m/s²
so magnitude of acceleration is 22.98 m/s²