Answer:
201.6 N
Explanation:
m = mass of disk shaped merry-go-round = 125 kg
r = radius of the disk = 1.50 m
w₀ = Initial angular speed = 0 rad/s
w = final angular speed = 0.700 rev/s = (0.700) (2π) rad/s = 4.296 rad/s
t = time interval = 2 s
α = Angular acceleration
Using the equation
w = w₀ + α t
4.296 = 0 + 2α
α = 2.15 rad/s²
I = moment of inertia of merry-go-round
Moment of inertia of merry-go-round is given as
I = (0.5) m r² = (0.5) (125) (1.50)² = 140.625 kgm²
F = constant force applied
Torque equation for the merry-go-round is given as
r F = I α
(1.50) F = (140.625) (2.15)
F = 201.6 N
Answer:
If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops, then <u>the specific heats of both objects will be equal.</u>
Explanation:
If the temperature of the colder object rises by the same amount as the temperature of the hotter object drops when the two<u> objects of same mass</u> are brought into contact, then their specific heat capacity is equal.
<u>We can prove this by the equation of heat for the two bodies:</u>
<em>According to given condition,</em>


<em>when there is no heat loss from the system of two bodies then </em>


- Thermal conductivity is ultimately affects the rate of heat transfer, however the bodies will attain their final temperature based upon their mass and their specific heat capacities.
The temperature of the colder object will rise twice as much as the temperature of the hotter object only in two cases:
- when the specific heat of the colder object is half the specific heat of the hotter object while mass is equal for both.
OR
- the mass of colder object is half the mass of the hotter object while their specific heat is same.
Answer:
The required angular speed the neutron star is 10992.32 rad/s
Explanation:
Given the data in the question;
mass of the sun M
= 1.99 × 10³⁰ kg
Mass of the neutron star
M
= 2( M
)
M
= 2( 1.99 × 10³⁰ kg )
M
= ( 3.98 × 10³⁰ kg )
Radius of neutron star R
= 13.0 km = 13 × 10³ m
Now, let mass of a small object on the neutron star be m
angular speed be ω
.
During rotational motion, the gravitational force on the object supplies the necessary centripetal force.
GmM
= / R
² = mR
ω
²
ω
² = GM
= / R
³
ω
= √(GM
= / R
³)
we know that gravitational G = 6.67 × 10⁻¹¹ Nm²/kg²
we substitute
ω
= √( ( 6.67 × 10⁻¹¹ )( 3.98 × 10³⁰ ) ) / (13 × 10³ )³)
ω
= √( 2.65466 × 10²⁰ / 2.197 × 10¹²
ω
= √ 120831133.3636777
ω
= 10992.32 rad/s
Therefore, The required angular speed the neutron star is 10992.32 rad/s
You might want to go to this website, http://www.indiana.edu/~p1013447/dictionary/mendel.htm
Welcome, And i hope this helps :P