Explanation:
c. if the vector is oriented at 0° from the X -axis.
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:
A. The electric field points to the left because the force on a negative charge is opposite to the direction of the field.
Explanation:
The electric force exerted on a charge by an electric field is given by:
where
F is the force
q is the charge
E is the electric field
We see that if the charge is negative, q contains a negative sign, so the force F and the electric field E will have opposite signs (which means they have opposite directions). This is due to the fact that the direction of the lines of an electric field shows the direction of the electric force experienced by a positive charge in that electric field: therefore, a negative charge will experience a force into opposite direction.
In order to form a real image using a concave mirror, the object must be placed beyond the center of curvature of the mirror. Therefore, the object must be further from the mirror than the focal point. The image that will form will be real, but it will also be inverted and its magnification will be less than 1, meaning it will be smaller than the actual object.
Absolute Zero, hopefully that helps :)
