Answer:
a
The radial acceleration is 
b
The horizontal Tension is 
The vertical Tension is 
Explanation:
The diagram illustrating this is shown on the first uploaded
From the question we are told that
The length of the string is 
The mass of the bob is 
The angle made by the string is 
The centripetal force acting on the bob is mathematically represented as
Now From the diagram we see that this force is equivalent to
where T is the tension on the rope and v is the linear velocity
So
Now the downward normal force acting on the bob is mathematically represented as
So
=> 
=> 
The centripetal acceleration which the same as the radial acceleration of the bob is mathematically represented as
=> 
substituting values
The horizontal component is mathematically represented as
substituting value
The vertical component of tension is
substituting value
The vector representation of the T in term is of the tension on the horizontal and the tension on the vertical is
substituting value
![T = [(0.3294) i + (3.3712)j ] \ N](https://tex.z-dn.net/?f=T%20%20%3D%20%5B%280.3294%29%20i%20%20%2B%20%283.3712%29j%20%5D%20%5C%20%20N)
20/40=0.5 g/cm^3 becuase, mass/volume=density.
The answers to your questions are as written below:
- The objects that represents a negatively charged particle is : Image B
- The object that represents a positively charged molecule is : Image A
- The object that represents an uncharged molecule is : Image C
- The object the will not move when in an electric fied is : Image C
<h3>Different types of charges molecules</h3>
A negatively charged molecule move inwards when placed in an electric field while positively charged molecule placed in a electric field will move outwards the electric field.
A neutral/uncharged molecule will remains still when placd in an elctric field due to the absence of charges.
Hence we can concude that the answers to your questions are as listed above.
Learn more about electric charges :brainly.com/question/857179
#SPJ4
attached below is the missing image
Answer:
time required after impact for a puck is 2.18 seconds
Explanation:
given data
mass = 30 g = 0.03 kg
diameter = 100 mm = 0.1 m
thick = 0.1 mm = 1 ×
m
dynamic viscosity = 1.75 ×
Ns/m²
air temperature = 15°C
to find out
time required after impact for a puck to lose 10%
solution
we know velocity varies here 0 to v
we consider here initial velocity = v
so final velocity = 0.9v
so change in velocity is du = v
and clearance dy = h
and shear stress acting on surface is here express as
= µ 
so
= µ 
............1
put here value
= 1.75× × 
= 0.175 v
and
area between air and puck is given by
Area = 
area = 
area = 7.85 × 
m²
so
force on puck is express as
Force = × area
force = 0.175 v × 7.85 × 
force = 1.374 × v
and now apply newton second law
force = mass × acceleration
- force = 
- 1.374 × v = 
t = 
time = 2.18
so time required after impact for a puck is 2.18 seconds
