Answer:
C.The car appears to be moving 30 km/hr in the opposite direction of the bus.
Explanation:
There are two reference systems involved in this situation:
- Reference system S: this is the reference system where velocities are measured with respect to the ground. In this reference system, the car is parked, so its velocity is 
- Reference system S': this is the reference system moving with the bus. This reference system is moving with a velocity of 
with respect to the reference system S
Calling 
the velocity of the car in the reference system S', we have:
From which
and the negative sign means that a passanger in the bus observes the car moving in the opposite direction.
Answer:
<em>Elevator That Is Moving Downwards At A Constant Speed Of 4.9 M/S. What Is The Magnitude Of The Net Force Acing On The Student?</em>
<em>This problem has been solved!</em>
<em>This problem has been solved!See the answer</em>
<em>This problem has been solved!See the answerA student weighs 1200N. They are standing in an elevator that is moving downwards at a constant speed of </em><em>4.9 m/s. What is the magnitude of the net force acing on the student?</em>
The truth is both of them are more reactive metal.
The frequency of the human ear canal is 2.92 kHz.
Explanation:
As the ear canal is like a tube with open at one end, the wavelength of sound passing through this tube will propagate 4 times its length of the tube. So wavelength of the sound wave will be equal to four times the length of the tube. Then the frequency can be easily determined by finding the ratio of velocity of sound to wavelength. As the velocity of sound is given as 339 m/s, then the wavelength of the sound wave propagating through the ear canal is
Wavelength=4*Length of the ear canal
As length of the ear canal is given as 2.9 cm, it should be converted into meter as follows:
Then the frequency is determined as
f=c/λ=339/0.116=2922 Hz=2.92 kHz.
So, the frequency of the human ear canal is 2.92 kHz.
In order to compute the torque required, we may apply Newton's second law for circular motion:
Torque = moment of inertia * angular acceleration
For this, we require the angular acceleration, α. We may calculate this using:
α = Δω/Δt
The time taken to achieve rotational speed may be calculated using:
time = 1 revolution * 2π radians per revolution / 3.5 radians per second
time = 1.80 seconds
α = (3.5 - 0) / 1.8
α = 1.94 rad/s²
The moment of inertia of a thin disc is given by:
I = MR²/2
I = (0.21*0.1525²)/2
I = 0.002
τ = 1.94 * 0.002
τ = 0.004
The torque is 0.004
