Answer:
The original speed of the mess kit is 5.65 m/s.
Explanation:
Given that,
Mass of kit =7.0 kg
Mass of first = 3.5 kg
Speed= 5.5 m/s
Speed of other particle part = 7.5 m/s
Angle = 30°
We need to calculate the momentum of first part
Using formula of momentum
Put the value into the formula
We need to calculate the momentum of second part
Using formula of momentum
We need to calculate the total momentum
We need to calculate the original speed of the mess kit
Using formula of momentum
Put the value into the formula
Hence, The original speed of the mess kit is 5.65 m/s.
Answer:
a= 0.22 m/s²
Explanation:
Given that
M = 3.5 kg
θ = 30°
m = 1 kg
μ= 0.3
The force due to gravity
F₁= M g sinθ
F₁=3.5 x 10 x sin 30
F₁= 17.5 N
F₂ = m g
F₂ = 1 x 10 = 10 N
The maximum value of the friction force on the incline plane
Fr = μ M g cosθ
Fr = 0.3 x 2.5 x 10 cos30°
Fr= 6.49 N
Lets take acceleration of the system is a m/s²
F₁ - F₂ - Fr = (M+m) a
17.5 - 10 - 6.49 = (3.5+1)a
a= 0.22 m/s²
Answer:
Well..
Explanation:
That's impossible. I know because I once weighed 11.1 kN, and I was temporarily immobile. It's probably the same for a car, and therefore it can not be "traveling" anywhere at all.. unless you put the car on an airplane or a boat or something.
Answer:
μ = 0.604
Explanation:
For the cat to stay in place on the merry go round, the maximum static frictional force must be equal in magnitude to that of the centripetal force.
Now, Centripetal force is given as;
Fc = mv²/r
Where r is radius and v is tangential speed and m is mass.
We also know that maximum static frictional force is given by;
F_static = μmg
Where μ is coefficient of friction
Now, equating both forces, we have;
mv²/r = μmg
Divide through by m;
v²/r = μg
Now, tangential speed can be expressed as;
v = circumference/period
Thus, v = 2πr/T
Where T is period of rotation and
2πr is the circumference of the merry go round.
Thus,
v²/r = μg is now;
(2πr/T)²/r = μg
Making μ the subject, we have;
(2πr/T)²/rg = μ
μ = [(2π x 5.4)/6]²/(5.4 x 9.8)
μ = 0.604
Answer:
a) w = 25.1 rad/s, b) θ = 0.9599 rad
, c) α = 328.1 rad/s² d) t= 0.0765 s
Explanation: Let's work on this exercise with the equations of angular kinematics
a) The angular velocity is
w = 4.00 rev / s (2π rad / 1 rev)
w = 25.1 rad/s
b) let's reduce the angle of degrees to radians
θ = 55 ° (π rad / 180 °)
θ = 0.9599 rad
c) Let's use the initial angular velocity as the system part of the rest is zero
w² = w₀² + 2 α θ
α = w² / 2 θ
α = 25.1²/2 0.9599
α = 328.1 rad / s²
d)
w = w₀ + α t
t = w / α
t = 25.1 / 328.1
t= 0.0765 s