The equation of the path of a projectile is y = x(1 - x). Projectile motion is the movement of an object in a parabolic trajectory as a result of being propelled or released under the influence of gravity. A projectile, in simple terms, is any object that is launched into the air, such as a bullet, a baseball, or a rock. It's important to note that the parabolic trajectory is due to the force of gravity acting on the object.
To better understand projectile motion, we must first examine the horizontal and vertical components of motion. The horizontal component of motion is constant, indicating that there is no acceleration in that direction. The vertical component, on the other hand, has acceleration because of gravity. The parabolic trajectory is the result of these two components of motion.As the projectile is launched, it travels a certain distance horizontally before beginning to descend as a result of gravity's influence, resulting in a parabolic path.The general formula for the trajectory of a projectile in two dimensions is given by:y = xtanθ - (gx²) / 2(v₀cosθ)²Where:y is the vertical distance covered by the projectilex is the horizontal distance covered by the projectileθ is the angle of projectiong is the acceleration due to gravityv₀ is the initial velocity of the projectile In the case of the given equation, y = x(1 - x), the path of the projectile will be a parabolic trajectory with a vertex at x = 0.5 and y = 0.25. The equation represents the projectile's vertical distance, y, as a function of its horizontal distance, x. It is crucial to note that the projectile's initial velocity and angle of projection are not considered in this equation.For such more question on acceleration
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17. The time taken by a 100 g stone to reach the ground when dropped from the top of (A) your school building has been given. Write a suitable formula for the calculation of ght of your school building.
A new ride being built at an amusement park includes a vertical drop of 79.8 meters. Starting from rest, the ride vertically drops that distance before the track curves forward. If friction is neglected, what would be the speed of the roller coaster at the bottom of the drop?
Answer:
Approximately [tex]39.6\; {\rm m\cdot s^{-1}}[/tex] (assuming that [tex]g = 9.81\; {\rm m\cdot s^{-2}}[/tex].)
Explanation:
Under the assumptions, vertical acceleration of the vehicle during the ride would be equal to the gravitational field strength: [tex]a = g = 9.81\; {\rm m\cdot s^{-2}}[/tex].
Apply the following SUVAT equation to find the velocity of the vehicle at the bottom of the drop:
[tex]v^{2} - u^{2} = 2\, a\, x[/tex],
Where:
[tex]v[/tex] is the final velocity at the bottom of the drop;[tex]u[/tex] is the initial velocity at the beginning of the drop; [tex]u = 0\; {\rm m\cdot s^{-1}}[/tex] since the vehicle started from rest;[tex]a = g = 9.81\; {\rm m\cdot s^{-2}}[/tex] is the vertical acceleration of the vehicle during the drop;[tex]x = 79.8\; {\rm m}[/tex] is the vertical displacement of the vehicle during the drop.Rearrange this equation to find [tex]v[/tex]:
[tex]\begin{aligned}v &= \sqrt{u^{2} + 2\, a\, x} \\ &\approx \sqrt{0^{2} + 2\, (9.81)\, (79.8)} \; {\rm m\cdot s^{-1}} \\ &\approx 39.6\; {\rm m\cdot s^{-1}}\end{aligned}[/tex].
Hence, the speed of this vehicle at the bottom of the drop would be approximately [tex]39.6\; {\rm m\cdot s^{-1}}[/tex].
A rigid circular loop has a radius of 0.20 m and is in the xy-plane. The loop carries a
clockwise current I. The magnitude of the magnetic moment of the loop is 0.75 A.m2
.
A uniform external magnetic field, B = 0.20 T is directed parallel to the plane of the
loop. An external torque changes the orientation of the loop from one of lowest potential energy to one of highest potential energy. Calculate the work done by this external
torque.
A rigid circular loop having a radius of 0.20 m and is in the xy-plane, carries a uniform external magnetic field. B = 0.20 T and is directed parallel to the plane of the loop. The torque exerted on the loop is given by the formula as follows;τ = NABsin θ
Where τ is the torque, N is the number of turns, A is the area of the loop, B is the magnetic field strength, and θ is the angle between the normal to the loop and the magnetic field direction. The direction of the torque is given by the right-hand rule and depends on the direction of the magnetic field.To obtain the magnitude of the torque, we need to evaluate all the terms in the formula, τ = NABsinθ.A = πr² = π (0.20 m)² = 0.1257 m²The angle between the normal to the loop and the magnetic field direction is 0 degrees since the field is parallel to the loop.θ = 0°Substituting the values in the formula, we have;τ = NABsinθ= N (0.1257 m²) (0.20 T) (sin 0°)τ = 0 NmThe torque exerted on the loop is zero since the angle between the normal to the loop and the magnetic field direction is zero. The torque only exists when there is an angle between the normal and the magnetic field direction. Therefore, there is no net force on the current loop.For such more question on torque
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A 50.0-kg wagon is pulled with a constant force of 380 N. Neglecting friction, the wagon's acceleration will be
A 50.0-kg wagon is pulled with a constant force of 380 N. Neglecting friction, the wagon's acceleration will be 7.6 m/s².
When a constant force acts on a wagon, it causes the wagon to accelerate We can calculate the wagon's acceleration using Newton's second law, which states that the force acting on an object is directly proportional to its acceleration, and the acceleration is inversely proportional to the mass of the object.A = F/mHere,A = AccelerationF = Force acting on the wagon m = mass of the wagon Substituting the given values, we getA = 380 N/50.0 kgA = 7.6 m/s²Therefore, the acceleration of the wagon is 7.6 m/s² when it is pulled with a constant force of 380 N, neglecting friction.For such more question on Newton's second law
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