Answer:
Explanation:
Given that,
Mass of star M(star) = 1.99×10^30kg
Gravitational constant G
G = 6.67×10^−11 N⋅m²/kg²
Diameter d = 25km
d = 25,000m
R = d/2 = 25,000/2
R = 12,500m
Weight w = 690N
Then, the person mass which is constant can be determined using
W =mg
m = W/g
m = 690/9.81
m = 70.34kg
The acceleration due to gravity on the surface of the neutron star is can be determined using
g(star) = GM(star)/R²
g(star) = 6.67×10^-11 × 1.99×10^30 / 12500²
g (star) = 8.49 × 10¹¹ m/s²
Then, the person weight on neutron star is
W = mg
Mass is constant, m = 70.34kg
W = 70.34 × 8.49 × 10¹¹
W = 5.98 × 10¹³ N
The weight of the person on neutron star is 5.98 × 10¹³ N
Assuming that the object starts at rest, we know the following values:
distance = 25m
acceleration = 9.81m/s^2 [down]
initial velocity = 0m/s
we want to find final velocity and we don't know the time it took, so we will use the kinematics equation without time in it:
Velocity final^2 = velocity initial^2 + 2 × acceleration × distance
Filling everythint in, we have:
Vf^2 = 0^2 + (2)(-9.81)(-25)
The reason why the values are negative is because they are going in the negative direction
Vf^2 = 490.5
Take the square root of that
Final velocity = 22.15m/s which is answer c
Answer: critical angle, sin^-1 (n2/n1)
Explanation: the angle of incidence at which the retracted ray makes an angle of 90° with the normal is known as the critical angle.
Snell's law defined refraction mathematically as shown below
n1 sin θi = n2 sin θr
n1 = refractive index of the first medium
n2 = refractive index of the second medium
θi = angle of incidence
θr = angle of refraction
When the refrafted ray is perpendicular to the normal, the angle of refraction (θr) is 90° hence making the angle of incidence (θi) the critical angle θc
By substituting these conditions into the Snell's law, we have that
n1 sin θc = n2 sin 90
According to trigonometry, the value of sin 90 is 1, hence we have that
n1 sin θc =n2
sin θc = n2/n1
θc = sin^-1 (n2/n1)
Answer:
center of mass = −0.50 m
Explanation:
given data
mass m1 = 3.04 kg
distance xm = -8 m
mass m2 = 5.61 kg
distance xM = 3.56 m
solution
we get here center of mass for n mass of system that is express as
center of mass = 
......................1
but we have only 2 particle system so we will get
center of mass = 
.................2
put here value and we will get
center of mass = 
solve it we will get
center of mass = −0.50 m
