As you run toward a source of sound, you perceive the frequency of that sound to decrease.
<u>Explanation:</u>
Doppler's effect is a principle used to describe the frequency and the intensity of sound and wavelengths of a source and observer with the two possibilities.
(i) Stationary sound source and moving observer.
(ii) Moving sound source and a stationary observer. It is a relative motion.
Consider when the observer is moving towards a source, the frequency of the sound will be higher and when moving away from the source, the frequency will decrease.
Answer:
Water and power come from external sources.
Explanation:
Answer:
30 cm
Explanation:
To solve this problem, we use the lens equation:

where
f is the focal length of the lens
p is the distance of the object from the lens
q is the distance of the image from the lens
In this problem, we know that
q = -30 cm is the distance of the image from the lens; it is negative because the image is formed in front of the lens, so it is a virtual image
We also know that the size of the image is twice that of the object, so the magnification is 2:

where M is the magnification of the lens. Solving this equation for p,

So, the distance of the object from the lens is 15 cm.
Now we can finally solve the lens equation to find f, the focal length:

Answer: 43s
Explanation:
The stored energy across a plate can be measured by calculating the electric charge and the potential difference. Also, the quantity of electricity can be measured by multiplying the current in the circuit with the time taken.
Given
Potential difference, V = 120 v
Current, I = 11.5 A
Energy, E = 59 kJ
Remember,
E = QV, so that
Q = E/V
Q = 59000/120
Q = 491.67 C
Q = IT
491.67 = 11.5*t
t = 491.67/11.5
t = 42.75s ~ 43s
Time taken to cook the 3 hot dogs simultaneously is 43s
Answer:
v_max = (1/6)e^-1 a
Explanation:
You have the following equation for the instantaneous speed of a particle:
(1)
To find the expression for the maximum speed in terms of the acceleration "a", you first derivative v(t) respect to time t:
(2)
where you have use the derivative of a product.
Next, you equal the expression (2) to zero in order to calculate t:
![a[(1)e^{-6t}-6te^{-6t}]=0\\\\1-6t=0\\\\t=\frac{1}{6}](https://tex.z-dn.net/?f=a%5B%281%29e%5E%7B-6t%7D-6te%5E%7B-6t%7D%5D%3D0%5C%5C%5C%5C1-6t%3D0%5C%5C%5C%5Ct%3D%5Cfrac%7B1%7D%7B6%7D)
For t = 1/6 you obtain the maximum speed.
Then, you replace that value of t in the expression (1):

hence, the maximum speed is v_max = ((1/6)e^-1)a