Answer:
a.) The main scale reading is 10.2cm
b.) Division 7 = 0.07
c.) 10.27 cm
d.) 10.31 cm
e.) 10.24 cm
Explanation:
The figure depicts a vernier caliper readings
a.) The main scale reading is 10.2 cm
The reading before the vernier scale
b.) Division 7 = 0.07
the point where the main scale and vernier scale meet
c.) The observed readings is
10.2 + 0.07 = 10.27 cm
d.) If the instrument has a positive zero error of 4 division
correct reading = 10.27 + 0.04 = 10.31cm
e.) If the instrument has a negative zero error of 3 division
correct reading = 10.27 - 0.03 = 10.24cm
If the bubble travels 10 meters per second and it takes 10 seconds, then just multiply the distance per second by the total seconds to get the total depth.
10 • 10 = 100
The lake is 100 meters deep.
Think of it this way to clarify the answer:
It takes a bubble traveling at a speed of 10 meters per second 10 seconds to travel 100 meters.
Look up pulleys problem through Khan academy and a video should pop up with a problem similar and you should be able to walk through it .
Answer:
The kinetic energy of the merry-goround after 3.62 s is 544J
Explanation:
Given :
Weight w = 745 N
Radius r = 1.45 m
Force = 56.3 N
To Find:
The kinetic energy of the merry-go round after 3.62 = ?
Solution:
Step 1: Finding the Mass of merry-go-round


m = 76.02 kg
Step 2: Finding the Moment of Inertia of solid cylinder
Moment of Inertia of solid cylinder I =
Substituting the values
Moment of Inertia of solid cylinder I
=>
=> 
=> 
Step 3: Finding the Torque applied T
Torque applied T =
Substituting the values
T = 
T = 81.635 N.m
Step 4: Finding the Angular acceleration
Angular acceleration ,
Substituting the values,


Step 4: Finding the Final angular velocity
Final angular velocity ,
Substituting the values,


Now KE (100% rotational) after 3.62s is:
KE = 
KE =
KE = 544J
I think the answer is A
.
im not the best with physics but i think its right