Complete Question
A field mouse trying to escape a hawk runs east for 5.0m, darts southeast for 3.0m, then drops 1.0m down a hole into its burrow. What is the magnitude of the net displacement of the mouse?
Answer:
The values is 
Explanation:
From the question we are told that
The distance it travels eastward is 
The distance it travel towards the southeast is 
The distance it travel towards the south is 
Let x-axis be east
y-axis south
z-axis into the ground
The angle made between east and south is 
The displacement toward x-axis is
The displacement toward the y-axis is
Now the overall displacement of the rat is mathematically evaluated as
Answer:
c) 10.7m/s
Explanation:
From the exercise we know that at 5m the ball is traveling at 4m/s
To calculate its initial velocity we need to solve the following equation:
Since the initial height is 0
Solving for 
Microscopes with life science and physics
Answer: 16.3 seconds
Explanation: Given that the
Initial velocity U = 80 ft/s
Let's first calculate the maximum height reached by using third equation of motion.
V^2 = U^2 - 2gH
Where V = final velocity and H = maximum height.
Since the toy is moving against the gravity, g will be negative.
At maximum height, V = 0
0 = 80^2 - 2 × 9.81 × H
6400 = 19.62H
H = 6400/19.62
H = 326.2
Let's us second equation of motion to find time.
H = Ut - 1/2gt^2
Let assume that the ball is dropped from the maximum height. Then,
U = 0. The equation will be reduced to
H = 1/2gt^2
326.2 = 1/2 × 9.81 × t^2
326.2 = 4.905t^2
t^2 = 326.2/4.905
t = sqrt( 66.5 )
t = 8.15 seconds
The time it will take for the rocket to return to ground level will be 2t.
That is, 2 × 8.15 = 16.3 seconds
The rubber protects him from being electrocuted by the flow of current going through the plug.
Hope this helped!!
