Answer:
F = 183.153 N
Explanation:
given,
mass of the toothpick = 0.12 g = 0.00012 kg
initial velocity = 227 m/s
final velocity = 0 m/s
penetration depth = 16 mm = 0.016 m
using the equation of motion
v² - u² = 2 a s
0 - u² = 2 a s
- 221² = 2 × a × 0.016
a = 1526281.25 m/s²
Force is equal to
F = m a
= 0.00012 × 1526281.25
F = 183.153 N
Displacement is a vector quantity. So, you incorporate the vector calculations when you try to determine the resultant vector. This is the shortest path from the starting point to the endpoint. If they are moving on one axis only, you use sign conventions. For motions moving to the left, use the negative sign. If it's moving to the right, then use the positive sign. Now, it the object moves 2 km to the left, and 2 km also to the right, the displacement is zero.
Displacement = 2 km - 2km = 0
Generally, the equation is:
<span>Displacement = Distance of motion to the right - Distance of motion to the left</span>
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
Answer:

Explanation:
d = Diameter of spot = 30 μm
r = Radius of spot = 
P = Power of the laser = 
A = Area = 
Intensity is given by

The light intensity within this spot is 
Answer:
hmmmmm ill get back later
Explanation: