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A wave with a large amplitude
a wave check
Answer:
25.08m/s
Explanation:
mgh1 + 0.5mv1² = mgh2 + 0.5mv2²
h1 = 0m
v1 = u
h2 = 5m
v2 = 23m/s
putting the values into the formula above;
m(10)(0) + 0.5m(u²) = m(10)(5) + 0.5m(23²)
0 + 0.5mu² = 50m + 264.5m
0.5mu² = 314.5m
dividing through by m
0.5u² = 314.5
u² = 629
u = <u>2</u><u>5</u><u>.</u><u>0</u><u>8</u><u>m</u><u>/</u><u>s</u>
<u>Theref</u><u>ore</u><u>,</u><u> </u><u>the</u><u> </u><u>init</u><u>ial</u><u> </u><u>speed</u><u> </u><u>"</u><u>u</u><u>"</u><u> </u><u>=</u><u> </u><u>2</u><u>5</u><u>m</u><u>/</u><u>s</u>
If the two waves combine to produce ANY wave that smaller
than either of the originals, that's destructive interference.
Answer:
(a) m = 33.3 kg
(b) d = 150 m
(c) vf = 30 m/s
Explanation:
Newton's second law to the block:
∑F = m*a Formula (1)
∑F : algebraic sum of the forces in Newton (N)
m : mass s (kg)
a : acceleration (m/s²)
Data
F= 100 N
a= 3.0 m/s²
(a) Calculating of the mass of the block:
We replace dta in the formula (1)
F = m*a
100 = m*3
m = 100 / 3
m = 33.3 kg
Kinematic analysis
Because the block moves with uniformly accelerated movement we apply the following formulas:
d= v₀t+ (1/2)*a*t² Formula (2)
vf= v₀+a*t Formula (3)
Where:
d:displacement in meters (m)
t : time interval in seconds (s)
v₀: initial speed in m/s
vf: final speed in m/s
a: acceleration in m/s²
Data
a= 3.0 m/s²
v₀= 0
t = 10 s
(b) Distance the block will travel if the force is applied for 10 s
We replace dta in the formula (2):
d= v₀t+ (1/2)*a*t²
d = 0+ (1/2)*(3)*(10)²
d =150 m
(c) Calculate the speed of the block after the force has been applied for 10 s
We replace dta in the formula (3):
vf= v₀+a*t
vf= 0+(3*(10)
vf= 30 m/s
