Answer:
a) T = 608.22 N
b) T = 608.22 N
c) T = 682.62 N
d) T = 533.82 N
Explanation:
Given that the mass of gymnast is m = 62.0 kg
Acceleration due to gravity is g = 9.81 m/s²
Thus; The weight of the gymnast is acting downwards and tension in the string acting upwards.
So;
To calculate the tension T in the rope if the gymnast hangs motionless on the rope; we have;
T = mg
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs the rope at a constant rate tension in the string is
= (62.0 kg)(9.81 m/s²)
= 608.22 N
When the gymnast climbs up the rope with an upward acceleration of magnitude
a = 1.2 m/s²
the tension in the string is T - mg = ma (Since acceleration a is upwards)
T = ma + mg
= m (a + g )
= (62.0 kg)(9.81 m/s² + 1.2 m/s²)
= (62.0 kg) (11.01 m/s²)
= 682.62 N
When the gymnast climbs up the rope with an downward acceleration of magnitude
a = 1.2 m/s² the tension in the string is mg - T = ma (Since acceleration a is downwards)
T = mg - ma
= m (g - a )
= (62.0 kg)(9.81 m/s² - 1.2 m/s²)
= (62.0 kg)(8.61 m/s²)
= 533.82 N
Answer:
c) At a distance greater than r
Explanation:
If G= Gravitational constant
M= Mass of earth
r= distance from earth center
then orbital speed is ;
v = 
==> v²=GM/r
If speed of first satellite = V₁
==> V₁² = GM/r
==> r = GM/V₁²
If speed of second satellite say V₂ is less than V₁ then square of V₂ will be less than square of V₁ , and hence GM will be divided by less number in case of second satellite, and hence will give greater value of r as compared to first satellite.
So our answer is c
Answer:
An incident ray is a ray of light that strikes a surface. The angle between this ray and the perpendicular or normal to the surface is the angle of incidence.
The reflected ray corresponding to a given incident ray, is the ray that represents the light reflected by the surface. The angle between the surface normal and the reflected ray is known as the angle of reflection. The Law of Reflection says that for a specular (non-scattering) surface, the angle of reflection always equals the angle of incidence.
If you have a string that is fixed on both ends the amplitude of the oscillation must be zero at the beginning and the end of the string. Take a look at the pictures I have attached. It is clear that our fundamental harmonic will have the wavelength of:
All the higher harmonics are just multiples of the fundamental:
Three longest wavelengths are:
<span>It is traveling at 10m/s.</span>
