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MissTica
1 year ago
10

5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5

Physics
1 answer:
Anvisha [2.4K]1 year ago
3 0

5 What is the angular displacement at the end of the 25-mm-diameter shaft and the linear displacement of point A of Figure P5.5

<h3>What is displacement ?</h3>

A displacement is a vector in geometry and mechanics that has a length equal to the shortest distance between a point P's initial and final positions. It calculates the length and angle of the net motion, or total motion, in a straight line from the starting point to the destination of the point trajectory. The translation that links the starting point and the ending point can be used to spot a displacement.

The final location xf of a point relative to its beginning position xi, or a relative position (derived from the motion), is another way to express a displacement. The difference between the end and beginning positions can be used to define the equivalent displacement vector

To learn more about displacement  from the given link:

brainly.com/question/321442

#SPJ4

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Bradley drops a rock in a well. It falls for 12 seconds. How deep is the well?
erik [133]
It’s 12 seconds long
3 0
3 years ago
A 4 kg textbook sits on a desk. It is pushed horizontally with a 50 N applied force against a 15 N frictional force.
GarryVolchara [31]

a) See free-body diagram in attachment

b) The book is stationary in the vertical direction

c) The net horizontal force is 35 N in the forward direction

d) The net force on the book is 35 N in the forward horizontal direction

e) The acceleration is 8.75 m/s^2 in the forward direction

Explanation:

a)

The free-body diagram of a body represents all the forces acting on the body using arrows, where the length of each arrow is proportional to the magnitude of the force and points in the same direction.

From the diagram of this book, we see there are 4 forces acting on the book:

- The applied force, F = 50 N, pushing forward in the horizontal direction

- The frictional force, F_f = 15 N, pulling backward in the horizontal direction (the frictional force always acts in the direction opposite to the motion)

- The weight of the book, W=mg, where m is the mass of the book and g=9.8 m/s^2 is the acceleration of gravity, acting downward. We can calculate its magnitude using the mass of the book, m = 4 kg:

W=(4)(9.8)=39.2 N

- The normal reaction exerted by the desk on the book, N, acting upward, and balancing the weight of the book

b)

The book is in equilibrium in the vertical direction, therefore there is no motion.

In fact, the magnitude of the normal reaction (N) exerted by the desk on the book is exactly equal to the weight of the book (W), so the equation of motion along the vertical direction is

N-W=ma

where a is the acceleration; however, since N = W, this becomes

a=0

And since the book is initially at rest on the desk, this means that there is no motion.

c)

We said there are two forces acting in the horizontal direction:

- The applied force, F = 50 N, forward

- The frictional force, F_f = 15 N, backward

Since they act along the same line, we can calculate their resultant as

\sum F = F - F_f = 50 - 15 = 35 N

and therefore the net force is 35 N in the forward direction.

d)

The net force is obtained as the resultant  of the net forces in the horizontal and vertical direction. However, we have:

- The net force in the horizontal direction is 35 N

- The net force in the vertical direction is zero, because the weight is balanced by the normal reaction

Therefore, this means that the total net force acting on the book is just the net force acting on the horizontal direction, so 35 N forward.

e)

The acceleration of the book can be calculated by using Newton's second law:

\sum F = ma

where

\sum F is the net force

m is the mass

a is the acceleration

Here we have:

\sum F = 35 N (in the forward direction)

m = 4 kg

Therefore, the acceleration is

a=\frac{\sum F}{m}=\frac{35}{4}=8.75 m/s^2 (forward)

Learn more about forces, weight and Newton's second law:

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8 0
3 years ago
How does the electric force between two charged participles change if one particle’s charge is reduced by a factor of 3?
Eva8 [605]

Answer:

ick

Explanation:

Me down daddy

3 0
3 years ago
A child is riding in a child-restraint chair, securely fastened to the seat of a car. Assume the car has speed 47 km/h when it h
Elina [12.6K]

Answer: F = 1235 N

Explanation: Newton's Second Law of Motion describes the effect of mass and net force upon acceleration: F_{net}=m.a

Acceleration is the change of velocity in a period of time: a=\frac{\Delta v}{\Delta t}

Velocity of the car is in km/h. Transforming it in m/s:

v=\frac{47.10^{3}}{36.10^{2}}

v = 13 m/s

At the moment the car decelerates, acceleration is

a=\frac{13}{0.2}

a = 65 m/s²

Then, force will be

F_{net}=19(65)

F_{net} = 1235 N

The horizontal net force the straps of the restraint chair exerted on the child to hold her is 1235 newtons.

5 0
2 years ago
If a 10.0 kg object is on a surface that is inclined 30o and the coefficient of static friction is 0.65, what is the force of st
Alexxandr [17]
  fraction equation is<span>
                     F =µR 
 F=friction,µ=coefficient , R=reaction = mg 

use same equation for b part, but the reaction is no longer mg because the plain is now inclined. Draw a forces diagram and you will see that the reaction force can be calculated from the weight of the object and inclination of the plain using trigonometry.</span>
6 0
3 years ago
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