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3 years ago

Explain why a book placed on a table does not move?

Physics

2 answers:

N76 [4]3 years ago

5 0

Answer:

4th: "The force that the book exerts down on the table is equal to the force the table exerts upward"

Explanation:

Basically, a book on a table is balanced in terms of forces. Gravity is opposed by the table and there are no forces acting to move the book in any direction. If there was a breeze that was not strong enough to move the book, it is because the force of friction is equal to that provided by the breeze.

kifflom [539]3 years ago

4 0

Answer:

The force that the book exerts down on the table is equal to the force the table exerts upward.

Explanation:

I am 100% positive. That results in balanced forces which is why it does not move. If you did want it to move, you would apply unbalanced forces.

You might be interested in

Find the frequency of a wave of wavelength 2.5m and speed 400 m/s

Sedaia [141]

Answer:

The frequency of wave is 160Hz.

Explanation:

Given that the formula of speed is V = f×λ where V represents speed, f is frequency and λ is wavelength.

So first thing, you have to make frequency the subject by dividing wavelength on both sides :

$v = f \times λ \:$

$v \div λ = f \times λ \div λ$

$f = \frac{v}{λ}$

Next you have to substitute the value of v and f into the formula :

Let λ = 2.5m,

Let v = 400m/s,

$f = \frac{400}{2.5}$

$f = 160Hz$

7 0

2 years ago

Se golpea una pelota de golf de manera que su velocidad inicial forma un ángulo de 45° con la horizontal. La pelota alcanza el s

nordsb [41]

Answer:

42m/s

6.06s

Explanation:

To find the initial velocity and time in which the ball is fling over the ground you use the following formulas:

$x_{max}=\frac{v_o^2sin(2\theta)}{g}\\\\x_{max}=vt_{max}$

θ: angle = 45°

vo: initial velocity

g: gravitational constant = 9.8m/s^2

x_max: max distance = 180 m

t_max: max time

by replacing the values of the parameters and do vo the subject of the first formula you obtain:

$v_o=\sqrt{\frac{gx_{max}}{sin(2\theta)}}\\\\v_o=\sqrt{\frac{(9.8m/s^2)(180m)}{sin(2(45\°))}}=42\frac{m}{s}$

with this value of vo you calculate the max time:

$t_{max}=\frac{x_{max}}{v}=\frac{x_{max}}{v_ocos(45\°)}\\\\t_{max}=\frac{180m}{(42m/s)cos(45\°)}=6.06s$

hence, the initial velocity of the ball is 42m/s and the time in which the ball is in the air is 6.06s

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TRANSLATION:

Para encontrar la velocidad inicial y el tiempo en el que la pelota está volando sobre el suelo, use las siguientes fórmulas:

θ: ángulo = 45 °

vo: velocidad inicial

g: constante gravitacional = 9.8m / s ^ 2

x_max: distancia máxima = 180 m

t_max: tiempo máximo

reemplazando los valores de los parámetros y haciendo el tema de la primera fórmula que obtiene:

con este valor de vo usted calcula el tiempo máximo:

por lo tanto, la velocidad inicial de la pelota es de 42 m / sy el tiempo en que la pelota está en el aire es de 6.06 s

4 0

2 years ago

5. What is the speed of a wave in a spring if it has a wavelength of 10 cm and a period of 0.2s

eimsori [14]

Explanation:

speed of wave

v = wavelength x frequency

since frequency is f = 1/Period then

v = wavelength : Period

v = 10 cm/ 0.2 s = 50 cm/s

v = 0.5 m/s

7 0

3 years ago

A steel ball bearing with a radius of 1.5 cm forms an image of an object that has been placed 1.1 cm away from the bearing’s sur

Nonamiya [84]

Answer:

Check the explanation

Explanation:

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

The image is virtual

The image is upright

given

R = 1.5 cm

object distance, u = 1.1 cm

focal length of the ball, f = -R/2

= -1.5/2

= -0.75 cm

let v is the image distance

use, 1/u + 1/v = 1/f

1/v = 1/f - 1/u

1/v = 1/(-0.75) - 1/(1.1)

v = -0.446 cm <<<<<---------------Answer

magnification, m = -v/u

= -(-0.446)/1.1

= 0.405 <<<<<<<<<---------------Answer

Kindly check the diagram in the attached image below.

5 0

3 years ago

The right-hand rule predicts the direction of the force on a positively charged object moving in a gravitational field true fals

eimsori [14]

The statement above is FALSE.
The right hand rule is used in physics to predict the direction of the force on a charged object moving in a MAGNETIC FIELD. The right hand rule is used to relate the relationship between the magnetic field and the forces that are exerted on the moving objects in the field. Using the right hand rule, for a positively charged object that is moving in an electric field, the pointer finger will point in the direction the charged object is moving, the middle finger will point in the direction of the magnetic field and the thumb will point in the direction of the magnetic force that is pushing the charged object.

3 0

3 years ago