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arlik [135]
3 years ago
10

If you have a book "sitting" on a table, the two forces acting on the book will be its

Physics
2 answers:
9966 [12]3 years ago
5 0
Yes there would be forces acting in the book
pishuonlain [190]3 years ago
3 0

Answer:Yes, there are forces acting on the book.

If the book is kept on a flat table then the forces acting on the book are Balanced in nature.

The weight of the book(=mg) acting downwards and the Normal Reaction acting upwards. The balanced nature of these 2 forces causes book to stay still on the table!

Explanation:

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Find the weight of a 2000 kg elaphant
Vika [28.1K]

19,600 Newtons (about 4,400 pounds).

On Earth only.

Different in other places.

5 0
3 years ago
When charges qa, qb, and qc are placed respectively at the corners a, b, and c of a right triangle, the potential at the midpoin
earnstyle [38]

Answer:

8v

Explanation:

First we apply super position principle

Vt= v1 + v2+ v3

Remove qa

But vt= 20v

So V = v2+v3

V1= 20-15

= 5v

Remove qb

V= v1+v3

V=8v

So the potential when qa and qc are remove is the potential due to qb

Which is 8v

7 0
3 years ago
What's the third prong on an electric plug called??
zaharov [31]
It is an extension of the grounding system
5 0
3 years ago
a nonrelativistic proton is confined to a length of 2.0 pm on the x-axis. what is the kinetic energy of the proton if its speed
Elina [12.6K]

Answer:

K = 1.29eV

Explanation:

In order to calculate the kinetic energy of the proton you first take into account the uncertainty principle, which is given by:

\Delta x \Delta p\geq \frac{h}{4\pi}      (1)

Δx : uncertainty of position = 2.0pm = 2.0*10^-12m

Δp: uncertainty of momentum = ?

h: Planck's constant = 6.626*10^-34 J.s

You calculate the minimum possible value of Δp from the equation (1):

\Delta p=\frac{h}{4\pi \Delta x}=\frac{6.626*10^{-34}J.s}{4\pi(2.0*10^{-12}m)}\\\\\Delta p=2.63*10^{-23}kg.\frac{m}{s}

The minimum kinetic energy is calculated by using the following formula:

k=\frac{(\Delta p)^2}{2m}       (2)

m: mass of the proton = 1.67*10^{-27}kg

k=\frac{(2.63*10^{-23}kgm/s)^2}{2(1.67*10^{-27}kg)}=2.08*10^{-19}J

in eV you have:

2.08*10^{-19}J*\frac{6.242*10^{18}eV}{1J}=1.29eV

The kinetic energy of the proton is 1.29eV

7 0
3 years ago
A group of science and engineering students embarks on a quest to make an electrostatic projectile launcher. For their first tri
vekshin1

Electric charge on the plastic cube: 1.3\cdot 10^{-7}C

Explanation:

The electric potential around a charged sphere (such as the Van der Graaf) generator is given by

V(r)=\frac{kQ}{r}

where

k is the Coulomb's constant

Q is the charge on the sphere

r is the distance from the centre of the sphere

Here we have:

V = 200,000 V on the surface of the sphere, so at r = 12.0 cm

We need to find the voltage V' at 2.0 cm from the edge of the sphere, so at

r' = 12.0 + 2.0 = 14.0 cm

Since the voltage is inversely proportional to r, we can use:

Vr=V'r'\\V'=\frac{Vr}{r'}=\frac{(200,000)(12.0)}{14.0}=171,429 V

This is the potential at the location of the plastic cube.

Now we can use the law of conservation of energy, which states that the initial electric potential energy of the cube is totally converted into kinetic energy when the plastic cube is at infinite distance from the generator. So we can write:

qV' = \frac{1}{2}mv^2

where:

q is the charge on the plastic cube

V' is the potential at the location of the cube

m = 5.0 g = 0.005 kg is the mass of the cube

v = 3.0 m/s is the final speed of the cube

Solving for q, we find the charge on the cube:

q=\frac{mv^2}{2V'}=\frac{(0.005)(3.0)^2}{2(171,429)}=1.3\cdot 10^{-7}C

Learn more about electric fields:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
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