Answer:
0.68 m
Explanation:
We know that the speed of sound in air is a product of frequency and wavelength. Taking speed of sound in air as 340 m/s
V=frequency*wavelength
Then wavelength is given by 350/500=0.68 m
Therefore, to repeat constructive interference at the listener's ear, a distance of 0.68 m is needed
Answer:
Option C. 5,000 kg m/s
Explanation:
<u>Linear Momentum on a System of Particles
</u>
Is defined as the sum of the momenta of each particles in a determined moment. The individual momentum is the product of the mass of the particle by its speed
P=mv
The question refers to an 100 kg object traveling at 50 m/s who collides with another object of 50 kg object initially at rest. We compute the moments of each object


The sum of the momenta of both objects prior to the collision is


Answer:
a) Osmolarity measures the moles of solute per liter of solution.
Explanation:
Osmolarity is defined as the number of moles of solute that contribute to the osmotic pressure, per liter of solution, of solution. That is, the measurement of the solute concentration. The prefix "osmo-" indicates the possible variation of the osmotic pressure in the cells, which will occur when the solution is introduced into the body.
Here we will the speed of seagull which is v = 9 m/s
this is the speed of seagull when there is no effect of wind on it
now in part a)
if effect of wind is in opposite direction then it travels 6 km in 20 min
so the average speed is given by the ratio of total distance and total time


now since effect of wind is in opposite direction then we can say



Part b)
now if bird travels in the same direction of wind then we will have


now we can find the time to go back



Part c)
Total time of round trip when wind is present


now when there is no wind total time is given by


So due to wind time will be more
PLS HELP ME AS QUICK AS POSSIBLE,
THANKS :)) I'm a bit confused
Can you answer 1 and 2, then confirm 3 :))))