Answer:
(a) A = 0.0800 m, λ = 20.9 m, f = 11.9 Hz
(b) 250 m/s
(c) 1250 N
(d) Positive x-direction
(e) 6.00 m/s
(f) 0.0365 m
Explanation:
(a) The standard form of the wave is:
y = A cos ((2πf) t ± (2π/λ) x)
where A is the amplitude, f is the frequency, and λ is the wavelength.
If the x term has a positive coefficient, the wave moves to the left.
If the x term has a negative coefficient, the wave moves to the right.
Therefore:
A = 0.0800 m
2π/λ = 0.300 m⁻¹
λ = 20.9 m
2πf = 75.0 rad/s
f = 11.9 Hz
(b) Velocity is wavelength times frequency.
v = λf
v = (20.9 m) (11.9 Hz)
v = 250 m/s
(c) The tension is:
T = v²ρ
where ρ is the mass per unit length.
T = (250 m/s)² (0.0200 kg/m)
T = 1250 N
(d) The x term has a negative coefficient, so the wave moves to the right (positive x-direction).
(e) The maximum transverse speed is Aω.
(0.0800 m) (75.0 rad/s)
6.00 m/s
(f) Plug in the values and find y.
y = (0.0800 m) cos((75.0 rad/s) (2.00 s) − (0.300 m⁻¹) (1.00 m))
y = 0.0365 m
Answer:
17.6 N
Explanation:
The force exerted by the punter on the football is equal to the rate of change of momentum of the football:
where
is the change in momentum of the football
is the time elapsed
The change in momentum can be written as
where
m = 0.55 kg is the mass of the football
u = 0 is the initial velocity (the ball starts from rest)
v = 8.0 m/s is the final velocity
Combining the two equations and substituting the values, we find the force exerted on the ball:
Answer:
the earth orbits a star called the sun
<h2>
The seagull's approximate height above the ground at the time the clam was dropped is 4 m</h2>
Explanation:
We have equation of motion s = ut + 0.5 at²
Initial velocity, u = 0 m/s
Acceleration, a = 9.81 m/s²
Time, t = 3 s
Substituting
s = ut + 0.5 at²
s = 0 x 3 + 0.5 x 9.81 x 3²
s = 44.145 m
The seagull's approximate height above the ground at the time the clam was dropped is 4 m
