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
Tranquilizers
Because they are meant to put you at ease which may cause drowsiness and mental clouding
Answer:
70 + 50 = 120 km/hr
Explanation:
The driver of either car would see the other car approaching or departing at 120 kph
Answer:
hope this helps!
Explanation:
Volume of the air bubble, V1=1.0cm3=1.0×10−6m3
Bubble rises to height, d=40m
Temperature at a depth of 40 m, T1=12oC=285K
Temperature at the surface of the lake, T2=35oC=308K
The pressure on the surface of the lake: P2=1atm=1×1.103×105Pa
The pressure at the depth of 40 m: P1=1atm+dρg
Where,
ρ is the density of water =103kg/m3
g is the acceleration due to gravity =9.8m/s2
∴P1=1.103×105+40×103×9.8=493300Pa
We have T1P1V1=T2P2V2
Where, V2 is the volume of the air bubble when it reaches the surface.
V2=
Explanation:
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>A</u><u>:</u>
Let the x-axis be (+) towards the right and y-axis be (+) in the upward direction. We can write the net forces on mass
as


Substituting (2) into (1), we get

where
, the frictional force on
Set this aside for now and let's look at the forces on 
<u>Forces</u><u> </u><u>on</u><u> </u><u>Block</u><u> </u><u>B</u><u>:</u>
Let the x-axis be (+) up along the inclined plane. We can write the forces on
as


From (5), we can solve for <em>N</em> as

Set (6) aside for now. We will use this expression later. From (3), we can see that the tension<em> </em><em>T</em><em> </em> is given by

Substituting (7) into (4) we get

Collecting similar terms together, we get

or
![a = \left[ \dfrac{m_B\sin30 - \mu_km_A}{(m_A + m_B)} \right]g\:\:\:\:\:\:\:\:\:(8)](https://tex.z-dn.net/?f=a%20%3D%20%5Cleft%5B%20%5Cdfrac%7Bm_B%5Csin30%20-%20%5Cmu_km_A%7D%7B%28m_A%20%2B%20m_B%29%7D%20%5Cright%5Dg%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%5C%3A%288%29)
Putting in the numbers, we find that
. To find the tension <em>T</em>, put the value for the acceleration into (7) and we'll get
. To find the force exerted by the inclined plane on block B, put the numbers into (6) and you'll get 