<h2><u>Answer:</u></h2>
The simulation kept track of the variables and automatically recorded data on object displacement, velocity, and momentum. If the trials were run on a real track with real gliders, using stopwatches and meter sticks for measurement, the data compared by the following statements:
1. (There would be variables that would be hard to control, leading to less reliable data.)
3. (Meter sticks may lack precision or may be read incorrectly.)
4. (Real glider data may vary since real collisions may involve loss of energy.)
5. (Human error in recording or plotting the data could be a factor.)
The calculation of the centripetal acceleration of an object following a circular path is based on the equation,
a = v² / r
where a is the acceleration, v is the velocity, and r is the radius.
Substituting the known values from the given above,
4.4 m/s² = (15 m/s)² / r
The value of r from the equation is 51.14 m.
Answer: 51.14 m
Answer:
(a): The frequency of the waves is f= 0.16 Hz
Explanation:
T/4= 1.5 s
T= 6 sec
f= 1/T
f= 0.16 Hz (a)
There are no choices on the list you provided that make such a statement,
and it's difficult to understand what is meant by "the following".
That statement is one way to describe the approach to 'forces of gravity'
taken by the theory of Relativity.
Answer:
(I). The motional emf induced between the ends of the segment is 
(II). The motional emf is zero.
Explanation:
Given that,
Magnetic field = 0.080 T
Velocity of wire segment = 78 m/s
Component in x direction = 18 m/s
Component in y direction = 24 m/s
Component in z direction = 72 m/s
Length = 0.50 m
We need to calculate the motional emf induced between the ends of the segment
Using formula of emf
Put the value into he formula
(II). If the wire segment is parallel to the y-axis then the angle between B and v is zero.
We need to calculate the motional emf
Using formula of emf
Here, 
Hence, (I). The motional emf induced between the ends of the segment is
(II). The motional emf is zero.
