Answer:
![y = 0.14 Cos\left ( 2.512x-34.66t \right )](https://tex.z-dn.net/?f=y%20%3D%200.14%20Cos%5Cleft%20%28%202.512x-34.66t%20%5Cright%20%29)
Explanation:
wavelength, λ = 2.5 m
speed, v = 13.8 m/s
Amplitude, A = 0.14 m
The general equation of the transverse harmonic wave which is travelling right is given by
![y = A Sin\left ( \frac{2\pi }{\lambda } (x - vt)+\phi \right )](https://tex.z-dn.net/?f=y%20%3D%20A%20Sin%5Cleft%20%28%20%5Cfrac%7B2%5Cpi%20%7D%7B%5Clambda%20%7D%20%28x%20-%20vt%29%2B%5Cphi%20%5Cright%20%20%29)
where, Ф is phase
At t = 0, x = 0 , y = 0.14 m
0.14 = 0.14 Sin Ф
Ф = π/2
So, the equation is
![y = 0.14Sin\left ( \frac{2\pi }{2.5 } (x - 13.8t)+\frac{\pi }{2} \right )](https://tex.z-dn.net/?f=y%20%3D%200.14Sin%5Cleft%20%28%20%5Cfrac%7B2%5Cpi%20%7D%7B2.5%20%7D%20%28x%20-%2013.8t%29%2B%5Cfrac%7B%5Cpi%20%7D%7B2%7D%20%5Cright%20%20%29)
![y = 0.14 Cos\left ( 2.512x-34.66t \right )](https://tex.z-dn.net/?f=y%20%3D%200.14%20Cos%5Cleft%20%28%202.512x-34.66t%20%5Cright%20%29)
The charge of the object must be ![1.11 \times e^{-5} \text { coulomb }](https://tex.z-dn.net/?f=1.11%20%5Ctimes%20e%5E%7B-5%7D%20%5Ctext%20%7B%20coulomb%20%7D)
Answer: Option C
<u>Explanation:</u>
Suppose an electric charge can be represented by the symbol Q. This electric charge generates an electric field; Because Q is the source of the electric field, we call this as source charge. The electric field strength of the source charge can be measured with any other charge anywhere in the area. The test charges used to test the field strength.
Its quantity indicated by the symbol q. In the electric field, q exerts an electric, either attractive or repulsive force. As usual, this force is indicated by the symbol F. The electric field’s magnitude is simply defined as the force per charge (q) on Q.
![Electric field, E=\frac{\text { Force }(F)}{q}](https://tex.z-dn.net/?f=Electric%20field%2C%20E%3D%5Cfrac%7B%5Ctext%20%7B%20Force%20%7D%28F%29%7D%7Bq%7D)
Here, given E = 4500 N/C and F = 0.05 N.
We need to find charge of the object (q)
By substituting the given values, we get
![q=\frac{F}{E}=\frac{0.05 N}{4500 \mathrm{N} / \mathrm{c}}=1.11 \times e^{-5} \text { coulomb }](https://tex.z-dn.net/?f=q%3D%5Cfrac%7BF%7D%7BE%7D%3D%5Cfrac%7B0.05%20N%7D%7B4500%20%5Cmathrm%7BN%7D%20%2F%20%5Cmathrm%7Bc%7D%7D%3D1.11%20%5Ctimes%20e%5E%7B-5%7D%20%5Ctext%20%7B%20coulomb%20%7D)
Explanation:
It is given that,
Mass of the tackler, m₁ = 120 kg
Velocity of tackler, u₁ = 3 m/s
Mass, m₂ = 91 kg
Velocity, u₂ = -7.5 m/s
We need to find the mutual velocity immediately the collision. It is the case of inelastic collision such that,
![v=\dfrac{m_1u_1+m_2u_2}{m_1+m_2}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7Bm_1u_1%2Bm_2u_2%7D%7Bm_1%2Bm_2%7D)
![v=\dfrac{120\ kg\times 3\ m/s+91\ kg\times (-7.5\ m/s)}{120\ kg+91\ kg}](https://tex.z-dn.net/?f=v%3D%5Cdfrac%7B120%5C%20kg%5Ctimes%203%5C%20m%2Fs%2B91%5C%20kg%5Ctimes%20%28-7.5%5C%20m%2Fs%29%7D%7B120%5C%20kg%2B91%5C%20kg%7D)
v = -1.5 m/s
Hence, their mutual velocity after the collision is 1.5 m/s and it is moving in the same direction as the halfback was moving initially. Hence, this is the required solution.
The movement of the earth is so small, especially compared to you, that you don't realize it. This is because the earth's mass is so much greater than your mass.
Answer is option C. nitrogen as:-
OXYGEN- 21%
NITROGEN- 78%
CARBON DIOXIDE, ARGON, NEON and SEVERAL OTHER GASES- 1%