Answer:
A) g = 9.751 m/s², B) h = 2.573 10⁴ m
Explanation:
The angular velocity of a pendulum is
w = √ g / L
Angular velocity and frequency are related.
w = 2π f
f = 1 / 2π √ g / L
A) with the initial data we can look for the pendulum length
L = 1 /4π² g / f²
L = 1 /4π² 9,800 / 0.3204²
L = 2.4181 m
The length of the pendulum does not change, let's look for the value of g for the new location
g = 4π² f² L
g = 4π² 0.3196² 2.4181
g = 9.75096 m / s²
g = 9.751 m/s²
B) The value of the acceleration of gravity can be found with the law of universal gravitation
F = G m M / 
²
And Newton's second law
W = m g
W = F
G m M / ² = mg
g = G M / ²
² = G M / g
Let's calculate
² = 6.67 10⁻¹¹ 5.98 10²⁴ /9.75096
R = √ 4.0905 10¹³ = √ 40.9053 10¹²
R = 6.395726 10⁶ m
The height above sea level is
h = R - [tex]R_{e}[/tex
h = (6.395726 -6.37) 10⁶
h = 0.0257256 106
h = 2.573 10⁴ m
Answer:
1.68 s
Explanation:
From newton's equation of motion,
a = (v-u)/t.................................. Equation 1
Making t the subject of the equation
t =(v-u)g............................. Equation 2
Where t = time taken for the bowling pin to reach the maximum height, v = final velocity bowling pin, u = initial velocity of the bowling pin, g = acceleration due to gravity.
Note: Taking upward to be negative and down ward to be positive,
Given: v = 0 m/s ( at the maximum height), u = 8.20 m/s, g = -9.8 m/s²
t = (0-8.20)/-9.8
t = -8.20/-9.8
t = 0.84 s.
But,
T = 2t
Where T = time taken for the bowling pin to return to the juggler's hand.
T = 2(0.84)
T = 1.68 s.
T = 1.68 s
Horizontal distance covered by a projectile is X = Vix *T
where Vix is the initial horizontal component of velocity and T is time taken by the projectile
Vix = ViCos theta
In question they said that initial velocity and angle is same on earth and moon
so Vix would remains same
now let's see about time taken T
time taken to reach the highest point
Vfy = Viy +gt
at highest point vertical velocity become zero so Vfy =0
0 = Vi Sin theta + gt
t = Vi Sintheta /g
Total time taken to land will be twice of that
On earth
Te= 2t
Te = 2Sinθ/g
on moon g is one-sixth of g(earth)
Tm = 2Sinθ/(g/6)
Tm = 6(2Sinθ/g)
Tm = 6Te
so total time taken by the projectile on moon will be six times the time taken on earth
From first equation X = Vix*T
we can see that X will also be 6 times on moon than earth
so projectile will cover 6 times distance on moon than on earth
