Ask question

Ask question

All categories

3 years ago

8

What is the mass of one sheet of printer paper?

1 answer:

Sloan [31]3 years ago

4 0

A standard sheet of laser-printer paper of the usual thickness, either US "letter" size or international "A4" size, weighs 4.5 grams. That means 100 sheets weighs almost exactly one pound.

You might be interested in

How has technology impacted our society? Describe both positive and negative effects.

KIM [24]

Positive : The development and adoption of technology have helped societies raise productivity, inclusivity of services and improve overall well-being.

Negative: The negative impact of tech on society involves mass-made products, with most often blamed: social media.

8 0

2 years ago

Bending the stick stretches the chemical bonds holding the stick's atoms together, which provides a source of elastic energy.

Tema [17]

Answer:

true

Explanation:

3 0

3 years ago

Calculate the mass of air in a room if the floor dimensions are 10 mx

aleksklad [387]

<em>Answer:</em>

<em>604.8 kg</em>

<em>Explanation:</em>

<em>Hi there ! </em>

<em>d = m/V => m = d×V</em>

<em>V = l×w×h = 10m×12m×4m = 480 m³</em>

<em>m = 1.26kg/m³×480m³ = 604.8 kg</em>

<em>Good luck !</em>

3 0

3 years ago

A long straight wire carries a 40.0 A current in the +x direction. At a particular instant, an electron moving at 1.0  107 m/s

ser-zykov [4K]

Answer:

$F=-1.28\times 10^{-16}\ N$

Explanation:

Given that,

Current flowing in the wire, I = 40 A (+x direction)

Speed of the electron, $v=10^7\ m/s$ (+y direction)

Distance from the wire, r = 0.1 m

Let F is the electric force on the electron. It is given by :

$F=qvB\ sin\theta$

Here, $\theta=90$

$F=qvB$

Here, $B=\dfrac{\mu_oI}{2\pi r}$

$F=qv\dfrac{\mu_oI}{2\pi r}$

$F=-1.6\times 10^{-19}\times 10^7\times \dfrac{4\pi\times 10^{-7}\times 40}{2\pi \times 0.1}$

$F=-1.28\times 10^{-16}\ N$

So, the force on the electron at this instant is $-1.28\times 10^{-16}\ N$ . Hence, this is the required solution.

5 0

4 years ago

Suppose you performed the experiment in atmosphere of Argon at 25 deg. C, (viscosity of argon is 2.26X10^-5 N.s/m^2 at that temp

yuradex [85]

Answer:

2*10^9electrons

Explanation:

Remember that the net force will be zero at terminal voltege so

Mg = 6πrng

At 35v

We have

qvr = 6πrng

q= 6 x 3.142* nx 2.6*10^-5/35

q,= 3.2x 10^ - 10C

So using n= q/e

= 3.2x 10^ - 10C/1.6*10-19

= 2*10^9electrons

7 0

3 years ago