For a system released from rest at θ = 0° when a constant couple moment M = 100 N.m is applied, the speed of the 10-kg block is mathematically given as
V=4.33m/s
<h3>What is the speed of the 10-kg block?</h3>
Generally, the equation for the workdone is mathematically given as
W=T tehta<dtheta
W=100(90*\pi/180)
W=1.5707*100
W=1.57Nm
The change in potential energy across the pulley
dP=mgh
dp=10*9.81*111.8
dp=10.97J
For the thrid position, potential energy is
dP=mg(0.3)
dP=17.658J
dP'=17.658J-13.125
dP'=-4.532J
For 2nd position dP=0
The change in Kinectic energy across the pulley\
dK.E=0.5mv^2
For 1st
dk.E=0.5m(10)^2
2nd
dK.E=0.5Iw^2
dK.E=0.5(7.5*10^-3)(v^2/0.05)^2
3rd
2nd=3rd
In conclusion,
157.07=dKE+dP.E
157.07=5v^2+ (7.5*10^-3)(v^2/0.05)^2+10.97-4.53
V=4.33m/s
Read more about Kinetic energy
brainly.com/question/999862
Answer:
The other angle is 75⁰
Explanation:
Given;
velocity of the projectile, v = 10 m/s
range of the projectile, R = 5.1 m
angle of projection, 15⁰
The range of a projectile is given as;
To find another angle of projection to give the same range;
<u>Check: </u>
sin(2θ) = sin(2 x 75) = sin(150) = 0.5
sin(2θ) = sin(2 x 15) = sin(30) = 0.5
F = 130 revs/min = 130/60 revs/s = 13/6 revs/s
t = 31s
wi = 2πf = 2π × 13/6 = 13π/3 rads/s
wf = 0 rads/s = wi + at
a = -wi/t = -13π/3 × 1/31 = -13π/93 rads/s²
wf² - wi² = 2a∅
-169π²/9 rads²/s² = 2 × -13π/93 rads/s² × ∅
∅ = 1209π/18 rads
n = ∅/2π = (1209π/18)/(2π) = 1209/36 ≈ 33.5833 revolutions.
Answer:
<h2>17.1 N</h2>
Explanation:
The force acting on an object given it's mass and acceleration can be found by using the formula
force = mass × acceleration
From the question
3800 g = 3.8 kg
We have
force = 3.8 × 4.5
We have the final answer as
<h3>17.1 N</h3>
Hope this helps you
Hello!
The Correct Answer would 100% be:
Option "C".
"People in location C would complain about foul taste in water".
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