The ball took half of the total time ... 4 seconds ... to reach its highest
point, where it began to fall back down to the point of release.
At its highest point, its velocity changed from upward to downward.
At that instant, its velocity was zero.
The acceleration of gravity is 9.8 m/s². That means that an object that's
acted on only by gravity gains 9.8 m/s of downward speed every second.
-- If the object is falling downward, it moves 9.8 m/s faster every second.
-- If the object is tossed upward, it moves 9.8 m/s slower every second.
The ball took 4 seconds to lose all of its upward speed. So it must have
been thrown upward at (4 x 9.8 m/s) = 39.2 m/s .
(That's about 87.7 mph straight up. Somebody had an amazing pitching arm.)
Answer:
As given in the problem statement
frequency=1 KHz=1*10^3 Hz
V(t) is represented as
v(t) = 5sin(2 \pi 1000t) + 0.05sin(2 \pi 3000t)
v ( t ) = 5 s i n ( 2 π 1000 t ) + 0.05 s i n ( 2 π 3000 t )
Total harmonic distortion will be 234 Pi
Answer:
Sarah is right
Explanation:
This is an exercise that differentiates between scalars and vectors.
A scalar is a number, instead a vector is a number that represents the module in addition to direction and sense.
In this case, the distance (scalar) traveled is a number, which is why it is worth 1500m, but the displacement is a vector and since the point where it leaves is the same point where the vector's modulus arrives is zero, so the DISPLACEMENT VECTOR is zero
consequently Sarah is right
Given Information:
Pendulum 1 mass = m₁ = 0.2 kg
Pendulum 2 mass = m₂ = 0.6 kg
Pendulum 1 length = L₁ = 5 m
Pendulum 2 length = L₂ = 1 m
Required Information:
Affect of mass on the frequency of the pendulum = ?
Answer:
The mass of the ball will not affect the frequency of the pendulum.
Explanation:
The relation between period and frequency of pendulum is given by
f = 1/T
The period of pendulum is given by
T = 2π√(L/g)
Where g is the acceleration due to gravity and L is the length of the string
As you can see the period (and frequency too) of pendulum is independent of the mass of the pendulum. Therefore, the mass of the ball will not affect the frequency of the pendulum.
Bonus:
Pendulum 1:
T₁ = 2π√(L₁/g)
T₁ = 2π√(5/9.8)
T₁ = 4.49 s
f₁ = 1/T₁
f₁ = 1/4.49
f₁ = 0.22 Hz
Pendulum 2:
T₂ = 2π√(L₂/g)
T₂ = 2π√(1/9.8)
T₂ = 2.0 s
f₂ = 1/T₂
f₂ = 1/2.0
f₂ = 0.5 Hz
So we can conclude that the higher length of the string increases the period of the pendulum and decreases the frequency of the pendulum.
