Answer:
9.6 m
Explanation:
This is a case of motion under variable acceleration . So no law of motion formula will be applicable here. We shall have to integrate the given equation .
a = 3.6 t + 5.6
d²x / dt² = 3.6 t + 5.6
Integrating on both sides
dx /dt = 3.6 t² / 2 + 5.6 t + c
where c is a constant.
dx /dt = 1.8 t² + 5.6 t + c
when t = 0 , velocity dx /dt is zero
Putting these values in the equation above
0 = 0 +0 + c
c = 0
dx /dt = 1.8 t² + 5.6 t
Again integrating on both sides
x = 1.8 t³ / 3 + 5.6 x t² /2 + c₁
x = 0.6 t³ + 2.8 t² + c₁
when t =0, x = 0
c₁ = 0
x = 0.6 t³ + 2.8 t²
when t = 1.6
x = .6 x 1.6³ + 2.8 x 1.6²
= 2.4576 + 7.168
= 9.6256
9.6 m
Answer:
D=387.28m
Explanation:
At the moment where the toss is made 
, so we need both equations:
For the red car:
With initial speed of 0 and acceleration of 6.12m/s^2.
For the green car:
With 
and Xo = 200m
Since both positions will be the same:
Solving for t:
t1 = -5.8s and t1 =11.25s
Replacing t = 11.25 on either equation to find the displacement:
When hard stabilization structures such as groins are used to stabilize a shoreline, the change in the longshore current results <u>deposition of sediment. </u>
On the upcurrent side of the barrier, sediment is deposited as the longshore current slows.
What is Hard stabilization?
- Hard stabilization is the prevention of erosion through the use of artificial barriers.
- Other hard stabilization structures, such as breakwaters and seawalls, are built parallel to the beach to protect the coast from the force of waves.
- Hard stabilization structures, such as groins, are built at right angles to the shore to prevent the movement of sand down the coast and maintain the beach.
- These constructions are made to last for many years, but because they detract from the visual splendor of the beach, they are not always the ideal answer.
- Additionally, they affect the habitats and breeding sites of native shoreline species, interfering with the ecosystem's natural processes.
Learn more about the Hard stabilization with the help of the given link:
brainly.com/question/16022736
#SPJ4
(a)
The formula is:
∑ F = Weight + T = mass * acceleration
as the elevator and lamp are moving downward, I choose downward forces to be
positive.
Weight is pulling down = +(9.8 * mass)
Tension is pulling up, so T = -63
Acceleration is upward = -1.7 m/s^2
(9.8 * mass) + -63 = mass * -1.7
Add +63 to both sides
Add (mass * 1.7) to both sides
(9.8 * mass) + (mass * 1.7) = 63
11.5 * mass = 63
mass = 63 / 11.5
Mass = 5.48 kg
(b)
Since the elevator and lamp are going upward, I choose upward forces to be
positive.
Weight is pulling down = -(9.8 * 5.48) = -53.70
Acceleration is upward, so acceleration = +1.7
-53.70 + T = 5.48 * 1.7
T = 53.70 + 9.316 = approx 63 N
The Tension is still the same - 63 N since the same mass, 5.48 kg, is being accelerated
upward at the same rate of 1.7 m/s^2
Answer: a) E= 6.63x10^-19J
E= 3.97×10^2KJ/mol
b) E = 3.31×10^-19J
E= 18.8×10^4 KJ/mol
C) E = 1.32×10^-33J
E= 8.01×10^-10KJ/mol
Explanation:
a) E = h ×f
h= planks constant= 6.626×10^-34
E=(6.626×10^-34)×(1.0×10^15)
E=6.63×10^-19J
1mole =6.02×10^23
E=( 6.63×10^-19)×(6.02×10^23)
E=3.97×10^2KJ/mol
b) E =(6.626×10^-34)/(1.0×10^15)
E=3.13×10^-19J
E= 3.13×10^-19) ×(6.02×10^23)
E= 18.8×10^3KJ/MOL
c) E= (6.626×10^-34) /0.5
E= 1.33×10^-33J
E= (1.33×10^-33) ×(6.02×10^23)
E= 8.01×10^-10KJ/mol
