Explanation:
c. if the vector is oriented at 0° from the X -axis.
Answer:
a) 145.6kgm^2
b) 158.4kg-m^2/s
c) 0.76rads/s
Explanation:
Complete qestion: a) the rotational inertia of the merry-go-round about its axis of rotation
(b) the magnitude of the angular momentum of the child, while running, about the axis of rotation of the merry-go-round and
(c) the angular speed of the merry-go-round and child after the child has jumped on.
a) From I = MK^2
I = (160Kg)(0.91m)^2
I = 145.6kgm^2
b) The magnitude of the angular momentum is given by:
L= r × p The raduis and momentum are perpendicular.
L = r × mc
L = (1.20m)(44.0kg)(3.0m/s)
L = 158.4kg-m^2/s
c) The total moment of inertia comprises of the merry- go - round and the child. the angular speed is given by:
L = Iw
158.4kgm^2/s = [145kgm^2 + ( 44.0kg)(1.20)^2]
w = 158.6/208.96
w = 0.76rad/s
Answer:
James Chadwick
Explanation:
In May 1932 James Chadwick announced that the core also contained a new uncharged particle, which he called the neutron
Answer: The height above the release point is 2.96 meters.
Explanation:
The acceleration of the ball is the gravitational acceleration in the y axis.
A = (0, -9.8m/s^)
For the velocity we can integrate over time and get:
V(t) = (9.20m/s*cos(69°), -9.8m/s^2*t + 9.20m/s^2*sin(69°))
for the position we can integrate it again over time, but this time we do not have any integration constant because the initial position of the ball will be (0,0)
P(t) = (9.20*cos(69°)*t, -4.9m/s^2*t^2 + 9.20m/s^2*sin(69°)*t)
now, the time at wich the horizontal displacement is 4.22 m will be:
4.22m = 9.20*cos(69°)*t
t = (4.22/ 9.20*cos(69°)) = 1.28s
Now we evaluate the y-position in this time:
h = -4.9m/s^2*(1.28s)^2 + 9.20m/s^2*sin(69°)*1.28s = 2.96m
The height above the release point is 2.96 meters.
Answer:
the required revolution per hour is 28.6849
Explanation:
Given the data in the question;
we know that the expression for the linear acceleration in terms of angular velocity is;
= rω²
ω² = / r
ω = √( / r )
where r is the radius of the cylinder
ω is the angular velocity
given that; the centripetal acceleration equal to the acceleration of gravity a = g = 9.8 m/s²
so, given that, diameter = 4.86 miles = 4.86 × 1609 = 7819.74 m
Radius r = Diameter / 2 = 7819.74 m / 2 = 3909.87 m
so we substitute
ω = √( 9.8 m/s² / 3909.87 m )
ω = √0.002506477 s²
ω = 0.0500647 ≈ 0.05 rad/s
we know that; 1 rad/s = 9.5493 revolution per minute
ω = 0.05 × 9.5493 RPM
ω = 0.478082 RPM
1 rpm = 60 rph
so
ω = 0.478082 × 60
ω = 28.6849 revolutions per hour
Therefore, the required revolution per hour is 28.6849
