The equation for Kinetic Energy is KE = 1/2 m v^2, where m is the mass, and v is the velocity. The velocity in this equation is squared, which means that it is exponential. That means that as the velocity increases, you will be multiplying by a bigger and bigger number! KE = 18 joules!
Answer:
vi = 4.77 ft/s
Explanation:
Given:
- The radius of the surface R = 1.45 ft
- The Angle at which the the sphere leaves
- Initial velocity vi
- Final velocity vf
Find:
Determine the sphere's initial speed.
Solution:
- Newton's second law of motion in centripetal direction is given as:
m*g*cos(θ) - N = m*v^2 / R
Where, m: mass of sphere
g: Gravitational Acceleration
θ: Angle with the vertical
N: Normal contact force.
- The sphere leaves surface at θ = 34°. The Normal contact is N = 0. Then we have:
m*g*cos(θ) - 0 = m*vf^2 / R
g*cos(θ) = vf^2 / R
vf^2 = R*g*cos(θ)
vf^2 = 1.45*32.2*cos(34)
vf^2 = 38.708 ft/s
- Using conservation of energy for initial release point and point where sphere leaves cylinder:
ΔK.E = ΔP.E
0.5*m* ( vf^2 - vi^2 ) = m*g*(R - R*cos(θ))
( vf^2 - vi^2 ) = 2*g*R*( 1 - cos(θ))
vi^2 = vf^2 - 2*g*R*( 1 - cos(θ))
vi^2 = 38.708 - 2*32.2*1.45*(1-cos(34))
vi^2 = 22.744
vi = 4.77 ft/s
Answer:
It's constant everywhere in its trajectory.
Explanation:
the projectile was launched with an initial velocity, the only acceleration that is affecting the projectile's velocity is gravity.
The acceleration of gravity is practically equal everywhere on earth, so during its trajectory, we have to take into consideration only the acceleration because of gravity.
This is only correct because the projectile was launched with an initial velocity and it's not accelerating from rest and then falls.
Acceleration = final velocity - inital / time
a = 75-10 / 7
a = 65 / 7
a = 9.29 m/s^2
<span>The primary reason a light bulb emits light is due to the heating of the resistance in the filament of the light bulb. In fact, the power dissipated in a resistor is given by
</span>
<span>where I is the current and R the resistance. The larger the resistance or the current in the resistor, the larger the power dissipated. Due to this dissipation of power, the temperature of the filament becomes very high, and the resistance becomes incandescent, emitting light.</span>
