Answer:
Option C
Explanation:
Hydrocarbon + Oxygen = Carbon dioxide + Water
is example of combustion reaction
Answer:
Answer: Kelvin ________________
Answer:
Explanation:
Given that,
Mass of counterweight m= 4kg
Radius of spool cylinder
R = 8cm = 0.08m
Mass of spool
M = 2kg
The system about the axle of the pulley is under the torque applied by the cord. At rest, the tension in the cord is balanced by the counterweight T = mg. If we choose the rotation axle towards a certain ~z, we should have:
Then we have,
τ(net) = R~ × T~
τ(net) = R~•i × mg•j
τ(net) = Rmg• k
τ(net) = 0.08 ×4 × 9.81
τ(net) = 3.139 Nm •k
The magnitude of the net torque is 3.139Nm
b. Taking into account rotation of the pulley and translation of the counterweight, the total angular momentum of the system is:
L~ = R~ × m~v + I~ω
L = mRv + MR v
L = (m + M)Rv
L = (4 + 2) × 0.08
L = 0.48 Kg.m
C. τ =dL/dt
mgR = (M + m)R dv/ dt
mgR = (M + m)R • a
a =mg/(m + M)
a =(4 × 9.81)/(4+2)
a = 6.54 m/s
Answer:
exoplanets is any planet beyond our solar system.
Answer:
Concepts and Principles
1- Kinetic Energy: The kinetic energy of an object is:
K=1/2*m*v^2 (1)
where m is the object's mass and v is its speed relative to the chosen coordinate system.
2- Gravitational potential energy of a system consisting of Earth and any object is:
U_g = -Gm_E*m_o/r*E-o (2)
where m_E is the mass of Earth (5.97x 10^24 kg), m_o is the mass of the object, and G = 6.67 x 10^-11 N m^2/kg^2 is Newton's gravitational constant.
Solution
The argument:
My friend thinks that escape speed should be greater for more massive objects than for less massive objects because the gravitational pull on a more massive object is greater than the gravitational pull for a less massive object and therefore the more massive object needs more speed to escape this gravitational pull.
The counterargument:
We provide a mathematical counterargument. Consider a projectile of mass m, leaving the surface of a planet with escape speed v. The projectile has a kinetic energy K given by Equation (1):
K=1/2*m*v^2 (1)
and a gravitational potential energy Ug given by Equation (2):
Ug = -G*Mm/R
where M is the mass of the planet and R is its radius. When the projectile reaches infinity, it stops and thus has no kinetic energy. It also has no potential energy because an infinite separation between two bodies is our zero-potential-energy configuration. Therefore, its total energy at infinity is zero. Applying the principle of energy consersation, we see that the total energy at the planet's surface must also have been zero:
K+U=0
1/2*m*v^2 + (-G*Mm/R) = 0
1/2*m*v^2 = G*Mm/R
1/2*v^2 = G*M/R
solving for v we get
v = √2G*M/R
so we see v does not depend on the mass of the projectile
