Answer:
Time, t = 0.23 seconds
Explanation:
It is given that,
Initial speed of the ranger, u = 52 km/h = 14.44 m/s
Final speed of the ranger, v = 0 (as brakes are applied)
Acceleration of the ranger, 
Distance between deer and the vehicle, d = 87 m
Let d' is the distance covered by the deer so that it comes top rest. So,
d' = 26.06 m
Distance between the point where the deer stops and the vehicle is :
D=d-d'
D=87 - 26.06 = 60.94 m
Let t is the maximum reaction time allowed if the ranger is to avoid hitting the deer. It can be calculated as :
t = 0.23 seconds
Hence, this is the required solution.
Answer:
The velocity 
of the cart with respect to the ground is
if we consider North the positive y-direction and East the positive x-direction.
Explanation:
We have for relative motion the following expression:
Where is the velocity of the cart with respect to the ground, 
is the velocity of the cart with respect to the plane and 
is the velocity of the plane with respect to the ground.
We find that:
Thus:
Cups
teaspoon
tablespoon
liters
milliliters
gallons
pints
tons
inches
Answer:
(a) 42 N
(b)36.7 N
Explanation:
Nomenclature
F= force test line (N)
W : fish weight (N)
Problem development
(a) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled in at constant speed
We apply Newton's first law of equlibrio because the system moves at constant speed:
∑Fy =0
F-W= 0
42N -W =0
W = 42N
(b) Calculating of weight of the heaviest fish that can be pulled up vertically, when the line is reeled with an acceleration whose magnitude is 1.41 m/s²
We apply Newton's second law because the system moves at constant acceleration:
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
∑Fy =m*a
m= W/g , m= W/9.8 , m:fish mass , W: fish weight g:acceleration due to gravity
F-W= ( W/9.8 )*a
42-W= ( W/9.8 )*1.41
42= W+0.1439W
42=1.1439W
W= 42/1.1439
W= 36.7 N
