Answer:
433 mph
Explanation:
We know that
...Eq(1)
Here,
Distance =1,592 miles
Time=3.68 hours
Putting the value of distance and Time in the Eq(1) We get
Velocity =

therefore Average velocity is 433 mph
Answer:
The magnitude of F1 is

The magnitude of F2 is

And the direction of F2 is

Explanation:
<u>Net Force
</u>
Forces are represented as vectors since they have magnitude and direction. The diagram of forces is shown in the figure below.
The larger pull F1 is directed 21° west of north and is represented with the blue arrow. The other pull F2 is directed to an unspecified direction (red arrow). Since the resultant Ft (black arrow) is pointed North, the second force must be in the first quadrant. We must find out the magnitude and angle of this force.
Following the diagram, the sum of the vector components in the x-axis of F1 and F2 must be zero:

The sum of the vertical components of F1 and F2 must equal the total force Ft

Solving for
in the first equation






The magnitude of F1 is

The magnitude of F2 is

And the direction of F2 is

Answer:
The average recoil force on the gun during that 0.40 s burst is 45 N.
Explanation:
Mass of each bullet, m = 7.5 g = 0.0075 kg
Speed of the bullet, v = 300 m/s
Time, t = 0.4 s
The change in momentum of an object is equal to impulse delivered. So,

For 8 shot burst, average recoil force on the gun is :

So, the average recoil force on the gun during that 0.40 s burst is 45 N.
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