Solve for the linear/tangential speed:
<em>a</em> = <em>v</em>²/<em>r</em>
where <em>a</em> = centripetal acceleration, <em>v</em> = speed, and <em>r</em> = radius.
4.7 m/s² = <em>v</em>²/(0.3 m)
<em>v</em>² = (0.3 m) (4.7 m/s²)
<em>v</em> ≈ 3.96 m/s
For every time the record completes one revolution, a fixed point on the edge of the record travels a distance equal to its circumference, which is 2<em>π</em> (0.3 m) ≈ 1.88 m. So if 1 rev ≈ 1.88 m, then the angular speed of the record is
(3.96 m/s) (1/1.88 rev/m) ≈ 7.46 rev/s
Take the reciprocal of this to get the period:
1 / (7.46 rev/s) ≈ 0.134 s/rev
So it takes the record about 0.134 seconds to complete one revolution.
Answer:
Explanation:
Let m be the mass of cylinder and r be the radius. It is moving with velocity v and angular velocity is ω. Let I be the moment of inertia of the cylinder.
I = 0.5 mr²
Total kinetic energy, T = 0.5 mv² + 0.5 Iω²
T = 0.5 (mv² + 0.5 mr²ω²)
v = rω
So, T = 0.5 (mv² + 0.5 mv²) = 0.75 mv²
Rotational kinetic energy is
R = 0.5 Iω² = 0.5 x 0.5 mr²ω²
R = 0.25 mv²
So, R / T = 0.25 / 0.75 = 1/3
Answer:
12.74 ms^-1 download
Explanation:
v=28.2, a=9.81
start from rest u=0
v=u+at=0+(9.81)t=28.2
t=2.875...
it reach 1.4 second before hitting the ground:
t=1.4, u=0, a=9.81
v=u+at=0+(9.81)(1.4)=12.74
Answer:
the correct result is r = 3.71 10⁸ m
Explanation:
For this exercise we will use the law of universal gravitation
F = 
We call the masses of the Earth M, the masses of the moon m and the masses of the rocket m ', let's set a reference system in the center of the Earth, the distance from the Earth to the moon is d = 3.84 108 m
rocket force -Earth
F₁ = - \frac{m' M }{r^2}
rocket force - Moon
F₂ = - \frac{m' m }{(d-r)^2}
in the problem ask for what point the force has the relation
2 F₁ = F₂
let's substitute
2
(d-r) ² =
r²
d² - 2rd + r² = \frac{m}{2M} r²
r² (1 -\frac{m}{2M}) - 2rd + d² = 0
Let's solve this quadratic equation to find the distance r, let's call
a = 1 - \frac{m}{2M}
a = 1 -
= 1 - 6.15 10⁻³
a = 0.99385
a r² - 2d r + d² = 0
r =
r = [2d ± 2d
] / 2a
r =
(1 ± √ (1.65 10⁻³)) =
(1 ± 0.04)
r₁ = \frac{d}{a} 1.04
r₂ = \frac{d}{a} 0.96
let's calculate
r₁ =
1.04
r₁ = 401.8 10⁸ m
r₂ = \frac{3.84 10^8}{0.99385} 0.96
r₂ = 3.71 10⁸ m
therefore the correct result is r = 3.71 10⁸ m
Right. You are true. The direction of the electric field is defined to be
the direction of the force on a small positive charge placed in the field.