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3 years ago

8

What is the high low and medium friction force on a box

1 answer:

dexar [7]3 years ago

7 0

Answer:

answer

hope it helps you

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BRAINLIESTTT!! HELP ME!!!

likoan [24]

The efficiency of the scissor is 200%.

<u>Explanation:</u>

Efficiency is defined as the ratio of output of any instrument or device or machine to the input supplied to it. So the greater the output the greater will be the efficiency of the device.

As here the work done by us on the system is said to be 10 J so this will be equal to the input work done on the system. And the work done by the system i.e., the scissor is 20 J, so this will be the output work.

So, the efficiency is the ratio of output to input as shown below.

Efficiency = $(\frac {20} {10} ) \times 100$ = 200

So, the efficiency of the scissor is 200%.

6 0

4 years ago

CAN SOMEONE DO THIS FOR ME!!!!!!!!!! PLEASEEEEEEEEEE

attashe74 [19]

We can’t see the attachment :(

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3 years ago

Read 2 more answers

In the simple Bohr model of the eighth excited state of the hydrogen atom, the electron travels in a circular orbit around a fix

Ksju [112]

Answer:

Net force on electron will be $12.568\times 10^{-12}N$

Explanation:

We have given that the orbit $r=4.28\times 10^{-9}m$

Speed of the electron $v=2.43\times 10^5m/sec$

Mass of the electron is given as $m=9.11\times 10^{-31}kg$

Centripetal acceleration is given by $a_c=\frac{v^2}{r}=\frac{(2.43\times 10^5)^2}{4.28\times 10^{-9}}=1.3796\times 10^{19}m/sec^2$

Force is given by F = ma

So force $F=9.11\times 10^{-31}\times 1.3796\times 10^{19}=12.568\times 10^{-12}N$

8 0

3 years ago

You are working at a company that manufactures electrical wire. Gold is the most ductile of all metals: it can be stretched into

devlian [24]

Answer: $R=1.424 M\Omega$

Explanation:

Given

mass of gold $m=1.6 gm$

Length of wire $L=2.2 km$

Resistivity of gold $\rho =2.44\times 10^{-8}$

density of gold $=19.3\times 10^3 kg/m^3$

and $mass=volume\times density$

$1.6\times 10^{-3}=volume\times 19.3\times 10^3$

$volume=8.29\times 10^{-8} m^3$

And $Resistance R=\frac{\rho L}{A}$

also be written as

$R=\frac{\rho L^2}{V}$

where $L=length$

$V=volume$

$\rho =resistivity\ of\ gold$

$R=\frac{2.44\times 10^{-8}\times (2200)^2}{8.29 \times 10^{-8}}$

$R=1.42\times 10^{6} \Omega$

$R=1.424 M\Omega$

5 0

3 years ago

I need SO much help please!!!!!!!!~!

bulgar [2K]

Hi,

I've found a link that should assist you or answer your question.
http://click.dji.com/ANbvbbP7bwUWtSACp6U_?pm=link&as=0004

Have a nice day!

3 0

4 years ago