Answer:
0.31
Explanation:
horizontal force, F = 750 N
mass of crate, m = 250 kg
g = 9.8 m/s^2
The friction force becomes applied force = 750 N
According to the laws of friction,
Friction force = μ x Normal reaction of the surface
here, μ be the coefficient of friction
750 = μ x m g
750 = μ x 250 x 9.8
μ = 0.31
Thus, the coefficient of static friction is 0.31.
Answer: The magnitude of the current in the second wire 2.67A
Explanation:
Here is the complete question:
Two straight parallel wires are separated by 7.0 cm. There is a 2.0-A current flowing in the first wire. If the magnetic field strength is found to be zero between the two wires at a distance of 3.0 cm from the first wire, what is the magnitude of the current in the second wire?
Explanation: Please see the attachments below
Answer:
The refractive index of glass, 
Solution:
Brewster angle is the special case of incident angle that causes the reflected and refracted rays to be perpendicular to each other or that angle of incident which causes the complete polarization of the reflected ray.
To determine the refractive index of glass:
(1)
where
= refractive index of glass
= refractive index of glass
Now, using eqn (1)
First
let us imagine the projectile launched at initial velocity V and at angle
θ relative to the horizontal. (ignore wind resistance)
Vertical component y:
The
initial vertical velocity is given as Vsinθ
The moment the projectile reaches the maximum
height of h, the vertical velocity
will be 0, therefore the time t taken to attain this maximum height is:
h = Vsinθ - gt
0 = Vsinθ - gt
t = (Vsinθ)/g
where
g is acceleration due to gravity
Horizontal component x:
The initial horizontal velocity is given as Vcosθ. However unlike
the vertical component, this horizontal velocity remains constant because this is unaffected by gravity. The time to travel the
horizontal distance D is twice the value of t times the horizontal velocity.
D = Vcosθ*[(2Vsinθ)/g]
D = (2V²sinθ cosθ)/g
D = (V²sin2θ)/g
In order for D (horizontal distance) to be
maximum, dD/dθ = 0
That is,
2V^2 cos2θ / g = 0
And since 2V^2/g must not be equal to zero, therefore cos(2θ) = 0
This is true when 2θ = π/2 or θ = π/4
Therefore it is now<span> shown that the maximum horizontal travelled is attained when
the launch angle is π/4 radians, or 45°.</span>
