The amplitude of a wave corresponds to its maximum oscillation of the wave itself.
In our problem, the equation of the wave is
![y(x,t)= (0.750cm)cos(\pi [(0.400cm-1)x+(250s-1)t])](https://tex.z-dn.net/?f=y%28x%2Ct%29%3D%20%280.750cm%29cos%28%5Cpi%20%5B%280.400cm-1%29x%2B%28250s-1%29t%5D%29)
We can see that the maximum value of y(x,t) is reached when the cosine is equal to 1. When this condition occurs,
and therefore this value corresponds to the amplitude of the wave.
Answer:
1. E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Explanation:
According to the problem, Q is the charge on the non conducting sphere of radius R. Let ρ be the volume charge density of the non conducting sphere.
As shown in the figure, let r be the radius of the sphere inside the bigger non conducting sphere. Hence, the charge on the sphere of radius r is :
Q₁ = ∫ ρ dV
Here dV is the volume element of sphere of radius r.
Q₁ = ρ x 4π x ∫ r² dr
The limit of integration is from 0 to r as r is less than R.
Q₁ = (4π x ρ x r³ )/3
But volume charge density, ρ = 
So, 
Applying Gauss law of electrostatics ;
∫ E ds = Q₁/ε₀
Here E is electric field inside the sphere and ds is surface element of sphere of radius r.
Substitute the value of Q₁ in the above equation. Hence,
E x 4πr² = ( Q x r³) / ( R³ x ε₀ )
Matter either loses or absorbs energy when it changes from one state to another. For example, when matter changes from a liquid to a solid, it loses energy. The opposite happens when matter changes from a solid to a liquid. For a solid to change to a liquid, matter must absorb energy from its surroundings.
Answer:
The minimum coefficient of friction is 0.544
Solution:
As per the question:
Radius of the curve, R = 48 m
Speed of the car, v = 16 m/s
To calculate the minimum coefficient of static friction:
The centrifugal force on the box is in the outward direction and is given by:
where
= coefficient of static friction
The net force on the box is zero, since, the box is stationary and is given by:
