<span>If an electrostatically charged balloon is brought near other objects,
then ...
-- It will be attracted to objects of the opposite charge.
and
-- It will be attracted to objects with a neutral charge.</span>
The distance an object falls from rest through gravity is
D = (1/2) (g) (t²)
Distance = (1/2 acceleration of gravity) x (square of the falling time)
We want to see how the time will be affected
if ' D ' doesn't change but ' g ' does.
So I'm going to start by rearranging the equation
to solve for ' t '. D = (1/2) (g) (t²)
Multiply each side by 2 : 2 D = g t²
Divide each side by ' g ' : 2 D/g = t²
Square root each side: t = √ (2D/g)
Looking at the equation now, we can see what happens to ' t ' when only ' g ' changes:
-- ' g ' is in the denominator; so bigger 'g' ==> shorter 't'
and smaller 'g' ==> longer 't' .--
They don't change by the same factor, because 1/g is inside the square root. So 't' changes the same amount as √1/g does.
Gravity on the surface of the moon is roughly 1/6 the value of gravity on the surface of the Earth.
So we expect ' t ' to increase by √6 = 2.45 times.
It would take the same bottle (2.45 x 4.95) = 12.12 seconds to roll off the same window sill and fall 120 meters down to the surface of the Moon.
Answer:
<em>The velocity of sound in the guitar string = 129.99 m/s</em>
Explanation:
Harmonics: A harmonic is a note with frequency equal to an integral multiple of that of the fundamental frequency.
<em>Note:</em> The harmonics of a string is the same as the harmonics of an open pipe.
f₆ = 3v/l ............................ Equation 1
making v the subject of the equation,
v = f₆l/3 ......................... Equation 2
Where f₆ = Is the sixth harmonics of the the string, v = velocity of sound in the guitar string, l = length of the guitar string.
<em>Given: </em>f₆ = 557.14 Hz, l = 70 cm = 0.7 m.
Substituting into equation 2
v = (557.14×0.7)/3
<em>v = 129.99 m/s</em>
<em>Thus the velocity of sound in the guitar string = 129.99 m/s</em>
Answer:
Angular momentum = 0.7 kg.m²/s
Angular velocity = 583.3 rad/s
Explanation:
1. The torque τ is related to the angular momentum L by the relation
τ = ΔL/Δt
ΔL = τΔt
τ = 10 N. m
Δt = 70 ms = 70 × 10⁻³s
ΔL = (10 N. m) × (70 × 10⁻³s) = 700 × 10⁻³ kg.m²/s = 0.7 kg.m²/s
2. The rotational inertia I relates the angular momentum L to the angular velocity w
L = Iw
w = L/I
L = 0.7 kg.m²/s
I = 1.2 × 10⁻³ kg.m²
w = (0.7 kg.m²/s)/(1.2 × 10⁻³ kg.m²) = 583.3 rad/s
Answer:
the diameter of the second pipe is 2.52 in
Explanation:
Given the data in the question;
We know that; the rate of flow is the same;
so
Av1 = Av2
v ∝ √h
= 
= √( 
)
( π/4.D1² / π/4.D2² ) = √( )
( D1² / D2² ) = √( 
) since second is double of first
so
( D1² / D2² ) = √( 
)
3² / D2² = √2
D2²√2 = 9
D2² = 9/√2
D2² = 9 / 1.4142
D2² = 6.364
D2 = √ 6.364
D2 = 2.52 in
Therefore, the diameter of the second pipe is 2.52 in
