I am pretty sure that the<span> word which is modified by an adjectival phrase in the following sentence is </span>being revealed by the second options from the box represented above : b. project. I choose this one as a correct answer because it is clear that this word is <span>modified by the phrase :"to the project manager". Hope it will help!</span>
Answer:
a) 156960 N/m
b) Yes the player can play on the team.
Explanation:
F = Force
a = Acceleration due to gravity = 9.81 m/s²
m = Mass
k = Spring constant
x = Deformation length of spring
F = ma
From Hooke's law

The effective spring constant is 156960 N/m
When x = 0.48 cm = 0.0048 m

The mass of the player is 76.8 kg which is less than the required mass limit of the players' which is 85 kg. So, the player is eligible to play.
Answer:
at y=6.29 cm the charge of the two distribution will be equal.
Explanation:
Given:
linear charge density on the x-axis, 
linear charge density of the other charge distribution, 
Since both the linear charges are parallel and aligned by their centers hence we get the symmetric point along the y-axis where the electric fields will be equal.
Let the neural point be at x meters from the x-axis then the distance of that point from the y-axis will be (0.11-x) meters.
<u>we know, the electric field due to linear charge is given as:</u>

where:
linear charge density
r = radial distance from the center of wire
permittivity of free space
Therefore,





∴at y=6.29 cm the charge of the two distribution will be equal.
Answer:
a principle stating that energy cannot be created or destroyed, but can be altered from one form to another.
Explanation:
Answer:
45 s .
Explanation:
The accelerator will first accelerate , then move with uniform velocity and at last it will decelerate to rest .
displacement s = ?
acceleration a = 1 m /s²
Final speed v = 5 m/s
initial speed u = 0
v² = u² + 2as
5² = 0 + 2 x 1 x s
s = 12.5 m
B) Let time of acceleration or deceleration be t
v = u + a t
5 = 0 + 1 t
t = 5 s
Similarly displacement during deceleration = 12.5 m
Total distance during uniform motion = 200 - ( 12.5 + 12.5 ) = 175 m .
velocity of uniform motion = 5 m /s
time during which there was uniform velocity = 175 / 5 = 35 s
Total time = 5 + 35 + 5 = 45 s .