Litmus paper and phenolphthalein
Answer:
factors that affect the force of gravity between objects include their distance apart and their mass.
Explanation:
Solution :
Given two resistors :
and
where
> ![$R_1$](https://tex.z-dn.net/?f=%24R_1%24)
When the resistors are connected in series :
Current,
.............(1)
When the resistors are connected in series :
![$10I_s=\frac{\epsilon}{\left(\frac{R_1R_2}{R_1+R_2}\right)}$](https://tex.z-dn.net/?f=%2410I_s%3D%5Cfrac%7B%5Cepsilon%7D%7B%5Cleft%28%5Cfrac%7BR_1R_2%7D%7BR_1%2BR_2%7D%5Cright%29%7D%24)
..................(2)
Therefore, dividing equation (2) by (1), we get
![$10=\frac{(R_1+R_2)^2}{R_1R_2}$](https://tex.z-dn.net/?f=%2410%3D%5Cfrac%7B%28R_1%2BR_2%29%5E2%7D%7BR_1R_2%7D%24)
Now, since
, we have ![$R_1=rR_2$](https://tex.z-dn.net/?f=%24R_1%3DrR_2%24)
∴ ![$10=\frac{(1+r)^2}{r}$](https://tex.z-dn.net/?f=%2410%3D%5Cfrac%7B%281%2Br%29%5E2%7D%7Br%7D%24)
![$\Rightarrow 1+r^2+2r=10r$](https://tex.z-dn.net/?f=%24%5CRightarrow%201%2Br%5E2%2B2r%3D10r%24)
![$\Rightarrow r^2-8r+1=0$](https://tex.z-dn.net/?f=%24%5CRightarrow%20r%5E2-8r%2B1%3D0%24)
Solving, we get, r = 0.127
Answer:
The angular frequency of the block is ω = 5.64 rad/s
Explanation:
The speed of the block v = rω where r = amplitude of the oscillation and ω = angular frequency of the oscillation.
Now ω = v/r since v = speed of the block = 62 cm/s and r = the amplitude of the oscillation = 11 cm.
The angular frequency of the oscillation ω is
ω = v/r
ω = 62 cm/s ÷ 11 cm
ω = 5.64 rad/s
So, the angular frequency of the block is ω = 5.64 rad/s
Solution :
From the Newton's second law of motion :
F = ma
![a=\frac{F}{m}](https://tex.z-dn.net/?f=a%3D%5Cfrac%7BF%7D%7Bm%7D)
![$\frac{dv}{dt}=\frac{F}{m}$](https://tex.z-dn.net/?f=%24%5Cfrac%7Bdv%7D%7Bdt%7D%3D%5Cfrac%7BF%7D%7Bm%7D%24)
![$\left(\frac{dv}{ds} \times \frac{ds}{dt}\right)=\frac{F}{m}$](https://tex.z-dn.net/?f=%24%5Cleft%28%5Cfrac%7Bdv%7D%7Bds%7D%20%5Ctimes%20%5Cfrac%7Bds%7D%7Bdt%7D%5Cright%29%3D%5Cfrac%7BF%7D%7Bm%7D%24)
![$v \frac{dv}{ds} = \frac{F}{m}$](https://tex.z-dn.net/?f=%24v%20%5Cfrac%7Bdv%7D%7Bds%7D%20%3D%20%5Cfrac%7BF%7D%7Bm%7D%24)
![$v dv =\frac{F}{m}\ ds$](https://tex.z-dn.net/?f=%24v%20dv%20%3D%5Cfrac%7BF%7D%7Bm%7D%5C%20ds%24)
Integrating above the expression by applying the limits :
![$\int_{v_i}^{v_f} v \ dv= \frac{F}{m} \int_0^s ds$](https://tex.z-dn.net/?f=%24%5Cint_%7Bv_i%7D%5E%7Bv_f%7D%20v%20%5C%20dv%3D%20%5Cfrac%7BF%7D%7Bm%7D%20%5Cint_0%5Es%20ds%24)
Here the diameter is s= D
![$\frac{v_f^2 - v_i^2}{2}=\frac{FD}{m}$](https://tex.z-dn.net/?f=%24%5Cfrac%7Bv_f%5E2%20-%20v_i%5E2%7D%7B2%7D%3D%5Cfrac%7BFD%7D%7Bm%7D%24)
The final speed of the particle after travelling distance D is
![$v_f = \sqrt{v_i^2 + \frac{2FD}{m}}$](https://tex.z-dn.net/?f=%24v_f%20%3D%20%5Csqrt%7Bv_i%5E2%20%2B%20%5Cfrac%7B2FD%7D%7Bm%7D%7D%24)
The kinetic energy of the particle of mass M is :
![$K_1=\frac{1}{2}Mv^2$](https://tex.z-dn.net/?f=%24K_1%3D%5Cfrac%7B1%7D%7B2%7DMv%5E2%24)
For M = 3M
![$K_2=\frac{1}{2}(3M)v^2$](https://tex.z-dn.net/?f=%24K_2%3D%5Cfrac%7B1%7D%7B2%7D%283M%29v%5E2%24)
![$=3(K_1)$](https://tex.z-dn.net/?f=%24%3D3%28K_1%29%24)
Thus the kinetic energy increases by a factor of 3.
The work done depends on the factor and the displacement of the body. Thus, the work done remains same even though the mass increases. Hence the work down increases by factor 1.