I believe it’s called Alluvium! It’s where the river mouth is build up of gravel,sand,silt, and clay!!
Answer:
865.08 m
Explanation:
From the question given above, the following data were obtained:
Initial velocity (u) = 243 m/s
Height (h) of the cliff = 62 m
Horizontal distance (s) =?
Next, we shall determine the time taken for the cannon to get to the ground. This can be obtained as follow:
Height (h) of the cliff = 62 m
Acceleration due to gravity (g) = 9.8 m/s²
Time (t) =?
h = ½gt²
62 = ½ × 9.8 × t²
62 = 4.9 × t²
Divide both side by 4.9
t² = 62/4.9
Take the square root of both side.
t = √(62/4.9)
t = 3.56 s
Finally, we shall determine the horizontal distance travelled by the cannon ball as shown below:
Initial velocity (u) = 243 m/s
Time (t) = 3.56 s
Horizontal distance (s) =?
s = ut
s = 243 × 3.56 s
s = 865.08 m
Thus, the cannon ball will impact the ground 865.08 m from the base of the cliff.
Answer:
a) x = 4.33 m
, b) w = 2 rad / s
, f = 0.318 Hz
, c) a = - 17.31 cm / s²,
d) T = 3.15 s, e) A = 5.0 cm
Explanation:
In this exercise on simple harmonic motion we are given the expression for motion
x = 5 cos (2t + π / 6)
they ask us for t = 0
a) the position of the particle
x = 5 cos (π / 6)
x = 4.33 m
remember angles are in radians
b) The general form of the equation is
x = A cos (w t + Ф)
when comparing the two equations
w = 2 rad / s
angular velocity and frequency are related
w = 2π f
f = w / 2π
f = 2 / 2pi
f = 0.318 Hz
c) the acceleration is defined by
a == d²x / dt²
a = - A w² cos (wt + Ф)
for t = 0
, we substitute
a = - 5,0 2² cos (π / 6)
a = - 17.31 cm / s²
d) El period is
T = 1/f
T= 1/0.318
T = 3.15 s
e) the amplitude
A = 5.0 cm
Answer:
160N
Explanation: When 80kg mass is one group . It's reaction force acting on a ground.
Weight of the object = 80*10
= 800 N
Here we are given cofficient of static friction its 0.2. It should be smaller than 1
Friction force = Reaction * Friction Cofficient
Reaction = 800N ( Considering Vertical Equilibrium )
F = 800* 0.2
F = 160N
Answer:
The half-life is 
Explanation:
Using the decay equation we have:

Where:
- λ is the decay constant
- A(0) the initial activity
- A is the activity at time t
We know the activity decrease by a factor of two in a one hour period (t = 1 h), it means that


Taking the natural logarithm on each side we have:


Now, the relationship between the decay constant λ and the half-life t(1/2) is:




I hope it helps you!