A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point
2 answers:
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Answer:
P = 5.22 Kg.m/s
Explanation:
given,
mass of the projectile = 1.8 Kg
speed of the target = 4.8 m/s
angle of deflection = 60°
Speed after collision = 2.9 m/s
magnitude of momentum after collision = ?
initial momentum of the body = m x v
= 1.8 x 4.8 = 8.64 kg.m/s
final momentum after collision
momentum along x-direction
P_x = m v cos θ
P_x = 1.8 x 2.9 x cos 60°
P_x = 2.61 kg.m/s
momentum along y-direction
P_y = m v sin θ
P_y = 1.8 x 2.9 x sin 60°
P_y = 4.52 kg.m/s
net momentum of the body
P = 5.22 Kg.m/s
momentum magnitude after collision is equal to P = 5.22 Kg.m/s
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)
Explanation:
The equation for the moment of inertia of a sphere is:
(1)
Where:
is the moment of inertia of the planet (assumed with the shape of a sphere)
is the mass of the planet
is the radius of the planet
Isolating from (1):
(2)
Solving:
(3)
Finally:
Therefore, the correct option is A.