Answer:
time taken with speed 23 km/h will be 1.8 hours or 1 hour 48 minutes
Explanation:
Given:
Time is inversely proportional to the speed
mathematically,
t ∝ (1/r)
let the proportionality constant be 'k'
thus,
t = k/r
therefore, for case 1
time = 3 hr
speed = 14 km/hr
3 = k/14
also,
for case 2
let the time be = t
r = 23 km/h
thus,
we have
t = k/23
on dividing equation 2 by 1
we get
or
or
t = 1.8 hr = or 1 hour 48 minutes ( 0.8 hours × 60 minutes/hour = 48 minutes)
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
The moon's orbital and rotational periods are identical or the same, I<span>ts rate of spin is done in unison with its rate of revolution (the time that is needed to complete one orbit). Thus, the moon rotates exactly once every time it circles the Earth.</span>
Answer:
V=14.9 m/s
Explanation:
In order to solve this problem, we are going to use the formulas of parabolic motion.
The velocity X-component of the ball is given by:
The motion on the X axis is a constant velocity motion so:
The whole trajectory of the ball takes 1.48 seconds
We know that:
Knowing the X and Y components of the velocity, we can calculate its magnitude by:
This would be typical of an elastic collision.
