The meter stick in the drawing can rotate about an axis located at the 20.0-cm mark. The axis is perpendicular to the screen. A force F acts at the left end; the force is perpendicular to the meter stick and has a magnitude of 175 N. A second force, either F1 or F2, acts at the 80.0-cm mark, as the drawing shows. The meter stick is in equilibrium. Which force, F1 or F2, acts on the meter stick, and what is its magnitude?

 F₁ = 43.75 N

Explanation:

In this exercise you fear a meter that at 20 cm has a force of 175 N applied and at 80 cm has another F1 force applied, so that the system is in equilibrium.

Let's use the equation of rotational equilibrium,

               τ₁ - τ₂ = 0

               F d = F₁ d₁

               F₁ = F d / d₁

Let's calculate

               F₁ = 175   0.2 /0.8

               F₁ = 43.75 N

Since the two forces are applied on the same bar, they must have the opposite direction for the torque to be canceled.

The answer is

The force F₂, acts on the meter stick, with a magnitude of 71.21 N

Explanation:

To solve the question, we note that the sum of moment about a point = 0 or ΣM =0

That is sum of clockwise moments = sum of anticlockwise moments

Also we are required to maintain the sign convention when calculating for the moment of the force thus

Clockwise moment = -ve and clockwise = +ve

Therefore taking moment about the axis at the 20.0 cm mark we have

-175 N × 20, 0 cm -F₁y × (80 - 20) cm = 0 where F₁y is the y component of the the force at the 80.0 cm mark

or -175 N × 20, 0 cm = F₁y × (80 - 20) cm

From the above equation, it is seen that F₁y = - (175 N × 20 cm)/(60 cm) = 58.33 N

F₁y = -58.33 N, Hence the correct force at the 80.0 cm mark is

58.33 N in the opposite direction as F₁y or F₂y

Hence the magnitude of F₂ can be found by resolving the force into components thus

y component of the force F₂ = 58.33 N angle to the vertical  is given by

90 - 55 or 35 ° therefore the y which is the vertical component = 58.33 N at an angle 35 ° to the vertical, we have

F₂ × cos (35 °) = 58.33 N

∴ F₂ = (58.33 N)/(cos (35 °))

= 71.21 N

Therefore the force F₂ acts on the meter stick with a magnitude of 71.21 N

answer: the exterior temperature of commercial jet planes as they leave the stratosphere and return to the ground increases steadily.

explanation:

the lower most layer of atmosphere is troposphere which extends from 0 to 12 km above the surface of the earth. the temperature initially decreases as distance is increased from the earth's surface. commercial jets fly in lower part of the stratosphere. the stratosphere extends upto 50 km. the temperature in stratosphere starts rising again due to presence of ozone which traps the uv rays.

thus, when the planes leave the stratosphere and return to the ground, the exterior temperature increases steadily.

i don't know that i agree with that but what's the question ?