Answer:
The number of charges is 1.25 × 10¹⁰
Explanation:
Current is the amount of charge flowing through a conductor per second. The formula for current (I) is given as:
I = Q/t
Where Q is the charge flowing in coulombs and t is the time taken in seconds.
Given that I = 2.0 nA = 2 × 10⁻⁹ A and t = 1 sec
I = Q / t
Q = It = 2 × 10⁻⁹ × 1 = 2 × 10⁻⁹ C
Since each charge = 1.6 x 10⁻¹⁹ C, therefore:
The number of charges = 2 × 10⁻⁹ C / 1.6 x 10⁻¹⁹ C = 1.25 × 10¹⁰
Answer:
The net force is 1.8N
Explanation:
Given that the formula for force is Force = mass×acceleration. So you have to substitute the values into the formula :
![force = mass \times acceleration](https://tex.z-dn.net/?f=force%20%3D%20mass%20%5Ctimes%20acceleration)
Let mass = 0.15kg,
Let acceleration = 12m/s²,
![force = 0.15 \times 12](https://tex.z-dn.net/?f=force%20%3D%200.15%20%5Ctimes%2012)
![force = 1.8](https://tex.z-dn.net/?f=force%20%3D%201.8)
The gravitational force between the spheres is
![F_{\rm g}=\dfrac{G(188\,\mathrm{kg})(0.93\,\mathrm{kg})}{(0.27\,\mathrm m)^2}\approx1.6\times10^{-7}\,\mathrm N](https://tex.z-dn.net/?f=F_%7B%5Crm%20g%7D%3D%5Cdfrac%7BG%28188%5C%2C%5Cmathrm%7Bkg%7D%29%280.93%5C%2C%5Cmathrm%7Bkg%7D%29%7D%7B%280.27%5C%2C%5Cmathrm%20m%29%5E2%7D%5Capprox1.6%5Ctimes10%5E%7B-7%7D%5C%2C%5Cmathrm%20N)
where <em>G</em> = 6.674 x 10⁻¹¹ N m²/kg².
The weight of the lighter sphere is
![F_{\rm w}=(0.93\,\mathrm{kg})g\approx9.1\,\mathrm N](https://tex.z-dn.net/?f=F_%7B%5Crm%20w%7D%3D%280.93%5C%2C%5Cmathrm%7Bkg%7D%29g%5Capprox9.1%5C%2C%5Cmathrm%20N)
where <em>g</em> = 9.80 m/s².
The ratio between the two forces is then
![\dfrac{F_{\rm g}}{F_{\rm w}}\approx1.8\times10^{-8}](https://tex.z-dn.net/?f=%5Cdfrac%7BF_%7B%5Crm%20g%7D%7D%7BF_%7B%5Crm%20w%7D%7D%5Capprox1.8%5Ctimes10%5E%7B-8%7D)
Answer:
0.4113772 s
Explanation:
Given the following :
Mass of bullet (m1) = 8g = 0.008kg
Initial horizontal Velocity (u1) = 280m/s
Mass of block (m2) = 0.992kg
Maxumum distance (x) = 15cm = 0.15m
Recall;
Period (T) = 2π√(m/k)
According to the law of conservation of momentum : (inelastic Collison)
m1 * u1 = (m1 + m2) * v
Where v is the final Velocity of the colliding bodies
0.008 * 280 = (0.008 + 0.992) * v
2.24 = 1 * v
v = 2.24m/s
K. E = P. E
K. E = 0.5mv^2
P.E = 0.5kx^2
0.5(0.992 + 0.008)*2.24^2 = 0.5*k*(0.15)^2
0.5*1*5.0176 = 0.5*k*0.0225
2.5088 = 0.01125k
k = 2.5088 / 0.01125
k = 223.00444 N/m
Therefore,
Period (T) = 2π√(m/k)
T = 2π√(0.992+0.008) / 233.0444
T = 2π√0.0042910
T = 2π * 0.0655059
T = 0.4113772 s
Answer:
20 Hz, 20000 Hz
0.0166 m, 16.6 m
Explanation:
The minimum frequency that a human ear can hear is 20 Hz
The maximum frequency that a human ear can hear is 20000 Hz.
v = Velocity of sound = 332 m/s
Wavelength is given by
![\lambda=\dfrac{v}{f}\\\Rightarrow \lambda=\dfrac{332}{20}\\\Rightarrow \lambda=16.6\ \text{m}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7Bv%7D%7Bf%7D%5C%5C%5CRightarrow%20%5Clambda%3D%5Cdfrac%7B332%7D%7B20%7D%5C%5C%5CRightarrow%20%5Clambda%3D16.6%5C%20%5Ctext%7Bm%7D)
The longest wavelength that can be heard by the human ear is 16.6 m
![\lambda=\dfrac{332}{20000}\\\Rightarrow \lambda=0.0166\ \text{m}](https://tex.z-dn.net/?f=%5Clambda%3D%5Cdfrac%7B332%7D%7B20000%7D%5C%5C%5CRightarrow%20%5Clambda%3D0.0166%5C%20%5Ctext%7Bm%7D)
The shortest wavelength that can be heard by the human ear is 0.0166 m.