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
<span>Igneous rocks which form by the crystallization of magma at a depth within the Earth are called intrusive rocks. Intrusive rocks are characterized by large crystal sizes, i.e., their visual appearance shows individual crystals interlocked together to form the rock mass. hope that helped</span>
(a) The momentum of the proton is determined as 5.17 x 10⁻¹⁸ kgm/s.
(b) The speed of the proton is determined as 3.1 x 10⁹ m/s.
<h3>
Momentum of the proton</h3>
The momentum of the proton is calculated as follows;
K.E = ¹/₂mv²
where;
- m is mass of proton = 1.67 x 10⁻²⁷ kg
- v is speed of the proton = ?
<h3>Speed of the proton</h3>
v² = 2K.E/m
v² = (2 x 50 x 10⁹ x 1.602 x 10⁻¹⁹ J)/(1.67 x 10⁻²⁷)
v² = 9.6 x 10¹⁸
v = 3.1 x 10⁹ m/s
<h3>Momentum of the proton</h3>
P = mv = (1.67 x10⁻²⁷ x 3.1 x 10⁹) = 5.17 x 10⁻¹⁸ kgm/s
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Answer:
Velocity = 4.33[m/s]
Explanation:
The total energy or mechanical energy is the sum of the potential energy plus the kinetic energy, as it is known the velocity and the height, we can determine the total energy.
All this energy will become kinetic energy and we can find the velocity.
A lean body mass that the cholesterol level in the body is low and that the excess fats in the body are getting rid off as much as one can. In this case, the exercises A.) cardiorespiratory fitness such as jogging and <span>C.) muscular strength and endurance such as body building, abs routines are employed. Answer is D. both A and C.</span>