Answer:
a.![\rm -1.49\ m/s^2.](https://tex.z-dn.net/?f=%5Crm%20-1.49%5C%20m%2Fs%5E2.)
b. ![\rm 50.49\ m.](https://tex.z-dn.net/?f=%5Crm%2050.49%5C%20m.)
Explanation:
<u>Given:</u>
- Velocity of the particle, v(t) = 3 cos(mt) = 3 cos (0.5t) .
<h2>
(a):</h2>
The acceleration of the particle at a time is defined as the rate of change of velocity of the particle at that time.
![\rm a = \dfrac{dv}{dt}\\=\dfrac{d}{dt}(3\cos(0.5\ t ))\\=3(-0.5\sin(0.5\ t.))\\=-1.5\sin(0.5\ t).](https://tex.z-dn.net/?f=%5Crm%20a%20%3D%20%5Cdfrac%7Bdv%7D%7Bdt%7D%5C%5C%3D%5Cdfrac%7Bd%7D%7Bdt%7D%283%5Ccos%280.5%5C%20t%20%29%29%5C%5C%3D3%28-0.5%5Csin%280.5%5C%20t.%29%29%5C%5C%3D-1.5%5Csin%280.5%5C%20t%29.)
At time t = 3 seconds,
![\rm a=-1.5\sin(0.5\times 3)=-1.49\ m/s^2.](https://tex.z-dn.net/?f=%5Crm%20a%3D-1.5%5Csin%280.5%5Ctimes%203%29%3D-1.49%5C%20m%2Fs%5E2.)
<u>Note</u>:<em> The arguments of the sine is calculated in unit of radian and not in degree.</em>
<h2>
(b):</h2>
The velocity of the particle at some is defined as the rate of change of the position of the particle.
![\rm v = \dfrac{dr}{dt}.\\\therefore dr = vdt\Rightarrow \int dr=\int v\ dt.](https://tex.z-dn.net/?f=%5Crm%20v%20%3D%20%5Cdfrac%7Bdr%7D%7Bdt%7D.%5C%5C%5Ctherefore%20dr%20%3D%20vdt%5CRightarrow%20%5Cint%20dr%3D%5Cint%20v%5C%20dt.)
For the time interval of 2 seconds,
![\rm \int\limits^2_0 dr=\int\limits^2_0 v\ dt\\r(t=2)-r(t=0)=\int\limits^2_0 3\cos(0.5\ t)\ dt](https://tex.z-dn.net/?f=%5Crm%20%5Cint%5Climits%5E2_0%20dr%3D%5Cint%5Climits%5E2_0%20v%5C%20dt%5C%5Cr%28t%3D2%29-r%28t%3D0%29%3D%5Cint%5Climits%5E2_0%203%5Ccos%280.5%5C%20t%29%5C%20dt)
The term of the left is the displacement of the particle in time interval of 2 seconds, therefore,
![\Delta r=3\ \left (\dfrac{\sin(0.5\ t)}{0.05} \right )\limits^2_0\\=3\ \left (\dfrac{\sin(0.5\times 2)-sin(0.5\times 0)}{0.05} \right )\\=3\ \left (\dfrac{\sin(1.0)}{0.05} \right )\\=50.49\ m.](https://tex.z-dn.net/?f=%5CDelta%20r%3D3%5C%20%5Cleft%20%28%5Cdfrac%7B%5Csin%280.5%5C%20t%29%7D%7B0.05%7D%20%5Cright%20%29%5Climits%5E2_0%5C%5C%3D3%5C%20%5Cleft%20%28%5Cdfrac%7B%5Csin%280.5%5Ctimes%202%29-sin%280.5%5Ctimes%200%29%7D%7B0.05%7D%20%5Cright%20%29%5C%5C%3D3%5C%20%5Cleft%20%28%5Cdfrac%7B%5Csin%281.0%29%7D%7B0.05%7D%20%5Cright%20%29%5C%5C%3D50.49%5C%20m.)
It is the displacement of the particle in 2 seconds.
Answer:
The force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Explanation:
F₂₁ = ![\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|}](https://tex.z-dn.net/?f=%5Cfrac%7BK%7Cq_1%7Cq_2%7C%7D%7Br%5E2%7D.%5Cfrac%7Br_2_1%7D%7B%7Cr_2_1%7C%7D)
Where;
F₂₁ is the vector force on q₁ due to q₂
K is the coulomb's constant = 8.99 X 10⁹ Nm²/C²
r₂₁ is the unit vector
|r₂₁| is the magnitude of the unit vector
|q₁| is the absolute charge on point charge one
|q₂| is the absolute charge on point charge two
r₂₁ = [(9-5)i +(7.4-(-4))j] = (4i + 11.5j)
|r₂₁| = ![\sqrt{(4^2)+(11.5^2)} = \sqrt{148.25}](https://tex.z-dn.net/?f=%5Csqrt%7B%284%5E2%29%2B%2811.5%5E2%29%7D%20%3D%20%5Csqrt%7B148.25%7D)
(|r₂₁|)² = 148.25
![F_2_1=\frac{K|q_1|q_2|}{r^2}.\frac{r_2_1}{|r_2_1|} = \frac{8.99X10^9(14X10^{-6})(60X10^{-6})}{148.25}.\frac{(4i + 11.5j)}{\sqrt{148.25} }](https://tex.z-dn.net/?f=F_2_1%3D%5Cfrac%7BK%7Cq_1%7Cq_2%7C%7D%7Br%5E2%7D.%5Cfrac%7Br_2_1%7D%7B%7Cr_2_1%7C%7D%20%3D%20%5Cfrac%7B8.99X10%5E9%2814X10%5E%7B-6%7D%29%2860X10%5E%7B-6%7D%29%7D%7B148.25%7D.%5Cfrac%7B%284i%20%2B%2011.5j%29%7D%7B%5Csqrt%7B148.25%7D%20%7D)
= 0.050938(0.19107i + 0.54933j) N
= (0.00973i + 0.02798j) N
Therefore, the force on q₁ due to q₂ is (0.00973i + 0.02798j) N
Before answering this question, first we have to understand the effect of ratio of surface area to volume on the rate of diffusion.
The rate of diffusion for a body having larger surface area as compared to the ratio of surface area to volume will be more than a body having less surface area. Mathematically it can written as-
V∝ R [ where v is the rate of diffusion and r is the ratio of surface area to volume]
As per the question,the ratio of surface area to volume for a sphere is given ![0.08m^{-1}](https://tex.z-dn.net/?f=0.08m%5E%7B-1%7D)
The surface area to volume ratio for right circular cylinder is given ![2.1m^{-1}](https://tex.z-dn.net/?f=2.1m%5E%7B-1%7D)
Hence, it is obvious that the ratio is more for right circular cylinder.As the rate diffusion is directly proportional to the surface area to volume ratio,hence rate of diffusion will be more for right circular cylinder.
Hence the correct option is B. The rate of diffusion would be faster for the right cylinder.
Answer:
Kinetic energy is energy possessed by a body by virtue of its movement. Potential energy is the energy possessed by a body by virtue of its position or state. While kinetic energy of an object is relative to the state of other objects in its environment, potential energy is completely independent of its environment.
Both energies are related to motion.
Explanation:
Fossil fuel are collected and turned into oils like car gas and many other products have fossil fuels in them and we don't even realize it idk oil was made out of fossils till i studied fossils in class Hope this helps:)