Answer:
Waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.
Explanation:
To understand why high-frequency waves work better than low frequency waves for successful echolocation, first we have to understand the relation between frequency and wavelength.
The relation between frequency and wavelength is given by
λ = c/f
Where λ is wavelength, c is the speed of light and f is the frequency.
Since the speed of light is constant, the wavelength and frequency are inversely related.
So that means high frequency waves have shorter wavelengths, which is the very reason for the successful echolocation because waves having shorter wavelength are more likely to reach and hit the target and then reflect back to the dolphin to form an image of the object.
Thus, waves with high frequencies have shorter wavelengths that work better than low frequency waves for successful echolocation.
Answer:
75degree don't forget wind and gravity force pulling down
Answer:
The magnitude of magnetic field at given point = 
× 
T
Explanation:
Given :
Current passing through both wires = 5.0 A
Separation between both wires = 8.0 cm
We have to find magnetic field at a point which is 5 cm from any of wires.
From biot savert law,
We know the magnetic field due to long parallel wires.
⇒ 
Where 
magnetic field due to long wires, 
, 
perpendicular distance from wire to given point
From any one wire 
5 cm, 
3 cm
so we write,
∴ 
![B =\frac{ 4\pi \times10^{-7} \times5}{2\pi } [\frac{1}{0.03} + \frac{1}{0.05} ]](https://tex.z-dn.net/?f=B%20%3D%5Cfrac%7B%204%5Cpi%20%5Ctimes10%5E%7B-7%7D%20%5Ctimes5%7D%7B2%5Cpi%20%7D%20%5B%5Cfrac%7B1%7D%7B0.03%7D%20%2B%20%5Cfrac%7B1%7D%7B0.05%7D%20%5D)
Therefore, the magnitude of magnetic field at given point = 
Answer:
The vapor pressure at 60.6°C is 330.89 mmHg
Explanation:
Applying Clausius Clapeyron Equation
![ln(\frac{P_2}{P_1}) = \frac{\delta H}{R}[\frac{1}{T_1}- \frac{1}{T_2}]](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7BP_1%7D%29%20%3D%20%5Cfrac%7B%5Cdelta%20H%7D%7BR%7D%5B%5Cfrac%7B1%7D%7BT_1%7D-%20%5Cfrac%7B1%7D%7BT_2%7D%5D)
Where;
P₂ is the final vapor pressure of benzene = ?
P₁ is the initial vapor pressure of benzene = 40.1 mmHg
T₂ is the final temperature of benzene = 60.6°C = 333.6 K
T₁ is the initial temperature of benzene = 7.6°C = 280.6 K
ΔH is the molar heat of vaporization of benzene = 31.0 kJ/mol
R is gas rate = 8.314 J/mol.k
![ln(\frac{P_2}{40.1}) = \frac{31,000}{8.314}[\frac{1}{280.6}- \frac{1}{333.6}]\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.003564 - 0.002998)\\\\ln(\frac{P_2}{40.1}) = 3728.65 (0.000566)\\\\ln(\frac{P_2}{40.1}) = 2.1104\\\\\frac{P_2}{40.1} = e^{2.1104}\\\\\frac{P_2}{40.1} = 8.2515\\\\P_2 = (40.1*8.2515)mmHg = 330.89 mmHg](https://tex.z-dn.net/?f=ln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%20%5Cfrac%7B31%2C000%7D%7B8.314%7D%5B%5Cfrac%7B1%7D%7B280.6%7D-%20%5Cfrac%7B1%7D%7B333.6%7D%5D%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%280.003564%20-%200.002998%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%203728.65%20%20%280.000566%29%5C%5C%5C%5Cln%28%5Cfrac%7BP_2%7D%7B40.1%7D%29%20%3D%202.1104%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%20e%5E%7B2.1104%7D%5C%5C%5C%5C%5Cfrac%7BP_2%7D%7B40.1%7D%20%3D%208.2515%5C%5C%5C%5CP_2%20%3D%20%2840.1%2A8.2515%29mmHg%20%3D%20330.89%20mmHg)
Therefore, the vapor pressure at 60.6°C is 330.89 mmHg
For the second question you’re solving for resistance. resistance= voltage/ current. 120/0.5= 240. the answer is 240 ohms
for the third question you would do 2*4 since it’s asking for voltage, the answer is 8 volts :)
