Set up the problem with the conversion rates as fractions where when you multiply the units cancel out leaving the desired units behind.
Velocity of a machine is always greater than mechanical advantage? why
1 answer:
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Answer:
2,87 *
Explanation:
When the bullets meet at the center and collide, since momentum is a vectoral quantity, their momentum vectors even up and are sumof zero. Formula of momentum is P = m.v , where m is mass and v is velocity. Let’s name the first two bullets as x,y and the one which mass is unknown as z. Then calculate momentum of x and y:
Px= 5,30 * * 301 = 1,5953 kg*m/s
Py= 5,30 * * 301 = 1,5953 kg*m/s
The angle between x and y bullets is 120°, and we know that if the angle between two equal magnitude vectors is 120°, the magnitude of the resultant vector will be equal to first two and placed in exact middle of two vectors. So we can say total momentum of x and y (Px+Py) equals to 1,5953 kg*m/s as well (Shown in the figure).
For z bullet to equalize the total momentum of x and y bullets, it needs to have the same amount of momentum in the opposite way.
Pz = 1,5953 = m * 554
m = 2,87 * kg
Answer:a) 629,5851 km/h in magnitude b)629,5851 km/h at 84,2 degrees from east pointing south direction or in vector form 626,6396 km/h south + 63,6396km/h east. c) 16,5 km NE of the desired position
Explanation:
Since the plane is flying south at 690 km/h and the wind is blowing at assumed constant speed of 90 km/h from SW, we get a triangle relation where
see fig 1
Then we can decompose those 90 km/h into vectors, one north and one east, both of the same magnitude, since the angle is 45 degrees with respect to the east, that is direction norhteast or NE, then
90 km/h NE= 63,6396 km/h north + 63,6396 km/h east,
this because we have an isosceles triangle, then the cathetus length is
hypotenuse/
using Pythagoras, here the hypotenuse is 90, then the cathetus are of length
90/ km/h= 63,6396 km/h.
Now the total speed of the plane is
690km/h south + 63,6396 km/h north +63,6396 km/h east,
this is 626,3604 km/h south + 63,6396 km/h east, here north is as if we had -south.
then using again Pythagoras we get the magnitude of the total speed it is
,
the direction is calculated with respect to the south using trigonometry, we know the
sin x= cathetus opposed / hypotenuse,
then
x= =5,801 degrees from South as reference (0 degrees) in East direction or as usual 84,2 degrees from east pointing south or in vector form
626,6396 km/h south + 63,6396km/h east.
Finally since the detour is caused by the west speed component plus the slow down caused by the north component of the wind speed, we get
Xdetour{east}= 63,6396 km/h* (11 min* h)/(60 min)=11,6672 km=Xdetour{north} ,
since 11 min=11/60 hours=0.1833 hours.
Then the total detour from the expected position, the one it should have without the influence of the wind, we get
Xdetour=[/tex]\sqrt{2* 11,6672x^{2} }[/tex] = 16,5km at 45 degrees from east pointing north
The situation is sketched as follows see fig 2
To solve this problem it is necessary to apply the concepts related to Newton's second law, the definition of density and sum of forces in bodies.
From Newton's second law we understand that
Gravity at this case)
Where,
m = mass
a= acceleration
Also we know that
Part A) The buoyant force acting on the balloon is given as
As mass is equal to the density and Volume and acceleration equal to Gravity constant
PART B) The forces acting on the balloon would be given by the upper thrust force given by the fluid and its weight, then
PART C) The additional mass that can the balloon support in equilibrium is given as
Complete Question
The complete question is shown on the first uploaded image
Answer:
Force on each hand is 196.22 N
Force on each foot is 95.8 N
Explanation:
In order to get a better understanding of this question let us explain some concepts
Normal Force:
We can define normal force Fn as that type of force which makes a 90 degree angle with the surface on which it is exerted.
Torque:
We can define torque as the moment of forces that tends to produce or cause rotation
From the question we are given that
Weight of body is (W) = 584 N
The normal force on both hands (Ha) = ?
The normal force on both legs (Lg) = ?
Looking at the diagram the person is at equilibrium so
584 = Ha + Lg
an also this mean that torques acting on the body is balanced
So, 0.410 Ha = 0.840 Lg
Making Lg the subject of formula in the equation above we
Lg = 0.4881 Ha
Considering the first equation and replacing Lg with this recent equation we have
584 = Ha + 0.4881 Ha
Therefore Ha = 392.44 N
This value obtained is for both hands for each hand we divide by 2
Therefore we have for each hand =196.55 N
Since we have been able to get the force on both hands we can substitute it in to the equation where we made Lg the subject of formula and we have
Lg = 0.4881 × 392.44
= 191.22 N
The value above is the force on both legs to obtain the force on each leg we have
191.22/2 = 95.8 N.
