Answer:
22J
Explanation:
Given :
radius 'r'= 3cm
rotational inertia 'I'=4.5 x 
kgm²
mass on one side of rope '
'= 2kg
mass on other side of rope'
' =4kg
velocity'v' of mass ' = 2m/s
Angular velocity of the pulley is given by
ω = v /r => 2/ 3x 
ω = 66.67 rad/s
For the rotating body, we have
KE = 
I ω²
= 10J
Next is to calculate kinetic energy of the blocks :
=12J
Therefore, the total kinetic energy will be
KE = 
=10 + 12
KE= 22J
It's true IF ' m ' stands for mass and ' v ' stands for acceleration. Otherwise it's false.
Finding acceleration= final speed-initial speed/time taken (or A=V-U\T)
Finial speed= 27.8s
Initial speed= 0s
Time taken= 5.15
So..
27.8-0/5.15= 5.40m/s (rounded to two decimal places)
Answer:
354 m/s
Explanation:
For the second overtune (Third harmonic) of an open pipe,
λ = 2L/3................................ Equation 1
Where L = Length of the open pipe, λ = Wave length.
Given: L = 1.75 m.
Substitute into equation 1
λ = 2(1.75)/3
λ = 1.17 m.
From the question,
V = λf.......................... Equation 2
V = speed of sound in the room, f = frequency
Given: f = 303 Hz.
Substitute into equation 2
V = 1.17(303)
V = 353.5
V ≈ 354 m/s
Hence the right answer is 354 m/s
Answer:
138.3 days
Explanation:
Given that a Planet Ayanna has a radius of 6.2 X 10%m and orbits the star named Dayli in 98 days. A new neighboring planet Clayton J-21 has been discovered and has a radius of 7.8 X 10 meters.
The period of time for Clayton J-21 to orbit Dayli can be calculated by using Kepler law.
T^2 is proportional to r^3
That is,
T^2/r^3 = constant
98^2 / 62^3 = T^2 / 78^3
Make T^2 the subject of formula.
T^2 = 98^2 / 62^3 × 78^3
T^2 = 19123.2
T = sqrt ( 19123.2 )
T = 138.2867 days
Therefore, the period of time for Clayton J-21 to orbit Dayli is 138.3 days approximately.
