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Where A Particle Is Projected Along A Linear Plane. The Horizontal Component Of The Velocity Is Twice The Vertical Component What Is The Initial Magnitude Of The Velocity And The Angle At Which It Is?
Wiki User
∙ 9y ago
The initial velocity is sqrt(5) times the vertical component, and its angle relative to the horizontal direction, is 0.46 radians (26.6 degrees).
Wiki User
∙ 9y ago
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Can particle with constant velocity be accelerating?
No. Velocity implies both a magnitude and a direction.No. Velocity implies both a magnitude and a direction.No. Velocity implies both a magnitude and a direction.No. Velocity implies both a magnitude and a direction.
Path of a particle moving under influence of force fixed in magnitude and direction?
Parabolic.
Give an example in which the distance traversed by a particle is larger than the magnitude of its displacement in the same time?
This may happen when the particle moves back and forth.
Find the magnitude and direction of the resultant if forces of 10N 12N 6N act on a particle in the direction due west S30E N30E respectively?
Sorry we can't sketch a pictorial of the question. 1). The 10N force is all in the -x direction. 2). The 12N force has a component in the +x direction = 12sin(30)and a component in the -y direction = 12cos(30) 3). The 6N force has a component in the +x direction = 6sin(30)and a component in the +y direction = 6cos(30) The horizontal components are: -10 + 12sin(30) + 6sin(30) = -10 + 18sin(30) = -10 + 9 = -1N The vertical components are: -12cos(30) + 6cos(30) = -6cos(30) = -(6/2)sqrt(3) = -3sqrt(3) = -5.19615N The (magnitude)2 of the resultant = (-1)2 + (3sqrt(3))2 = 1 + 27 = 28The magnitude of the resultant = sqrt(28) = 5.2915N (rounded) The direction of the resultant = S [tan-1(1/5.19615)] W = S30W (just a little bit rounded)
If the acceleration of a particle is constant in magnitude but not in direction what type of path the particle follow?
You are mad 😂😂 you don't know this much easy answer 😆😆
A particle is projected at 60 degree to the horizontal with a kinetic energy k the kinetic energy at highest point is?
k/4
How is motion of charged particle parallel to unifrom electric field is equivalent to the body falling freely under gravity?
If the charge of the particle is such that it is attracted to the E-field then it's motion is analogous to gravitational free fall in a couple of ways. - Both forces have no horizontal component, therefore no horizontal acceleration. - Both forces interact at a magnitude inversely proportional to the square of their distance. - In both cases, Newton's third law applies, despite the nearly unmeasurable effects that the particle/body has on the field/earth.
What happens to a particle when heated?
The magnitude of the vibration of its molecules gets increased.
What is the relation of the time in the air to the height reached with a constant horizontal distance and applied force?
I am assuming that this is to do with the trajectory that is simplified to that of a particle which does not incur air resistance. If I have understood the question correctly, the particle travels under the influence of a constant force - assumed to be gravity which acts downwards. The model can be extended to allow for a constant force acting at an angle but the calculations then become more complicated. The particle is projected upwards, with the initial velocity, u ms-1, which makes an angle P with the horizontal. u is a variable such that the horizontal range of the particle is a constant. The vertical component of the initial velocity is u*sin(P) ms-1. The gravitational force, acting downwards, is -g ms-2. When the particle returns to the ground level, the vertical component of its velocity is -u*sin(P) ms-1. So if the particle returns at time t seconds, then t = [u*sin(P) - -u*sin(P)] /g = 2*u*sin(P)/g sec. The horizontal component of the velocity of the particle is a constant u*cos(P) ms-1. So during the time in flight, it travels u*cos(P)*2*u*sin(P)/g m = 2*u2*sin(P)*cos(P)/g m. This horizontal distance is constant, which implies that 2*u2*sin(P)*cos(P)/g is constant so that u2 is inversely proportional to sin(P)*cos(P). So let u = sqrt[k/sin(P)*cos(P)] ms-1 for some constant k. then its vertical component is u*sin(P) = sqrt[k/sin(P)*cos(P)]*sin(P) ms-1 = sqrt[k*tan(P)] Then at time T, its height is sqrt[k*tan(P)]*T - 0.5g*T2 I just hope this is correct!
Can particle with constant velocity be accelerating?
No. Velocity implies both a magnitude and a direction.No. Velocity implies both a magnitude and a direction.No. Velocity implies both a magnitude and a direction.No. Velocity implies both a magnitude and a direction.
What is the relationship between porosity bulk density and particle density?
If you were to graph particle size and porosity, it would be a constant slope (horizontal line).Porosity is not affected by particle size.
Path of a particle moving under influence of force fixed in magnitude and direction?
Parabolic.
Give an example in which the distance traversed by a particle is larger than the magnitude of its displacement in the same time?
This may happen when the particle moves back and forth.
Find the magnitude and direction of the resultant if forces of 10N 12N 6N act on a particle in the direction due west S30E N30E respectively?
Sorry we can't sketch a pictorial of the question. 1). The 10N force is all in the -x direction. 2). The 12N force has a component in the +x direction = 12sin(30)and a component in the -y direction = 12cos(30) 3). The 6N force has a component in the +x direction = 6sin(30)and a component in the +y direction = 6cos(30) The horizontal components are: -10 + 12sin(30) + 6sin(30) = -10 + 18sin(30) = -10 + 9 = -1N The vertical components are: -12cos(30) + 6cos(30) = -6cos(30) = -(6/2)sqrt(3) = -3sqrt(3) = -5.19615N The (magnitude)2 of the resultant = (-1)2 + (3sqrt(3))2 = 1 + 27 = 28The magnitude of the resultant = sqrt(28) = 5.2915N (rounded) The direction of the resultant = S [tan-1(1/5.19615)] W = S30W (just a little bit rounded)
If the acceleration of a particle is constant in magnitude but not in direction what type of path the particle follows?
You are mad 😂😂 you don't know this much easy answer 😆😆
If the acceleration of a particle is constant in magnitude but not in direction what type of path the particle follow?
You are mad 😂😂 you don't know this much easy answer 😆😆
How you calculate the kinetic energy of the particle projected vertically upwards?
Use the classic formula e = 1/2 x mv2
