Answer:
The runner's deceleration is -23.33
Given:
Initial speed = 3.5
Final speed = 0
Time taken = 0.15 s
To determine:
Deceleration of the runner =?
Used Formula:
Using the first equation of motion,
v = u + at
Where, v = final speed
u = initial speed
a = deceleration
t = duration
Solution:
<pusing the="" first="" equation="" of="" motion="">v = u + at
Where, v = final speed
u = initial speed
a = deceleration
t = duration
0 = 3.5 + a (0.15)
-3.5 = 0.15 (a)
a =
a = -23.33
The negative sign indicates that this represents deceleration.
Hence, the deceleration of the runner is -23.33
Answer
Given:
Wavelength = λ = 18.7 cm
= 0.187 m
Amplitude, A = 2.34 cm
Velocity, v = 0.38 m/s
A) Calculate the angular frequency.
Angular frequency,
ω = 2π f
ω = 2π x 2.03
ω = 12.75 rad/s
B) Calculate the wave number:
C)
Since the wave is traveling in the -x direction, the sign is positive between x and t
y (x, t) = A sin(k x - ω t)
y (x, t) = 2.34 sin(33.59 x - 12.75 t)
Answer:
Jari
Explanation:
To determine who is traveling faster, we need to evaluate their gradients. A steeper slope indicates a higher speed.
For Jari's path, starting point is (0, 0) and (6, 7) is another point.
The gradient is the difference in y divided by the difference in x:
Change in y=7-0=7
Change in x=6-0=6
Thus, the slope equals 7/6.
For Jade, her first point is (0, 10) and another is (6, 16).
Change in y=16-10=6
Change in x=6-0=6
Thus, the slope equals 6/6=1.
It's evident that 7/6 exceeds 6/6 or 1, proving Jari is quicker than Jade.
The entire and thorough solution is included.
Answer:
529.15 m/s
Explanation:
h = Highest point = 70000 m
g = Gravitational acceleration = 2 m/s²
m = Sulfur's mass
Since both potential and kinetic energies are conserved
The velocity at which the liquid sulfur exited the volcano is 529.15 m/s