Answer:
B = 0.8 T
Explanation:
It is given that,
Radius of circular loop, r = 0.75 m
Current in the loop, I = 3 A
The loop may be rotated about an axis that passes through the center and lies in the plane of the loop.
When the orientation of the normal to the loop with respect to the direction of the magnetic field is 25°, the torque on the coil is 1.8 Nm.
We need to find the magnitude of the uniform magnetic field exerting this torque on the loop. Torque acting on the loop is given by :
B is magnetic field
So, the magnitude of the uniform magnetic field exerting this torque on the loop is 0.8 T.
Antoine-Laurent Lavoisier was the first person to report the four element classification system but also ended up including some compounds rather than elements.
Explanation:
Let magnitude of the two forces be x and y.
Resultant at right angle R1= √15N) and at
60 degrees be R2= √18N.
Now, R1 = √(x² + y²) = √15,
R2= √(x² + y² +2xycos50) = √18.
So x² + y² = 15,
and x² + y² + 1.29xy = 18,
therefore 1.29xy = 3,
y = 3/1.29x.
y = 2.33/x
Now, x2 + (2.33/x)2 = 15,
x² + 5.45/x² = 15
multiply through by x²
x⁴ + 5.45 = 15x²
x⁴ - 15x2 + 5.45 = 0
Now find the roots of the equation, and later y. The two values of x will correspond to the
magnitudes of the two vectors.
Good luck
Answer:
Theta1 = 12° and theta2 = 168°
The solution procedure can be found in the attachment below.
Explanation:
The Range is the horizontal distance traveled by a projectile. This diatance is given mathematically by Vo cos(theta) t. Where t is the total time of flight of the projectile in air. It is the time taken for the projectile to go from starting point to finish point. This solution assumes the projectile finishes uts motion on the same horizontal level as the starting point and as a result the vertical displacement is zero (no change in height).
In the solution as can be found below, the expression to calculate the range for any launch angle theta was first derived and then the required angles calculated from the equation by substituting the values of the the given quantities.
When dealing with multiple forces acting on a body, it is advisable to draw a free-body diagram like that shown in the picture. There are four forces acting on the box: weight (W) pointing straight down, normal force perpendicular to the slope denoted as Fn, force used to push the box upwards along the slope and the frictional force acting opposite to the direction of motion of the box denoted as Ff. Frictional force is equal to coefficient of kinetic friction (μk) multiplied with Fn.
∑Fy = Fn - mgcos30° = 0
Fn = (50)(9.81)(cos 16) = 471.5 N
When in motion, the net force is equal to mass times acceleration according to Newton's 2nd Law of Motion:
Fnet = F - μk*Fn - mgsin30° = ma
250 - (0.2)(471.5 N) - (50)(sin 16°) = (50)(a)
a = 2.84 m/s²
