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Question
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A particle, mass 0.25 kg is at a position (<em>-7i + 7j + 5k</em>) m, has a velocity (<em>6i - j + 4k</em>) m/s, and is subject to a force (<em>-5i + 0j - k</em>) N. What is the magnitude of the torque on the particle about the origin?
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Answer:</h2>
47.94Nm
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Explanation:</h2>
The torque (τ) on a particle subject to a force (represented as force vector F) at a position (represented as position vector r) about the origin is given by the cross product of the position vector r for the point of application of a force and the force F. i.e
τ = r x F
Given:
r = (-7i + 7j + 5k) m
F = (-5i + 0j - k) N
| i j k |
r x F = | -7 7 5 |
| -5 0 -1 |
r x F = i(-7 - 0) - j(7+25) + k(0+35)
r x F = i(-7) - j(32) + k(35)
r x F = -7i - 32j + 35k
Therefore the torque τ = -7i - 32j + 35k
The magnitude of the torque is therefore;
|τ| = ![\sqrt{(-7)^2 + (-32)^2 + (35)^2}](https://tex.z-dn.net/?f=%5Csqrt%7B%28-7%29%5E2%20%2B%20%28-32%29%5E2%20%2B%20%2835%29%5E2%7D)
|τ| = ![\sqrt{49 + 1024 + 1225}](https://tex.z-dn.net/?f=%5Csqrt%7B49%20%2B%201024%20%2B%201225%7D)
|τ| = ![\sqrt{2298}](https://tex.z-dn.net/?f=%5Csqrt%7B2298%7D)
|τ| = 47.94Nm
The magnitude of the torque on the particle about the origin is 47.94Nm
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Answer:
a) 0.64 b) 2.17m/s^2 c) 8.668joules
Explanation:
The block was on the ramp, the ramp was inclined at 20degree. A force of 5N was acting horizontal to the but not parallel to the ramp,
Frictional force = horizontal component of the weight of the block along the ramp + the applied force since the block was just about move
Frictional force = mgsin20o + 5N = 6.71+5N = 11.71
The force of normal = the vertical component of the weight of the block =mgcos20o = 18.44
Coefficient of static friction = 11.71/18.44= 0.64
Remember that g = acceleration due to gravity (9.81m/s^2) and m = mass (2kg)
b) coefficient of kinetic friction = frictional force/ normal force
Fr = 0.4* mgcos 20o = 7.375N
F due to motion = ma = total force - frictional force
Ma = 11.71 - 7.375 = 4.335
a= 4.335/2(mass of the block) = 2.17m/s^2
C) work done = net force *distance = 4.335*2= 8.67Joules
Answer:transverse- transfers energy perpendicular to the direction of the wave
Electromagnetic- dose not require a medium for transition
Mechanical - requires a medium for transmission
Longitudinal- transfers energy parallel to the direction of the wave
Explanation: just took the activity
Answer:
11.8 Joules
Explanation:
Given:-
- The height of the target ball, si = 0.860 m
- The mass of target and steel ball, m = 0.012 kg
- The target ball travels a distance ( x ) after being struck = 1.40 m
Find:-
What is the kinetic energy (in joules) of the target ball just after it is struck?
Solution:-
- We are given the initial distance of the target ball as 0.86 m above the floor which travels a distance ( x ) after being struck.
- We will employ the one dimensional kinematic equation of motion to determine the initial velocity ( vi ) of the target ball as follows:
![vf^2 = vi^2 - 2*g*x](https://tex.z-dn.net/?f=vf%5E2%20%3D%20vi%5E2%20-%202%2Ag%2Ax)
Where,
vf: The final velocity of target ball at maximum height = 0
g: The gravitational acceleration constant = 9.8 m/s^2
- Plug in the required parameters and evaluate the ( vi ) as follows:
![0^2 = vi^2 - 2*( 9.80 )*( 1.40 )\\\\vi^2 = 27.44\\\\vi = \sqrt{27.44} = 5.24 m/s](https://tex.z-dn.net/?f=0%5E2%20%3D%20vi%5E2%20-%202%2A%28%209.80%20%29%2A%28%201.40%20%29%5C%5C%5C%5Cvi%5E2%20%3D%2027.44%5C%5C%5C%5Cvi%20%3D%20%5Csqrt%7B27.44%7D%20%3D%205.24%20m%2Fs)
- The kinetic energy ( Ek ) of an object with mass ( m ) and initial velocity ( vi ) is expressed as:
![E_k = 0.5*m*(vi)^2\\\\E_k = 0.5*0.86*27.44\\\\E_k = 11.8 J](https://tex.z-dn.net/?f=E_k%20%3D%200.5%2Am%2A%28vi%29%5E2%5C%5C%5C%5CE_k%20%3D%200.5%2A0.86%2A27.44%5C%5C%5C%5CE_k%20%3D%2011.8%20J)
Answer: The kinetic energy of the target ball just after it is struck is 11.8 Joules.