2. Draw a ray diagram that shows the image

In order for work to be done, what three things are necessary?

andrey2020 [161]

The three things that are necessary are power, a force, and movement in the opposite direction of the applied force.

6 0

3 years ago

)If a force of 5.00 N is needed to open a 90.0 cm wide door when applied at the edge opposite the hinges, what force must be app

masha68 [24]

Answer:

A force of 12.857 newtons must be applied to open the door.

Explanation:

In this case, a force is exerted on the door, a moment is performed and the door is opened. If moment remains constant, the force is inversely proportional to distance respect to axis of rotation passing through hinges. That is:

$F \propto \frac{1}{r}$

$F = \frac{k}{r}$ (Eq. 1)

Where:

$F$ - Force, measured in newtons.

$k$ - Proportionality ratio, measured in newton-meters.

$r$ - Distance respect to axis of rotation passing through hinges, measured in meters.

From (Eq. 1) we get the following relationship and clear the final force within:

$F_{A}\cdot r_{A} = F_{B}\cdot r_{B}$

$F_{B}=\left(\frac{r_{A}}{r_{B}} \right)\cdot F_{A}$ (Eq. 2)

Where:

$F_{A}$ , $F_{B}$ - Initial and final forces, measured in newtons.

$r_{A}$ , $r_{B}$ - Initial and final distances, measured in meters.

If we know that $F_{A} = 5\,N$ , $r_{A} = 0.9\,m$ and $r_{B} = 0.35\,m$ , then final force is:

$F_{B}= \left(\frac{0.9\,m}{0.35\,m} \right)\cdot (5\,N)$

$F_{B} = 12.857\,N$

A force of 12.857 newtons must be applied to open the door.

3 0

3 years ago

Consumers that eat both producers and other consumers are known as

matrenka [14]

Omnivores...————Fhfhfjdhfh

3 0

3 years ago

If you rub an inflated balloon against your hair and place it against a door, by what mechanism does it stick? explain.

Snowcat [4.5K]

When you rub inflated balloon with your hair or your kitten's fur, charge is induced on all over the balloon's surface. This is called "charging by friction" because you developed charges by rubbing to bodies with each other. It will also stick on your wall you can check it out. This is because of "unlike charges attract each other". Rubbed balloon and wall possessed unlike charges which made them stick together.

8 0

4 years ago

A small box of mass m1 is sitting on a board of mass m2 and length L. The board rests on a frictionless horizontal surface. The

Nadusha1986 [10]

Answer:

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}$

Explanation:

The Newton’s second law states that the net force on an object is the product of mass of the object and final acceleration of the object. The expression of newton’s second law is,

$\sum {F = ma}$

Here, is the sum of all the forces on the object, mm is mass of the object, and aa is the acceleration of the object.

The expression for static friction over a horizontal surface is,

$F_{\rm{f}}} \leq {\mu _{\rm{s}}}mg$

Here, ${\mu _{\rm{s}}}$ is the coefficient of static friction, mm is mass of the object, and $g$ is the acceleration due to gravity.

Use the expression of static friction and solve for maximum static friction for box of mass ${m_1}$

Substitute for in the expression of maximum static friction ${F_{\rm{f}}} = {\mu _{\rm{s}}}mg$

${F_{\rm{f}}} = {\mu _{\rm{s}}}{m_1}g$

Use the Newton’s second law for small box and solve for minimum acceleration aa to pull the box out.

Substitute for , [/tex]{m_1}[/tex] for in the equation .

${F_{\rm{f}}} = {m_1}a$

Substitute ${\mu _{\rm{s}}}{m_1}g$ for ${F_{\rm{f}}}$ in the equation ${F_{\rm{f}}} = {m_1}a$

${\mu _{\rm{s}}}{m_1}g = {m_1}a$

Rearrange for a.

$a = {\mu _{\rm{s}}}g$

The minimum acceleration of the system of two masses at which box starts sliding can be calculated by equating the pseudo force on the mass with the maximum static friction force.

The pseudo force acts on in the direction opposite to the motion of the board and the static friction force on this mass acts in the direction opposite to the pseudo force. If these two forces are cancelled each other (balanced), then the box starts sliding.

Use the Newton’s second law for the system of box and the board.

Substitute for for in the equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right)a$

Substitute for in the above equation .

${F_{\min }} = \left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

The constant force with least magnitude that must be applied to the board in order to pull the board out from under the box is $\left( {{m_1} + {m_2}} \right){\mu _{\rm{s}}}g$

There is no friction between the board and the surface. So, the force required to accelerate the system with the minimum acceleration to slide the box over the board is equal to total mass of the board and box multiplied by the acceleration of the system.

5 0

3 years ago