Answer:
![F = 1.489*10^{-7} N](https://tex.z-dn.net/?f=F%20%3D%201.489%2A10%5E%7B-7%7D%20%20N)
Explanation: Weight of space probes on earth is given by:![W= m*g](https://tex.z-dn.net/?f=W%3D%20m%2Ag)
W= weight of the object( in N)
m= mass of the object (in kg)
g=acceleration due to gravity(9.81
)
Therefore,
![m_{1} = \frac{14500}{9.81}](https://tex.z-dn.net/?f=m_%7B1%7D%20%3D%20%5Cfrac%7B14500%7D%7B9.81%7D)
![m_{1} = 1478.08 kg](https://tex.z-dn.net/?f=m_%7B1%7D%20%3D%201478.08%20%20kg)
Similarly,
![m_{2} = \frac{4800}{9.81}](https://tex.z-dn.net/?f=m_%7B2%7D%20%3D%20%5Cfrac%7B4800%7D%7B9.81%7D)
![m_{2} = 489.29 kg](https://tex.z-dn.net/?f=m_%7B2%7D%20%3D%20489.29%20%20kg)
Now, considering these two parts as uniform spherical objects
Also, according to Superposition principle, gravitational net force experienced by an object is sum of all individual forces on the object.
Force between these two objects is given by:
![F = \frac{Gm_{1} m_{2}}{R^{2} }](https://tex.z-dn.net/?f=F%20%3D%20%20%5Cfrac%7BGm_%7B1%7D%20m_%7B2%7D%7D%7BR%5E%7B2%7D%20%7D)
G= gravitational constant (
)
= masses of the object
R= distance between their centres (in m)(18 m)
Substituiting all these values into the above formula
![F = 1.489*10^{-7} N](https://tex.z-dn.net/?f=F%20%3D%201.489%2A10%5E%7B-7%7D%20%20N)
This is the magnitude of force experienced by each part in the direction towards the other part, i.e the gravitational force is attractive in nature.
1. 1.59 s
The period of a pendulum is given by:
![T=2\pi \sqrt{\frac{L}{g}}](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7BL%7D%7Bg%7D%7D)
where L is the length of the pendulum and g the gravitational acceleration.
In this problem,
L = 0.625 m
g = 9.81 m/s^2
Substituting into the equation, we find
![T=2\pi \sqrt{\frac{(0.625 m)}{9.81 m/s^2}}=1.59 s](https://tex.z-dn.net/?f=T%3D2%5Cpi%20%5Csqrt%7B%5Cfrac%7B%280.625%20m%29%7D%7B9.81%20m%2Fs%5E2%7D%7D%3D1.59%20s)
2. 54,340 oscillations
The total number of seconds in a day is given by:
![t=24 h \cdot 60 min/h \cdot 60 s/min =86,400 s](https://tex.z-dn.net/?f=t%3D24%20h%20%5Ccdot%2060%20min%2Fh%20%5Ccdot%2060%20s%2Fmin%20%3D86%2C400%20s)
So in order to find the number of oscillations of the pendulum in one day, we just need to divide the total number of seconds per day by the period of one oscillation:
![N=\frac{t}{T}=\frac{86,400 s}{1.59 s}=54,340](https://tex.z-dn.net/?f=N%3D%5Cfrac%7Bt%7D%7BT%7D%3D%5Cfrac%7B86%2C400%20s%7D%7B1.59%20s%7D%3D54%2C340)
3. 0.842 m
We want to increase the period of the pendulum by 16%, so the new period must be
![T'=T+0.16T=1.16 T = 1.16 (1.59 s)=1.84 s](https://tex.z-dn.net/?f=T%27%3DT%2B0.16T%3D1.16%20T%20%3D%201.16%20%281.59%20s%29%3D1.84%20s)
Now we can re-arrange the equation for the period of the pendulum, using T=1.84 s, to find the new length of the pendulum that is required to produce this value of the period:
![L=g(\frac{T}{2\pi})^2=(9.81 m/s^2)(\frac{1.84 s}{2\pi})^2=0.842 m](https://tex.z-dn.net/?f=L%3Dg%28%5Cfrac%7BT%7D%7B2%5Cpi%7D%29%5E2%3D%289.81%20m%2Fs%5E2%29%28%5Cfrac%7B1.84%20s%7D%7B2%5Cpi%7D%29%5E2%3D0.842%20m)
Answer:
33% or 50% I hope that helped
Frequency heard by listener: fl = ?
frequency of source: fs = 300 Hz
velocity of listener: vl = 16 m/s
velocity of source: vs = 0
<span>velocity of sound in air: c = 344 m/s
</span><span>fl = fs [(c – vl)/c] = 300[(344-16)/344]
</span> =286.04 Hz
Answer:
3,295.348Joules
Explanation:
Potential Energy = mass * acceleration due to gravity * height
Given the following
Mass = 73.1kg
Acceleration due to gravity = 9.8m/s²
Distance (Height) = 4.6m
Substitute into the formula;
Potential Energy = 73.1 * 9.8 * 4.6
Potential Energy = 3,295.348Joules
Hence Tim gain 3,295.348Joules of energy