Answer:
Explanation:
Givens
The frequency is defined by
Where 
is the speed of the wave in the string and 
is its wave length.
The wave length is defined as 
Now, to find the speed, we need the tension of the wire and its linear mass density
Where 
and the tension is defined as 
Replacing this value, the speed is
Then, we replace the speed and the wave length in the first equation
Therefore, the frequency is
Answer:
6858.5712 m/s
Explanation:
Given that:
Radius, r
R = 3.20 * 10^3.
Normal force = 0.5 * normal weight
Normal force = Fn ; Normal weight = Fg
Fn = 0.5Fg
Recall:
mv² / R = Fn + Fg
Fn = 0.5Fg
mv² / R = 0.5Fg + Fg
mv² /R = 1.5Fg
mv² = 1.5Fg * R
F = mg
mv² = 1.5* mg * R
v² = 1.5gR
v = sqrt(1.5gR)
V = sqrt(1.5 * 9.8 * 3.2 * 10^3)
V = sqrt(47.04^3)
V = 6858.5712 m/s
For vertical motion, use the following kinematics equation:
H(t) = X + Vt + 0.5At²
H(t) is the height of the ball at any point in time t for t ≥ 0s
X is the initial height
V is the initial vertical velocity
A is the constant vertical acceleration
Given values:
X = 1.4m
V = 0m/s (starting from free fall)
A = -9.81m/s² (downward acceleration due to gravity near the earth's surface)
Plug in these values to get H(t):
H(t) = 1.4 + 0t - 4.905t²
H(t) = 1.4 - 4.905t²
We want to calculate when the ball hits the ground, i.e. find a time t when H(t) = 0m, so let us substitute H(t) = 0 into the equation and solve for t:
1.4 - 4.905t² = 0
4.905t² = 1.4
t² = 0.2854
t = ±0.5342s
Reject t = -0.5342s because this doesn't make sense within the context of the problem (we only let t ≥ 0s for the ball's motion H(t))
t = 0.53s
Answer:
An apple in free fall accelerates toward the Earth with a free fall acceleration, g. The force of the apple on the Earth also causes the Earth to accelerate toward the falling apple. By Newton's Third Law, the force of the Earth on the apple is exactly equal and opposite to the force of the apple on the Earth. By Newton,s Second law, the force of the Earth on the apple is equal to the mass of the apple times g , the accelerations due to gravity. And, the force of the the apple on the Earth is equal to the mass of the Earth times the acceleration of the Earth toward the apple. In conclusion, the magnitude of the forces are equal, or
F ( apple on the Earth) = F( the Earth on the apple) or
M( mass of the earth) x a( the acceleration of the earth toward the apple) = m(mass of the apple) x g( the acceleration of the apple toward the Earth) or
a = (m/M) g
Explanation:
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