Answer:
Explanation:
a )
initial velocity u = 45 m/s
acceleration a = - 5 m/s²
final velocity v = 0
v = u - at
0 = 45 - 5 t
t = 9 s
b )
s = ut - 1/2 at²
= 45 x 9 - .5 x5x 9²
405 - 202.5
202.5 m
2 )
a )
s = ut + 1/2 a t²
u = 0
s = 1/2 at²
= .5 x 9.54 x 6.5²
= 201.5 m
b )
v = u + at
= 0 + 9.54 x 6.5
= 62.01 m / s
3
a )
acceleration = (v - u) / t
= (34 - 42) / 2.4
= - 3.33 m /s²
b )
v² = u² - 2 a s
34² = 42² - 2 x 3.33² s
s = 27.41 m
c )
Average velocity
Total displacement / time
= 27.41 / 2.4
= 11.42 m /s
4 )
a )
v = u + at
v = 0 + 3 x 4
= 12 m /s
b )
s = ut + 1/2 a t²
= o + .5 x 3 x 4²
= 24 m
Answer:
Work done in both the cases will be same
Explanation:
As we know that the work done against gravity is given as

here we know that gravitational force is a conservative force and the work done against gravitational force is independent of the path
So here the work done by person to move the object between two different heights will be independent of the path they choose
So for the first person and second person will be same in both the cases because the height through which the boxes are transferred will be same in both the cases
Answer:
λ = 3 10⁻⁷ m, UV laser
Explanation:
The diffraction phenomenon is described by the expression
a sin θ = m λ
let's use trigonometry
tan θ = y / L
as in this phenomenon the angles are small
tan θ =
= sin θ
sin θ = y / L
we substitute
a y / L = m λ
let's apply this equation to the initial data
a 0.04 / L = 1 600 10⁻⁹
a / L = 1.5 10⁻⁵
now they tell us that we change the laser and we have y = 0.04 m for m = 2
a 0.04 / L = 2 λ
a / L = 50 λ
we solve the two expression is
1.5 10⁻⁵ = 50 λ
λ = 1.5 10⁻⁵ / 50
λ = 3 10⁻⁷ m
UV laser
Answer:
<em>1.228 x </em>
<em> mm </em>
<em></em>
Explanation:
diameter of aluminium bar D = 40 mm
diameter of hole d = 30 mm
compressive Load F = 180 kN = 180 x
N
modulus of elasticity E = 85 GN/m^2 = 85 x
Pa
length of bar L = 600 mm
length of hole = 100 mm
true length of bar = 600 - 100 = 500 mm
area of the bar A =
=
= 1256.8 mm^2
area of hole a =
=
= 549.85 mm^2
Total contraction of the bar =
total contraction =
==>
= <em>1.228 x </em>
<em> mm </em>
The value of the force, F₀, at equilibrium is equal to the horizontal
component of the tension in string 2.
Response:
- The value of F₀ so that string 1 remains vertical is approximately <u>0.377·M·g</u>
<h3>How can the equilibrium of forces be used to find the value of F₀?</h3>
Given:
The weight of the rod = The sum of the vertical forces in the strings
Therefore;
M·g = T₂·cos(37°) + T₁
The weight of the rod is at the middle.
Taking moment about point (2) gives;
M·g × L = T₁ × 2·L
Therefore;

Which gives;


F₀ = T₂·sin(37°)
Which gives;

<u />
Learn more about equilibrium of forces here:
brainly.com/question/6995192