Answer:
ax = 6.43m/s²
Explanation:
The acceleration is the time derivative of the velocity function ax = dvx(t)/dt
We have been given the velocity function v(t) and also the velocity v = 12.0m/s and we are requested to calculate the acceleration at this time which we don't know.
So the first step is to calculate the time at which the velocity =12.0m/s and with this time calculate the acceleration. Detailed solution can be found in the attachment below.
Answer:
The answer is 4N or B
Explanation:
Just the equation W = F x D.
We have W = 8 J and D = 2 m using algebra ....
8J/2m = F ... F = 4.
Answer:
Open- closed
Explanation:
It has Open-closed configuration of its end
Answer:
155.38424 K
2.2721 kg/m³
Explanation:
= Pressure at reservoir = 10 atm
= Temperature at reservoir = 300 K
= Pressure at exit = 1 atm
= Temperature at exit
= Mass-specific gas constant = 287 J/kgK
= Specific heat ratio = 1.4 for air
For isentropic flow
The temperature of the flow at the exit is 155.38424 K
From the ideal equation density is given by
The density of the flow at the exit is 2.2721 kg/m³
Answer:
In a tuning fork, two basic qualities of sound are considered, they are
1) The pitch of the waveform: This pitch depends on the frequency of the wave generated by hitting the tuning fork.
2) The loudness of the waveform: This loudness depends on the intensity of the wave generated by hitting the tuning fork.
Hitting the tuning fork harder will make it vibrate faster, increasing the number of vibrations per second. The number of vibration per second is proportional to the frequency, so hitting the tuning fork harder increase the frequency. From the explanation on the frequency above, we can say that by increasing the frequency the pitch of the tuning fork also increases.
Also, hitting the tuning fork harder also increases the intensity of the wave generated, since the fork now vibrates faster. This increases the loudness of the tuning fork.
