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lora16 [44]
2 years ago
8

A book is sitting on a table. Which of the following is true about the table? O A. It is pushing up on the book. O B. It exerts

no force on the book at all. O C. It is pulling down on the book. D. It can affect the mass of the book.​
Physics
1 answer:
777dan777 [17]2 years ago
5 0

Answer:

yes it does exert a force, it pushes it up

Explanation:

this is called normal force

if it didn't exert a force the book would keep going down

according to newton every force has an equal amd opposite reaction

so the book exerts a force on the table and vice versa

hope this helped

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An object of mass 30 kg is falling in air and experiences a force due to air resistance of 50 newtons. Calcuate the acceleration
rjkz [21]

Answer:

this answer you question

6 0
3 years ago
A plane electromagnetic wave travels northward. At one instant, its electric field has a magnitude of 9.6 V/m and points eastwar
lara31 [8.8K]

Answer:

The values is  B  = 3.2 *10^{-8} \  T

The  direction is out of the plane

Explanation:

From the question we are told that

  The  magnitude of the electric field is  E  =  9.6 \  V/m

 

The  magnitude of the magnetic field is mathematically represented as

       B  = \frac{E}{c}

where c is the speed of light with value

      B  = \frac{ 9.6}{3.0 *10^{8}}

     B  = 3.2 *10^{-8} \  T

Given that the direction off the electromagnetic wave( c ) is  northward(y-plane ) and  the electric field(E) is eastward(x-plane ) then the magnetic field will be acting in the out of the page  (z-plane  )

     

3 0
3 years ago
ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself o
shtirl [24]

Q: ken, 0.75 kg, moves toward a wall (his path normal to the wall) at 52 m/s. 13.0 ms after he touches the wall he pushes himself off in the opposite direction at 60 m/s. What is the magnitude of the average force the wall exerts on Ken during this rapid maneuver

Answer:

-6461.54 N

Explanation:

From Newton's Fundamental equation,

F = m(v-u)/t.................... Equation 1

Where F = Force exerted in sonic, m = mass of ken, v = final velocity, u = initial velocity, t = time.

Given: m = 0.75 kg, v = - 60 m/s (opposite direction), u = 52 m/s, t = 13 ms = 0.013 s

Substitute into equation 1

F = 0.75(-60-52)/0.013

F = 0.75(-112)/0.013

F = -84/0.013

F = -6461.54 N

Note: The negative sign tells that the force act in opposite direction to the initial motion of ken.

Hence the magnitude of the average force of the wall = -6461.54 N

4 0
3 years ago
A roller coaster has a vertical loop with radius 25.7 m. With what minimum speed should the roller-coaster car be moving at the
gogolik [260]

Answer:

15.88m/s

Explanation:

At the top of the roller coaster you will have three forces acting on the roller-coaster. See the image below. Fc is the centripetal force (for an object in circular motion), Fg is the gravitational force, and Fn is the normal force. To achieve the minimum speed we assume the roller-coaster is barely touching the vertical loop and so the normal force is zero. This leaves two acting forces.

F_g = F_c\\mg = \frac{m\times v^2}{r}\\v = \sqrt{gr} = \sqrt{9.81 \times 25.7}  = 15.88 m/s

7 0
3 years ago
A particle's trajectory is described by x = (0.5t^3-2t^2) meters and y = (0.5t^2-2t), where time is in seconds. What is the part
mestny [16]

Differentiate the components of position to get the corresponding components of velocity :

v_x = \dfrac{\mathrm dx}{\mathrm dt} = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) t^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)t

v_y = \dfrac{\mathrm dy}{\mathrm dt} = \left(1\dfrac{\rm m}{\mathrm s^2}\right)t-2\dfrac{\rm m}{\rm s}

At <em>t</em> = 5.0 s, the particle has velocity

v_x = \left(1.5\dfrac{\rm m}{\mathrm s^3}\right) (5.0\,\mathrm s)^2 - \left(4\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s) = 17.5\dfrac{\rm m}{\rm s}

v_y = \left(1\dfrac{\rm m}{\mathrm s^2}\right)(5.0\,\mathrm s)-2\dfrac{\rm m}{\rm s} = 3.0\dfrac{\rm m}{\rm s}

The speed at this time is the magnitude of the velocity :

\sqrt{{v_x}^2 + {v_y}^2} \approx \boxed{17.8\dfrac{\rm m}{\rm s}}

The direction of motion at this time is the angle \theta that the velocity vector makes with the positive <em>x</em>-axis, such that

\tan(\theta) = \dfrac{3.0\frac{\rm m}{\rm s}}{17.5\frac{\rm m}{\rm s}} \implies \theta = \tan^{-1}\left(\dfrac{3.0}{17.5}\right) \approx \boxed{9.73^\circ}

4 0
2 years ago
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