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Aleks04 [339]
2 years ago
15

If you try opening a door by pushing too close to the side where the hinges are, you may find it difficult to push open. Suppose

you are trying to open a door that is 75 cm wide. (Hint: To solve this one, remember the example problem in the lesson about the MA of the broom). a. If you push on the door at a point 15 cm away from the hinges, what is the mechanical advantage? b. What is the best mechanical advantage that you can obtain for any door that you simply push open? Your answer:
Physics
1 answer:
krok68 [10]2 years ago
8 0

Answer:

<em>a) Mechanical advantage is 0.2</em>

<em>b) maximum obtainable mechanical advantage is 1</em>

<em></em>

Explanation:

The opening of the door is only possible when a force is used to push the door, by generating a torque on the door around the hinge.

Since the door is 75 cm wide, torque generated will be the product of the force applied and the width of the door.

torque = F x 75 = 75F

If the door was pushed at point on the door 15 cm from the hinges, the torque that will be generated will be

torque = F x 15 = 15F

a)The mechanical advantage MA is therefore

MA = 15/75 = 0.2

b) <em>The best mechanical advantage that can be obtained from when you simply push it open is 1</em>

From the case above, if we assume we push it at its maximum width from the hinge, then

MA = 75/75 = 1

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two students are on a balcony 19.6 m above the street. one student throws a ball vertically downward at 14.7 m:ds. at the same i
NARA [144]

A. The difference in the two ball's time in the air is 3 seconds

B. The velocity of each ball as it strikes the ground is 24.5 m/s

C. The balls 0.500 s after they are thrown are 14.7 m apart

<h3>Further explanation</h3>

Acceleration is rate of change of velocity.

\large {\boxed {a = \frac{v - u}{t} } }

\large {\boxed {d = \frac{v + u}{2}~t } }

<em>a = acceleration ( m/s² )</em>

<em>v = final velocity ( m/s )</em>

<em>u = initial velocity ( m/s )</em>

<em>t = time taken ( s )</em>

<em>d = distance ( m )</em>

Let us now tackle the problem!

<u>Given:</u>

Initial Height = H = 19.6 m

Initial Velocity = u = 14.7 m/s

<u>Unknown:</u>

A. Δt = ?

B. v = ?

C. Δh = ?

<u>Solution:</u>

<h2>Question A:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

0 = 19.6 - 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 - 14.7t - 4.9t^2

4.9t^2 + 14.7t - 19.6 = 0

t^2 + 3t - 4 = 0

(t + 4)(t - 1) = 0

(t - 1) = 0

\boxed {t = 1 ~ second}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

0 = 19.6 + 14.7t - \frac{1}{2}(9.8)t^2

0 = 19.6 + 14.7t - 4.9t^2

4.9t^2 - 14.7t - 19.6 = 0

t^2 - 3t - 4 = 0

(t - 4)(t + 1) = 0

(t - 4) = 0

\boxed {t = 4 ~ seconds}

The difference in the two ball's time in the air is:

\Delta t = 4 ~ seconds - 1 ~ second

\large {\boxed {\Delta t = 3 ~ seconds} }

<h2>Question B:</h2><h3>First Ball</h3>

v^2 = u^2 - 2gH

v^2 = (-14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

<h3>Second Ball</h3>

v^2 = u^2 - 2gH

v^2 = (14.7)^2 + 2(-9.8)(-19.6)

v^2 = 600.25

v = \sqrt {600.25}

\boxed {v = 24.5 ~ m/s}

The velocity of each ball as it strikes the ground is 24.5 m/s

<h2>Question C:</h2><h3>First Ball</h3>

h = H - ut - \frac{1}{2}gt^2

h = 19.6 - 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 11.025 ~ m}

<h3>Second Ball</h3>

h = H + ut - \frac{1}{2}gt^2

h = 19.6 + 14.7(0.5) - \frac{1}{2}(9.8)(0.5)^2

\boxed {h = 25.725 ~ m}

The difference in the two ball's height after 0.500 s is:

\Delta h = 25.725 ~ m - 11.025 ~ m

\large {\boxed {\Delta h = 14.7 ~ m} }

<h3>Learn more</h3>
  • Velocity of Runner : brainly.com/question/3813437
  • Kinetic Energy : brainly.com/question/692781
  • Acceleration : brainly.com/question/2283922
  • The Speed of Car : brainly.com/question/568302

<h3>Answer details</h3>

Grade: High School

Subject: Physics

Chapter: Kinematics

Keywords: Velocity , Driver , Car , Deceleration , Acceleration , Obstacle

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2 years ago
Aldis is swinging a ball tied to the end of a string over his head. Suddenly, the string breaks and the ball flies away. Arrow b
prisoha [69]

Answer:

Straight line in the direction of the tangential velocity the ball had at the moment the string broke

Explanation:

After the string breaks, the ball now disconnected from the centripetal force that was exerted via the string, continues its travel in a straight line in the direction of the tangential velocity it had at the moment the string broke.

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2 years ago
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A 2.5 g marshmallow is placed in one end of a 40 cm pipe, as shown in the figure above. A person blows into the left end of the
frosja888 [35]

Answer:

Your answer would be

A person 40 cm- blows into the left end of the pipe to eject the marshmallow from the right end. ... A strain of sound waves is propagated along an organ pipe and gets reflected from an. play · like-icon ... The velocity of sound in air is 340ms^(-1). ... The two pipes are submerged in sea water, arranged as shown in figure. Pipe.Explanation:

I belive this is the answer sorry if im wrong!

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2 years ago
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Racing greyhounds are capable of rounding corners at very high speeds. A typical greyhound track has turns that are 45-m-diamete
Margarita [4]

Explanation:

It is given that,

Diameter of the semicircle, d = 45 m

Radius of the semicircle, r = 22.5 m      

Speed of greyhound, v = 15 m/s

The greyhound is moving under the action of centripetal acceleration. Its formula is given by :

a=\dfrac{v^2}{r}

a=\dfrac{(15)^2}{22.5}

a=10\ m/s^2

We know that, g=9.8\ m/s^2

a=\dfrac{10\times g}{9.8}

a=1.02\ g

Hence, this is the required solution.                                              

5 0
2 years ago
A 60.0-kg crate rests on a level floor at a shipping dock. The coefficients of static and kinetic friction are 0.760 and 0.410,
Alik [6]
  • The horizontal pushing force required to just start the crate moving is 447 N.
  • The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.

<u>Explanation</u>:

  • By the definition of the coefficient of static friction we have:

                                 μ_{s} = \frac{F_{appl} }{W}= \frac{F_{s} }{N},

where, F_{appl} is the horizontal pushing force,

            W = mg is the weight of the crate directed downward,

            F_{s} is  the static  friction  force-directed  opposite  to  the  horizontal  pushing force and equal to it,

            N is the force of reaction directed upward and equal to the weight of the crate.

From  this  formula  we  can  find the horizontal pushing  force  required to  just  start the crate moving:

                         F_{appl} = F_{s} = u_{s}N = u_{s}mg

                                                      = 0.760 \times 60 kg \times 9.8 m / s^2

                                                      = 447 N.

  • By the definition of the coefficient of kinetic friction we have:

                              u_{k} = \frac{F_{appl} }{W} = \frac{F_{k} }{N},

where, F_{appl} is the horizontal pushing force,

            W = mg is the weight of the crate directed downward,

            F_{k} is the kinetic friction force-directed opposite to the horizontal pushing force and equal to it,

            N is the force of reaction directed upward and equal to the weight of the crate.

From this formula we can find the horizontal pushing force required to slide the crate across the dock at a constant speed:

                              F_{appl} = F_{k} = u_{k}N = u_{k}mg

                                                            = 0.410 \times 60 \times 9.8

                                                            = 241 N.

  • The horizontal pushing force required to just start the crate moving is 447 N.
  • The horizontal pushing force required to slide the crate across the dock at a constant speed is 241 N.
6 0
2 years ago
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