eight points are in/on the circle of radius 10cm. show that distance between some two points is less than 1cm.
1 answer:
Here we have shown that the distance between some two points is less than 1cm, in the circle of radius of 10 cm.
considering the points be p₁,p₂,…,p₈, now placing the point p8 at the center of the circle and the other seven points on the circle's periphery, at equal distances. we can see we have seven points in the periphery and one single point in the center. after joining the seven points, we get a heptagon having its side equal in length. The length of the side of the heptagon is less than 2π/7, which is almost less than 1 cm. if we replace the position of any points come up with a pair of points that are apart from each from a length that is less than 1 cm
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Answer:
The shell hit at a distance of 1.9 x 10² km
The time of flight of the shell was 5.3 x 10² s
Explanation:
The position of the shell is given by the vector "r":
r = (x0 + v0 * t * cos α ; y0 + v0 * t * sin α + 1/2 g t²)
where:
x0 = initial horizontal position
v0 = magnitude of the initial velocity
t = time
α = launching angle
y0 = initial vertical position
g = acceleration of gravity
When the shell hit, the vertical component (ry) of the vector position r is 0. See figure.
Then:
ry = 0 = y0 + v0 * t * sin α + 1/2 g t²
Since the gun is at the center of our system of reference, y0 and x0 = 0
0 = t (v0 sin α + 1/2 g t)
t= 0 is discarded as solution
v0 sin α + 1/2 g t = 0
t = -2v0 sin α / g
t = (-2 * 2610 m/s * sin 81.9°)/ (-9.8 m/s²) = 5.3 x 10² s. This is the time of flight of the shell until it hit.
Then, the distance at which the shell hit is:
Distance = Module of r = ( x0 + v0 * t * cos α; 0) = x0 + v0 * t * cos α
Distance = 2.61 km/s * 5.3 x 10² s * cos 81.9 = 1.9 x 10² km
Answer:
a) 20 rad/s
b) 6 m/s
Explanation:
b) Force acting on the wheel is 200 N
mass of the wheel is 100 kg
From Newton's second law of motion, F = m × a
Where F is the net force acting on the body
m is mass of the body
a is the acceleration of the body
By substituting the values we get, a = 2 m/s²
As acceleration is constant, we can use the below formula for calculating the final velocity of the object
v = u + a × t
Where v is the final velocity
u is the initial velocity
a is the acceleration
t is the time taken
u = 0 (∵ it starts from rest)
By substituting the values we get
v = 0 + 2 × 3 = 6 m/s
∴ Speed of center of mass after 3 seconds = 6 m/s
a) As the wheel rotates about z-axis, radius of gyration will be the radius of wheel
∴ Radius of the wheel = 300 mm
Torque acting on the wheel about axis of rotation = 300 mm × 200 N =
60 N·m
Torque = (Moment of inertia) × (angular acceleration)
Assuming that the mass of spokes of the wheel to be negligible,
Moment of inertia of the wheel about axis of rotation = 100 × 300² × = 9 kg·m²
Then,
60 = 9 × (angular acceleration)
∴ angular acceleration ≈ 6·67 rad/s²
As angular acceleration of the wheel is constant, we can use the below formula for calculation of final angular speed
=
+ α × t
Where
is the final angular velocity
is the initial angular velocity
α is the angular acceleration
t is the time taken
is 0 (∵ initially it starts from rest)
By substituting the values we get
= 6·67 × 3 = 20 rad/s
∴ Angular velocity of the wheel after three seconds = 20 rad/s