Answer:
Of longitudinal waves
Explanation:
Depending on the direction of the oscillation, there are two types of waves:
- Transverse waves: in a transverse wave, the oscillations occur perpendicularly to the direction of propagation of the wave. Examples are electromagnetic waves.
- Longitudinal waves: in a longitudinal wave, the oscillations occur parallel to the direction of propagation of the wave. In such a wave, the oscillations are produced by alternating regions of higher density of particles, called compressions, and regions of lower density of particles, called rarefactions. Examples of longitudinal waves are sound waves.
Answer:
The attached diagram explains the system,
Sum of Fy = 0
N=9.81
N - mgCos60 = 0
F= ukN= (0.53)(9.81) =
F= 5.12 N
So
F.d= 1/2(mv.v) - mgdsin60
-5.12*0.5 = 0.5*v^2 - 2*(9.81)*(0.5*sin60)
(a) v = 2.436 m/s
For deflection
-F.x = 1/2(mv.v) - mgxsin60 + 1/2 (k*x*x)
by solving for with values of v, m, g, F, k
800x^2 - 11.87 x - 5.938 = 0
by solving the quadratic equation
x = 0.093, -0.079
(b) x = 0.093 m
correct Answer is 0.093m
Explanation:
Answer:
Wn = 9.14 x 10¹⁷ N
Explanation:
First we need to find our mass. For this purpose we use the following formula:
W = mg
m = W/g
where,
W = Weight = 675 N
g = Acceleration due to gravity on Surface of Earth = 9.8 m/s²
m = Mass = ?
Therefore,
m = (675 N)/(9.8 m/s²)
m = 68.88 kg
Now, we need to find the value of acceleration due to gravity on the surface of Neutron Star. For this purpose we use the following formula:
gn = (G)(Mn)/(Rn)²
where,
gn = acceleration due to gravity on surface of neutron star = ?
G = Universal Gravitational Constant = 6.67 x 10⁻¹¹ N.m²/kg²
Mn = Mass of Neutron Star = Mass of Sun = 1.99 x 10³⁰ kg
Rn = Radius of neutron Star = 20 km/2 = 10 km = 10000 m
Therefore,
gn = (6.67 x 10⁻¹¹ N.m²/kg²)(1.99 x 10³⁰ kg)/(10000)
gn = 13.27 x 10¹⁵ m/s²
Now, my weight on neutron star will be:
Wn = m(gn)
Wn = (68.88)(13.27 x 10¹⁵ m/s²)
<u>Wn = 9.14 x 10¹⁷ N</u>
I'm pretty sure it's B
I hope that helps