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3 years ago

Calculate the force when the moment is 40nm and distance is 4m

Physics

1 answer:

Tems11 [23]3 years ago

3 0

the answer is 10 because you divide 40 by 4

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Explanation:

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3 years ago

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3 years ago

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12. The diameter of a circle is 2.42m. Calculate its area in proper significant figure

Sever21 [200]

Answer:

A = 4.6 [m²]

Explanation:

The area of a circle can be calculated by means of the following equation.

$A=\frac{\pi }{4} *D^{2}$

where:

A = area [m²]

D = diameter = 2.42 [m]

Now replacing:

$A=\frac{\pi }{4} *(2.42)^{2} \\A = 4.6 [m^{2} ]$

7 0

2 years ago

Represent 7468 N with SI units having an appropriate prefix. Express your answer to four significant figures and include the app

sergeinik [125]

Answer:

7.468 kN

Explanation:

Here the force of 7468 Newton is given.

Some of the prefixes of the SI units are

kilo = 10³

Mega = 10⁶

Giga = 10⁹

The number is 7468.0

Here, the only solution where the number of significant figures is kilo

1 kilonewton = 1000 Newton

$1\ Newton=\frac{1}{1000}\ kilonewton$

$\\\Rightarrow 7468\ Newton=\frac{7468}{1000}\ kilonewton\\ =7.468\ kilonewton$

So 7468 N = 7.468 kN

3 0

3 years ago

Find the deBroglie wavelength of an electron with 4.0 eV.

pantera1 [17]

Explanation:

It is given that,

Voltage, $V=4 eV=4\times 1.6\times 10^{-19}\ V= 6.4\times 10^{-19}\ V$

De broglie wavelength in terms of voltage is given by :

$\lambda=\dfrac{h}{\sqrt{2meV} }$

m and e are the mass and charge on electron. So,

$\lambda=\dfrac{12.27}{\sqrt{V} }\ A$

$\lambda=\dfrac{12.27}{\sqrt{6.4\times 10^{-19}} }\ A$

$\lambda=1.53\times 10^{10}\ A$

$\lambda=1.53\ m$

So, the De broglie wavelength of an electron is 1.53 meters. Hence, this is the required solution.

3 0

3 years ago