Answer:
The speed the bat is gaining on its prey is 0.03m/s
Explanation:
Given;
speed of the bat, v₀ = 3.7 m/s
frequency of the bat, F₀ = 36 kHz
frequency of the source, Fs = 36.79
This is relative motion between a source of the sound and the observer. The phenomenon is known as Doppler effect.
Apply the following equation to determine the speed of the insect which is the source;
![F_0 = F_s[\frac{v+v_0}{v-v_s} ]\\\\\frac{F_0}{F_s} = [\frac{v+v_0}{v-v_s} ]\\\\\frac{36.79}{36} = \frac{340+3.7}{340-v_s}\\\\1.0219 = \frac{343.7}{340-v_s}\\\\ 340-v_s = \frac{343.7}{1.0219}\\\\340-v_s = 336.33\\\\v_s = 340-336.33\\\\v_s = 3.67 \ m/s](https://tex.z-dn.net/?f=F_0%20%3D%20F_s%5B%5Cfrac%7Bv%2Bv_0%7D%7Bv-v_s%7D%20%5D%5C%5C%5C%5C%5Cfrac%7BF_0%7D%7BF_s%7D%20%3D%20%5B%5Cfrac%7Bv%2Bv_0%7D%7Bv-v_s%7D%20%5D%5C%5C%5C%5C%5Cfrac%7B36.79%7D%7B36%7D%20%3D%20%5Cfrac%7B340%2B3.7%7D%7B340-v_s%7D%5C%5C%5C%5C1.0219%20%3D%20%5Cfrac%7B343.7%7D%7B340-v_s%7D%5C%5C%5C%5C%20%20340-v_s%20%3D%20%5Cfrac%7B343.7%7D%7B1.0219%7D%5C%5C%5C%5C340-v_s%20%3D%20336.33%5C%5C%5C%5Cv_s%20%3D%20340-336.33%5C%5C%5C%5Cv_s%20%3D%203.67%20%5C%20m%2Fs)
The speed the bat is gaining on its prey = 3.7m/s - 3.67m/s = 0.03 m/s
Therefore, the speed the bat is gaining on its prey is 0.03m/s
Answer:
Explanation:
1.)66.36
2.)11.4
3.)0.8104
4.)4158.315
i got some of them and try the rest good luck
As the gas cools down, particle movement slows down with it, and so does the temperature of the gas. This is because the more heat or thermal energy a gas has, the faster the particles move. This is why absolute zero, or 0 degrees Kelvin is defined as zero particle movement, because it has zero energy.
Hope this helps
GPE I am assuming is gravitational potential energy. I'll denote it as U for simplicity.
U = mgy
U = (70kg)(9.81m/s^2)(1m) = 686.7J
U = 686.7J
Hope this helps!
Answer:
The equation that will relate all the given parameters, in other to calculate the potential energy of charge A is:
∆V = ∆U/q, ∆V is potential at charge A position, q is magnitude of charge A, ∆U will be made the subject of the relation, which is the Potential Energy of charge A. The notation "∆" show, the quantities have both in values and final values, in the electric field.(Change in Electric potential and potential energy, due to the effect of the field)
Explanation:
The potential energy of a charged particle (Charge A) in an electric field depends on the magnitude of the charge(Known as stated in the question). However, the potential energy per unit charge has a unique value at any point in the electric field.
