Answer:
a) frequency = 0.1724 Hz
b) Period = 5.8 sec
c) speed = 7.04 m/s
d) acceleration = 7.62 m/s²
Explanation:
Given that;
radius = 6.5m
time period = 5.8 sec every circle
a) the frequency
frequency is the number of rotation in unit time
frequency = 1 / time period = 1/5.8
frequency = 0.1724 Hz
b) the period
period is time taken in one rotation
period = total time / rotation = 5.8 / 1
Period = 5.8 sec
c) the speed
speed = distance/time = circumference/time period = 2πr / t = (2π×6.5) / 5.8
speed = 7.04 m/s
d) acceleration
To find the acceleration we take the linear velocity squared divided by the radius of the circle.
so
acceleration = v² / r = (7.04)² / 6.5 = 49.5616 / 6.5
acceleration = 7.62 m/s²
Answer:
A
Explanation:
The answer is A because proton number is the same as atomic number
Answer:
Explanation:
Volume of the insulating shell is,

Charge density of the shell is,

Here, 

B)
The electric field is 
For 0 <r<R the electric field is zero, because the electric field inside the conductor is zero.
C)
For R <r <2R According to gauss law

substitute 

D)
The net charge enclosed for each r in this range is positive and the electric field is outward
E)
For r>2R
Charge enclosed is zero, so electric field is zero
Answer:
The force required to move the quarterback with linebacker is <u>1215 N</u>
Explanation:



Using Newton's second law, it is established that F = Ma
Where F is net force acting on the system, a is the acceleration and M is mass of the two object 
Now consider both
as a system, so net force acting on the system is 
Substitute the given values in the above formula,


Force = 1215 N
<u>1215 N </u>is the force required to move the quarterback with linebacker.