A distorted image. :3
hope i help
Answer:
<em>t=14.96 sec</em>
Explanation:
<u>Diagonal Launch
</u>
It's a physical event that happens where an object is thrown in free air (no friction) forming an angle with the horizontal reference. The object then describes a path called a parabola.
The object will reach its maximum height and then return to the height from which it was launched. The equation for the height is
:
Where vo is the initial speed, is the angle, t is the time and g is the acceleration of gravity
.
In this problem we'll assume the arrow was launched from the ground level (won't consider the archer's height). Thus , and:
The value of y is zero twice: when t=0 (at launching time) and in t= when it goes back to the ground. We need to find that time by making
Dividing by
Then we find the total flight time as
In other words a infinitesimal segment dV caries the charge
<span>dQ = ρ dV </span>
<span>Let dV be a spherical shell between between r and (r + dr): </span>
<span>dV = (4π/3)·( (r + dr)² - r³ ) </span>
<span>= (4π/3)·( r³ + 3·r²·dr + 3·r·(dr)² + /dr)³ - r³ ) </span>
<span>= (4π/3)·( 3·r²·dr + 3·r·(dr)² + /dr)³ ) </span>
<span>drop higher order terms </span>
<span>= 4·π·r²·dr </span>
<span>To get total charge integrate over the whole volume of your object, i.e. </span>
<span>from ri to ra: </span>
<span>Q = ∫ dQ = ∫ ρ dV </span>
<span>= ∫ri→ra { (b/r)·4·π·r² } dr </span>
<span>= ∫ri→ra { 4·π·b·r } dr </span>
<span>= 2·π·b·( ra² - ri² ) </span>
<span>With given parameters: </span>
<span>Q = 2·π · 3µC/m²·( (6cm)² - (4cm)² ) </span>
<span>= 2·π · 3×10⁻⁶C/m²·( (6×10⁻²m)² - (4×10⁻²m)² ) </span>
<span>= 3.77×10⁻⁸C </span>
<span>= 37.7nC</span>
The box is kept in motion at constant velocity by a force of F=99 N. Constant velocity means there is no acceleration, so the resultant of the forces acting on the box is zero. Apart from the force F pushing the box, there is only another force acting on it in the horizontal direction: the frictional force
which acts in the opposite direction of the motion, so in the opposite direction of F.
Therefore, since the resultant of the two forces must be zero,
so
The frictional force can be rewritten as
where
,
. Re-arranging, we can solve this equation to find
, the coefficient of dynamic friction:
Answer:
L = 3.35 m
Explanation:
given,
mass of the block = 0.2 Kg
spring constant = 1.40 kN/m
until spring compressed = 10 cm
inclination of ramp = 60°
now,
initial potential energy of spring = (work done by friction) + (final gravitational potential energy of block)
L = 3.35 m