The skater's final angular speed is equal to 12 rad/s.
When implemented to angular momentum, the regulation of conservation means that the momentum of a rotating item is no longer exchanged until some form of external torque is carried out. Torque, in this sense, can check with any outside pressure that acts upon the object for the purpose to twist or rotate.
The law of conservation of angular momentum states that once no external torque acts on an item, no trade of angular momentum will occur. The angular momentum of a machine is conserved as long as there may be no net external torque performing on the machine.
In angular kinematics, the conservation of angular momentum refers back to the tendency of a device to keep its rotational momentum inside the absence of outside torque. For a round orbit, the system for angular momentum is (mass) ×(pace) ×(radius of the circle): (angular momentum) = m × v × r.
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In collision that are categorized as elastic, the total kinetic energy of the system is preserved such that,
KE1 = KE2
The kinetic energy of the system before the collision is solved below.
KE1 = (0.5)(25)(20)² + (0.5)(10g)(15)²
KE1 = 6125 g cm²/s²
This value should also be equal to KE2, which can be calculated using the conditions after the collision.
KE2 = 6125 g cm²/s² = (0.5)(10)(22.1)² + (0.5)(25)(x²)
The value of x from the equation is 17.16 cm/s.
Hence, the answer is 17.16 cm/s.
Answer:
Explanation:
Velocity of plane relative to ground V_pg = ?
Given the velocity in vector form ,
velocity of plane relative to air V_pw = 120 cos30 i + 120sin30j
V_wg = 60 i
V_pg = V_pw +V_wg
= 120 cos30 i + 120sin30j + 60i
= 164 i + 60 j
magnitude
=251 km / h
=
Answer:
Explanation:
In order to calculate the angular momentum of the particle you use the following formula:
(1)
r is the position vector respect to the point (0 , 5.0), that is:
r = 0m i + 5.0m j (2)
p is the linear momentum vector and it is given by:
(3)
the direction of p comes from the fat that the particle is moving along the i + j direction.
Then, you use the results of (2) and (3) in the equation (1) and solve for L:
The angular momentum is -30 kgm^2/s ^k
The answer is "156.6 m/s".
This is how we calculate this;
-N + mg = ma = mv²/r
For "weightlessness" N = 0, so
0 = mg - mv²/r
g - v²/r = 0
v =√( gr)
g = 9.8 and r = 2.5km = 2500 m
v = √(9.8 x 2500)
= 156.6 m/s
