<span>The water is clearer, has higher oxygen levels, and freshwater fish such as trout and heterotrophs can be found. In the middle of the river, the width increases, so do the species of aquatic green plants and algae. Toward the mouth of the river, the water becomes murky from all the sediments that it has picked up upstream.<span>
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Either 175 N or 157 N depending upon how the value of 48° was measured from.
You didn't mention if the angle of 48° is from the lug wrench itself, or if it's from the normal to the lug wrench. So I'll solve for both cases and you'll need to select the desired answer.
Since we need a torque of 55 N·m to loosen the nut and our lug wrench is 0.47 m long, that means that we need 55 N·m / 0.47 m = 117 N of usefully applied force in order to loosen the nut. This figure will be used for both possible angles.
Ideally, the force will have a 0° degree difference from the normal and 100% of the force will be usefully applied. Any value greater than 0° will have the exerted force reduced by the cosine of the angle from the normal. Hence the term "cosine loss".
If the angle of 48° is from the normal to the lug wrench, the usefully applied power will be:
U = F*cos(48)
where
U = Useful force
F = Force applied
So solving for F and calculating gives:
U = F*cos(48)
U/cos(48) = F
117 N/0.669130606 = F
174.8537563 N = F
So 175 Newtons of force is required in this situation.
If the 48° is from the lug wrench itself, that means that the force is 90° - 48° = 42° from the normal. So doing the calculation again (this time from where we started plugging in values) we get
U/cos(42) = F
117/0.743144825 = F
157.4390294 = F
Or 157 Newtons is required for this case.
Using
V = Amplitude x angular frequency(omega)
But omega= 2πf
= 2πx875
=5498.5rad/s
So v= 1.25mm x 5498.5
= 6.82m/s
B. .Acceleration is omega² x radius= 104ms²
The momentum of ball is given by:
Since both have the same momentum, we have:
Number 3If you notice any mistake in my english, please let me know, because i am not native.
this can be solve using the formala of free fall
t = sqrt( 2y/ g)
where t is the time of fall
y is the height
g is the acceleration due to gravity
48.4 s = sqrt (2 (1.10e+02 m)/ g)
G = 0.0930 m/s2
The velocity at impact
V = sqrt(2gy)
= sqrt( 2 ( 0.0930 m/s2)( 1.10e+02 m)
V = 4.523 m/s
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