All categories

3 years ago

A book that weighs 19 Newtons sits on a table. With what force

Physics

1 answer:

iVinArrow [24]3 years ago

5 0

Answer:

We know there's two forces acting on a book while it sits on a table:the force of gravity pulling it down, and the normal force of the table acting upward on the book. The book isn't accelerating while it sits there. That's because the weight of the book is being counteracted by the normal force of the table.

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.

You might be interested in

A ray is incident on a film of thickness t and the index of refraction n=1.6 at an angle θ = 35 degrees. Find the angle of refra

lidiya [134]

Answer

Given,

refractive index of film, n = 1.6

refractive index of air, n' = 1

angle of incidence, i = 35°

angle of refraction, r = ?

Using Snell's law

n' sin i = n sin r

1 x sin 35° = 1.6 x sin r

r = 21°

Angle of refraction is equal to 21°.

Now,

distance at which refractive angle comes out

d = 2.5 mm

α be the angle with horizontal surface and incident ray.

α = 90°-21° = 69°

t be the thickness of the film.

So,

$tan \alpha = \dfrac{t}{d/2}$

$tan 61^0 = \dfrac{t}{2.5/2}$

t = 2.26 mm

Hence, the thickness of the film is equal to 2.26 mm.

4 0

3 years ago

can someone help me answer this question? A student connected an ammeter as shown in this picture. Did the student connect the a

nalin [4]

Answer:

No - It is connected in parallel instead of series

Explanation:

4 0

2 years ago

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton

pishuonlain [190]

Hello!

Using Hooke's law, F spring=k delta x, find the distance a spring with an elastic constant of 4 N/cm will stretch if a 2 newton force is applied to it.

Data:

Hooke represented mathematically his theory with the equation:

F = K * Δx

On what:

F (elastic force) = 2 N

K (elastic constant) = 4 N/cm

Δx (deformation or elongation of the elastic medium or distance from a spring) = ?

Solving:

$F = K * \Delta{x}$

$2\:N = 4\:N/cm*\Delta{x}$

$4\:N/cm*\Delta{x} = 2\:N$

$\Delta{x} = \dfrac{2\:\diagup\!\!\!\!\!N}{4\:\diagup\!\!\!\!\!N/cm}$

simplify by 2

$\Delta{x} = \dfrac{2}{4}\frac{\div2}{\div2}$

$\boxed{\boxed{\Delta{x} = \dfrac{1}{2}\:cm}}\Longleftarrow(distance)\end{array}}\qquad\checkmark$

Answer:

B.) 1/2 cm

_______________________

I Hope this helps, greetings ... Dexteright02! =)

7 0

3 years ago

A gun fires a bullet vertically into a 1.40-kg block of wood at rest on a thin horizontal sheet. If the bullet has a mass of 22.

Bogdan [553]

Answer:

A. 1.172 metres

B. 6.82 Ns

C. 4.796 m/s

Explanation:

The total initial momentum is gotten by multiplying the mass and initial velocity of the both bodies.

The 1.40 kg block is at rest so velocity is zero and has no momentum.

The bullet of mass 22 g = 0.022 kg with velocity of 310 m/s

Momentum = 310*0.022

Momentum = 6.82 Ns.

If the bullet gets embedded they will both have common velocity v

6.82 = (0.022+1.40)v

6.82 = 1.422v

V = 6.82/1.422

V = 4.796 m/s

How high the block will rise after the bullet is embedded is given by

H = (U²Sin²tita)/2g

Where tita is 90°

H = (4.796² * sin²(90))/(2*9.81)

H =( 23.001616*1)/19.62

H = 1.172 metres

6 0

3 years ago

As players speed up, their kinetic energy will?

igomit [66]

Kinetic energy is proportional to the square of the speed. So when anything or anybody speeds up, its kinetic energy increases.

6 0

3 years ago

Read 2 more answers