Answer:
![g_{moon}=1.67 [m/s^{2} ]](https://tex.z-dn.net/?f=g_%7Bmoon%7D%3D1.67%20%5Bm%2Fs%5E%7B2%7D%20%5D)
Explanation:
The weight of some mass is defined as the product of mass by gravitational acceleration. In this way using the following formula we can find the weight.
where:
w = weight [N]
m = mass = 0.06 [kg]
g = gravity acceleration = 10 [N/kg]
Therefore:
![w=0.06*10\\w=0.6[N]](https://tex.z-dn.net/?f=w%3D0.06%2A10%5C%5Cw%3D0.6%5BN%5D)
By Hooke's law we know that the force in a spring can be calculated by means of the following expression.
where:
k = spring constant [N/m]
x = deformed distance = 6 [cm] = 0.06 [m]
We can find the spring constant.
![k= F/x\\k=0.6/0.06\\k=10 [N/m]](https://tex.z-dn.net/?f=k%3D%20F%2Fx%5C%5Ck%3D0.6%2F0.06%5C%5Ck%3D10%20%5BN%2Fm%5D)
Since we use the same spring on the moon and the same mass, the constant of the spring does not change, the same goes for the mass.
![F_{moon}=k*0.01\\F = 10*0.01\\F=0.1[N]](https://tex.z-dn.net/?f=F_%7Bmoon%7D%3Dk%2A0.01%5C%5CF%20%3D%2010%2A0.01%5C%5CF%3D0.1%5BN%5D)
Since this force is equal to the weight, we can now determine the gravitational acceleration.
![F=m*g_{moon}\\g=F/m\\g = 0.1/0.06\\g_{moon} = 1.67[m/s^{2} ]](https://tex.z-dn.net/?f=F%3Dm%2Ag_%7Bmoon%7D%5C%5Cg%3DF%2Fm%5C%5Cg%20%3D%200.1%2F0.06%5C%5Cg_%7Bmoon%7D%20%3D%201.67%5Bm%2Fs%5E%7B2%7D%20%5D)
by the concept of momentum conservation we can say
if net force on a system of mass is ZERO then its momentum will remain conserved
Here a ball is projected upwards so if we take Ball + Earth as a system then total momentum of the system will remain conserved
Initially when ball is on the surface of earth the system has zero momentum and hence we can say after throwing the ball momentum of earth + ball must be zero
now using same equation we can say
given that
from above equation velocity of earth will be
so above will be the recoil speed of earth
<span>4.5 m/s
This is an exercise in centripetal force. The formula is
F = mv^2/r
where
m = mass
v = velocity
r = radius
Now to add a little extra twist to the fun, we're swinging in a vertical plane so gravity comes into effect. At the bottom of the swing, the force experienced is the F above plus the acceleration due to gravity, and at the top of the swing, the force experienced is the F above minus the acceleration due to gravity. I will assume you're capable of changing the velocity of the ball quickly so you don't break the string at the bottom of the loop.
Let's determine the force we get from gravity.
0.34 kg * 9.8 m/s^2 = 3.332 kg m/s^2 = 3.332 N
Since we're getting some help from gravity, the force that will break the string is 9.9 N + 3.332 N = 13.232 N
Plug known values into formula.
F = mv^2/r
13.232 kg m/s^2 = 0.34 kg V^2 / 0.52 m
6.88064 kg m^2/s^2 = 0.34 kg V^2
20.23717647 m^2/s^2 = V^2
4.498574938 m/s = V
Rounding to 2 significant figures gives 4.5 m/s
The actual obtainable velocity is likely to be much lower. You may handle 13.232 N at the top of the swing where gravity is helping to keep you from breaking the string, but at the bottom of the swing, you can only handle 6.568 N where gravity is working against you, making the string easier to break.</span>
Answer:
g = 11.2 m/s²
Explanation:
First, we will calculate the time period of the pendulum:
where,
T = Time period = ?
t = time taken = 135 s
n = no. of swings in given time = 98
Therefore,
T = 1.38 s
Now, we utilize the second formula for the time period of the simple pendulum, given as follows:
where,
l = length of pendulum = 54 cm = 0.54 m
g = acceleration due to gravity on the planet = ?
Therefore,
<u>g = 11.2 m/s²</u>
