A person of mass m stands on roller skates facing a wall.After pushing against the wall with a constant force F he moves away, r
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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.
We get the clearest image if there is no magnification. When we have no magnification the image and real object have the same size.
If we look at the diagram that I attached we can see that:
Two triangles that I marked are similar and from this we get:
The image and the object must have the same height so we get:
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is:
If we look at the diagram that I attached we can see that:
Two triangles that I marked are similar and from this we get:
The image and the object must have the same height so we get:
This tells how far the screen should be from the lens.
The position of the screen on the optical bench is: