To solve the problem it is necessary to use the concepts of Orbital Speed considering its density, and its angular displacement.
In general terms the Orbital speed is described as,
PART A) If the orbital speed of a star in this galaxy is constant at any radius, then,
PART B) This time we have
, where 
is the angular velocity (constant at this case)
PART C) If the total mass interior to any radius r is a constant,
Answer:
0.25
Explanation:
Magnification = image distance/object distance
mag = v/u.................. Equation 1
Given: f = -10 cm ( diverging lens) u = 30 cm.
Where can calculate for the value of v using
1/f = 1/u+1/v
make v the subject of the equation
v = fv/(u-f)..................... Equation 2
Substitute into equation 2
v = -30(10)/(30+10)
v = -300/40
v = -7.5 cm.
substituting into equation 1,
mag = 7.5/30
mag = 0.25
hence the magnification of the wretch = 0.25
Answer:
459.6J
Explanation:
Given parameters:
Angle of pull = 40°
Force applied = 30N
Distance moved = 20m
Unknown:
Work done by Kraig = ?
Solution:
To solve this problem;
Work done = F x dcosФ
d is the distance
F is the force
Ф is the angle given
Work done = 30 x 20cos40° = 459.6J
Answer:
The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight is approximately 19.81 m/s
Explanation:
The given parameters are;
The curvature of the hill, r = 200 m
Due to the velocity, v, the occupants weight = 20% less than the normal weight
The outward force of an object due to centripetal (motion) force is given by the following equation;
Where;
r = The radius of curvature of the hill = 200 m
Given that the weight of the occupants, W = m × g, we have;
v = √(0.2 × g × r)
By substitution, we have;
v = √(0.2 × 9.81 × 200) ≈ 19.81
The velocity of motion at which the occupants of the car appear to weigh 20% less than their normal weight ≈ 19.81 m/s.
Answer:
53.125m
Explanation:
The displacement of the car, denoted by S, can be calculated using the formula:
S = ut + 1/2at²
Where;
u = initial velocity/speed (m/s)
t = time (s)
a = acceleration (m/s²)
According to the information provided in this question, u = 10m/s, t = 5s, a = 0.25m/s², S = ?
S = ut + 1/2at²
S = (10 × 5) + 1/2 (0.25 × 5²)
S = 50 + 1/2 (0.25 × 25)
S = 50 + 1/2(6.25)
S = 50 + 3.125
S = 53.125m
