1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
Amanda [17]
3 years ago
8

A person is pushing a book lying on a table. Identify the different forces that act on

Physics
2 answers:
BaLLatris [955]3 years ago
3 0

Explanation:

There are two forces acting upon the book. One force - the Earth's gravitational pull - exerts a downward force. The other force - the push of the table on the book (sometimes referred to as a normal force) - pushes upward on the book.

sveticcg [70]3 years ago
3 0

There are mainly 3 forces which acts on a book.

Listed below are the forces and their points:-

• <u>Earth's Gravitational force of attraction</u>- Usually, we may not consider or notice this force, but when a book falls, we understand this, because the book falls due to this gravitational force.

• <u>Force exerted on book while pushing it</u>- While we push a book, we apply Push Force, due to which the book moves on the pushing it on a table.

• <u>Frict</u><u>ional</u><u> </u><u>Force</u>- We notice that while pushing a book, it does not continuously move in motion. At a certain distance, this Frictional force increases and therefore, the book comes to a halt.

<em>Location-</em> <em>Opposite of exerted force in all directions</em>

You might be interested in
A rocket is fired straight up. It contains two stages (Stage 1 and Stage 2) of solid rocket fuel that are designed to burn for 1
Archy [21]

Answer:

a)  y= 3.5 10³ m, b)   t = 64 s

Explanation:

a) For this exercise we use the vertical launch kinematics equation

Stage 1

          y₁ = y₀ + v₀ t + ½ a t²

          y₁ = 0 + 0 + ½ a₁ t²

Let's calculate

         y₁ = ½ 16 10²

         y₁ = 800 m

At the end of this stage it has a speed

        v₁ = vo + a₁ t₁

        v₁ = 0 + 16 10

        v₁ = 160 m / s

Stage 2

        y₂ = y₁ + v₁ (t-t₀) + ½ a₂ (t-t₀)²

        y₂ = 800 + 150 5 + ½ 11 5²

        y₂ = 1092.5 m

Speed ​​is

        v₂ = v₁ + a₂ t

        v₂ = 160 + 11 5

        v₂ = 215 m / s

The rocket continues to follow until the speed reaches zero (v₃ = 0)

         v₃² = v₂² - 2 g y₃

         0 = v₂² - 2g y₃

         y₃ = v₂² / 2g

         y₃ = 215²/2 9.8

         y₃ = 2358.4 m

The total height is

          y = y₃ + y₂

          y = 2358.4 + 1092.5

          y = 3450.9 m

           y= 3.5 10³ m

b) Flight time is the time to go up plus the time to go down

Let's look for the time of stage 3

          v₃ = v₂ - g t₃

          v₃ = 0

          t₃ = v₂ / g

          t₃ = 215 / 9.8

          t₃ = 21.94 s

The time to climb is

          t_{s} = t₁ + t₂ + t₃

          t_{s} = 10+ 5+ 21.94

          t_{s} = 36.94 s

The time to descend from the maximum height is

          y = v₀ t - ½ g t²

When it starts to slow down it's zero

         y = - ½ g t_{b}²

         t_{b}  = √-2y / g

         

        t_{b} = √(- 2 (-3450.9) /9.8)

        t_{b} = 26.54 s

Flight time is the rise time plus the descent date

        t = t_{s} + t_{b}

        t = 36.94 + 26.54

        t =63.84 s

        t = 64 s

3 0
3 years ago
What is numbers 2 and 4? Please help, and many thanks to those who do chose to help.
Elan Coil [88]

Answer:

2. 6 km/h

4. 150 miles/h

5 0
3 years ago
Two charged particles are located on the x axis. The first is a charge 1Q at x 5 2a. The second is an unknown charge located at
sergejj [24]

Answer:

Q_2 = +/- 295.75*Q

Explanation:

Given:

- The charge of the first particle Q_1 = +Q

- The second charge = Q_2

- The position of first charge x_1 = 2a

- The position of the second charge x_2 = 13a

- The net Electric Field produced at origin is E_net = 2kQ / a^2

Find:

Explain how many values are possible for the unknown charge and find the possible values.

Solution:

- The Electric Field due to a charge is given by:

                               E = k*Q / r^2

Where, k: Coulomb's Constant

            Q: The charge of particle

            r: The distance from source

- The Electric Field due to charge 1:

                               E_1 = k*Q_1 / r^2

                               E_1 = k*Q / (2*a)^2

                               E_1 = k*Q / 4*a^2

- The Electric Field due to charge 2:

                               E_2 = k*Q_2 / r^2

                               E_2 = k*Q_2 / (13*a)^2

                               E_2 = +/- k*Q_2 / 169*a^2

- The two possible values of charge Q_2 can either be + or -. The Net Electric Field can be given as:

                               E_net = E_1 + E_2

                               2kQ / a^2 = k*Q_1 / 4*a^2 +/- k*Q_2 / 169*a^2

- The two equations are as follows:

        1:                   2kQ / a^2 = k*Q / 4*a^2 + k*Q_2 / 169*a^2

                               2Q = Q / 4 + Q_2 / 169

                               Q_2 = 295.75*Q

        2:                    2kQ / a^2 = k*Q / 4*a^2 - k*Q_2 / 169*a^2

                               2Q = Q / 4 - Q_2 / 169

                               Q_2 = -295.75*Q

- The two possible values corresponds to positive and negative charge Q_2.

7 0
2 years ago
A spring on a horizontal surface can be stretched and held 0.5 m from its equilibrium position with a force of 60 N. a. How much
Lostsunrise [7]

Answer:

a)1815Joules b) 185Joules

Explanation:

Hooke's law states that the extension of a material is directly proportional to the applied force provided that the elastic limit is not exceeded. Mathematically;

F = ke where;

F is the applied force

k is the elastic constant

e is the extension of the material

From the formula, k = F/e

F1/e1 = F2/e2

If a force of 60N causes an extension of 0.5m of the string from its equilibrium position, the elastic constant of the spring will be ;

k = 60/0.5

k = 120N/m

a) To get the work done in stretching the spring 5.5m from its position,

Work done by the spring = 1/2ke²

Given k = 120N/m, e = 5.5m

Work done = 1/2×120×5.5²

Work done = 60× 5.5²

Work done = 1815Joules

b) work done in compressing the spring 1.5m from its equilibrium position will be gotten using the same formula;

Work done = 1/2ke²

Work done =1/2× 120×1.5²

Works done = 60×1.5²

Work done = 135Joules

8 0
2 years ago
Read 2 more answers
While driving on the highway, you have a tire blowout. after you have slowed down and regained control, you should?
Svetllana [295]
When a person has experienced a tire blow out, there are steps to follow in order to prevent harm to others and self. After following the required and helpful steps after a tire blow out, slowed down and regain control, it is best that the person should head to a stop road or at the safe side of the lane where they won't be a bother to the road or to people driving in the high way. After pulling over to the side, it is advisable to turn on the emergency flashers of the car. This will set as a signal that you are in need of help or had gone through a problem.
4 0
3 years ago
Other questions:
  • A wave has a frequency of 58 hz and a speed of 31 m/s. what is the wavelength of this wave?
    15·1 answer
  • ) A hot air balloon along with its cargo has a mass of 4.0 x 105 kg, and it holds 7.0 x 10 5 m3 of hot air. It is floating at a
    14·1 answer
  • A snowball is thrown with an initial x velocity of 7.5 m/s and an initial y velocity of 8.4 m/s . Part A How much time is requir
    13·1 answer
  • (a) The Eskimo pushes the same 50.0-kg sled over level ground with a force of 1.75 102 N exerted horizontally, moving it a dista
    10·1 answer
  • Three forces act on an object. a 3 n force acts due east and a 8 n force acts due north. if the net force on the object is zero,
    14·1 answer
  • (02.09 MC) An object starts at rest. Its acceleration over 30 seconds is shown in the graph below: An acceleration versus time g
    8·2 answers
  • While traveling on a horizontal road at speed vi, a driver sees a large rabbit ahead and slams on the brakes. The wheels lock an
    11·1 answer
  • A dog chases a skateboarder travelling at 10.4\,\dfrac{\text{m}}{\text{s}}10.4 sm ​ 10, point, 4, start fraction, start text, m,
    12·1 answer
  • A cosmic ray proton moving toward Earth at 5.00 x 107 m/s experiences a magnetic force of 1.7 x 10-16 N. What is the strength of
    7·1 answer
  • Please help ASAP....
    6·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!