Answer:
![[F]=[MLT^{-2}]](https://tex.z-dn.net/?f=%5BF%5D%3D%5BMLT%5E%7B-2%7D%5D)
Explanation:
Newton’s second law states that the acceleration a of an object is proportional to the force F acting on it is inversely proportional to its mass m. The mathematical expression for the second law of motion is given by :
F = m × a
F is the applied force
m is the mass of the object
a is the acceleration due to gravity
We need to find the dimensions of force. The dimension of force m and a are as follows :
![[m]=[M]](https://tex.z-dn.net/?f=%5Bm%5D%3D%5BM%5D)
![[a]=[LT^{-2}]](https://tex.z-dn.net/?f=%5Ba%5D%3D%5BLT%5E%7B-2%7D%5D)
So, the dimension of force F is,
. Hence, this is the required solution.
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:
7.1 m/s
Explanation:
First, find the time it takes for the fish to reach the water.
Given in the y direction:
Δy = 6.1 m
v₀ = 0 m/s
a = 9.8 m/s²
Find: t
Δy = v₀ t + ½ at²
6.1 m = (0 m/s) t + ½ (9.8 m/s²) t²
t = 1.12 s
Next, find the velocity needed to travel 7.9 m in that time.
Given in the x direction:
Δx = 7.9 m
a = 0 m/s²
t = 1.12 s
Find: v₀
Δx = v₀ t + ½ at²
7.9 m = v₀ (1.12 s) + ½ (0 m/s²) (1.12 s)²
v₀ = 7.1 m/s
Answer: I think the answer is D.
Positively charged particles.
Answer:
v = 1.15*10^{7} m/s
Explanation:
given data:
charge/ unit area
plate seperation = 1.69*10^{-2} m
we know that
electric field btwn the plates is
force acting on charge is F = q E
Work done by charge q id
this work done is converted into kinectic enerrgy

solving for v



v = 1.15*10^{7} m/s