Answer:
576 joules
Explanation:
From the question we are given the following:
weight = 810 N
radius (r) = 1.6 m
horizontal force (F) = 55 N
time (t) = 4 s
acceleration due to gravity (g) = 9.8 m/s^{2}
K.E = 0.5 x MI x ω^{2}
where MI is the moment of inertia and ω is the angular velocity
MI = 0.5 x m x r^2
mass = weight ÷ g = 810 ÷ 9.8 = 82.65 kg
MI = 0.5 x 82.65 x 1.6^{2}
MI = 105.8 kg.m^{2}
angular velocity (ω) = a x t
angular acceleration (a) = torque ÷ MI
where torque = F x r = 55 x 1.6 = 88 N.m
a= 88 ÷ 105.8 = 0.83 rad /s^{2}
therefore
angular velocity (ω) = a x t = 0.83 x 4 = 3.33 rad/s
K.E = 0.5 x MI x ω^{2}
K.E = 0.5 x 105.8 x 3.33^{2} = 576 joules
Answer:
The magnetic force points in the positive z-direction, which corresponds to the upward direction.
Option 2 is correct, the force points in the upwards direction.
Explanation:
The magnetic force on any charge is given as the cross product of qv and B
F = qv × B
where q = charge on the ball thrown = +q (Since it is positively charged)
v = velocity of the charged ball = (+vî) (velocity is in the eastern direction)
B = Magnetic field = (+Bj) (Magnetic field is in the northern direction; pointing forward)
F = qv × B = (+qvî) × (Bj)
F =
| î j k |
| qv 0 0|
| 0 B 0
F = i(0 - 0) - j(0 - 0) + k(qvB - 0)
F = (qvB)k N
The force is in the z-direction.
We could also use the right hand rule; if we point the index finger east (direction of the velocity), the middle finger northwards (direction of the magnetic field), the thumb points in the upward direction (direction of the magnetic force). Hence, the magnetic force is acting upwards, in the positive z-direction too.
Hope this Helps!!!
<h2>
Answer: x=125m, y=48.308m</h2>
Explanation:
This situation is a good example of the projectile motion or parabolic motion, in which we have two components: x-component and y-component. Being their main equations to find the position as follows:
x-component:
(1)
Where:
is the projectile's initial speed
is the angle
is the time since the projectile is launched until it strikes the target
is the final horizontal position of the projectile (the value we want to find)
y-component:
(2)
Where:
is the initial height of the projectile (we are told it was launched at ground level)
is the final height of the projectile (the value we want to find)
is the acceleration due gravity
Having this clear, let's begin with x (1):
(3)
(4) This is the horizontal final position of the projectile
For y (2):
(5)
(6) This is the vertical final position of the projectile
As we know that KE and PE is same at a given position
so we will have as a function of position given as
also the PE is given as function of position as
now it is given that
KE = PE
now we will have
so the position is 0.707 times of amplitude when KE and PE will be same
Part b)
KE of SHO at x = A/3
we can use the formula
now to find the fraction of kinetic energy
now since total energy is sum of KE and PE
so fraction of PE at the same position will be
Answer:
7.00 m
Explanation:
Given:
v₀ = 2.00 m/s
v = 5.00 m/s
a = 1.50 m/s²
Find: Δx
v² = v₀² + 2aΔx
(5.00 m/s)² = (2.00 m/s)² + 2(1.50 m/s²)Δx
Δx = 7.00 m