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2 years ago

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Physics

2 answers:

baherus [9]2 years ago

8 0

$\huge\underline\mathtt\colorbox{cyan}{G=}$

$6.673 \times {10}^{ - 11}$

And unit is Nm^2/kg^2

crimeas [40]2 years ago

6 0

Explanation:

The attractive force between two bodies when multiplied by the product of the masses of two bodies and divided by the square of the distance between them.

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The CERN particle accelerator is circular with a circumference of 7.0 km.

Contact [7]

Answer:

$a_c=2.0196\times 10^{13}\ m/s^2$

$F=3.37273\times 10^{-14}\ N$

Explanation:

m = Mass of proton = $1.67\times 10^{-27}\ kg$

v = Speed of proton = 0.5c = $0.5\times 3\times 10^8=1.5\times 10^8\ m/s$

Circumference of the colider is 7 km

$P=2\pi r\\\Rightarrow r=\frac{P}{2\pi}\\\Rightarrow r=\frac{7000}{2\pi}\ m$

$a_c=\frac{v^2}{r}\\\Rightarrow a_c=\frac{\left(1.5\times 10^8\right)^2}{\frac{7000}{2\pi}}\\\Rightarrow a_c=2.0196\times 10^{13}\ m/s^2$

Centripetal acceleration is $2.0196\times 10^{13}\ m/s^2$

$F_c=ma_c\\\Rightarrow F_c=1.67\times 10^{-27}\times 2.0196\times 10^{13}\\\Rightarrow F=3.37273\times 10^{-14}\ N$

Force on protons is $3.37273\times 10^{-14}\ N$

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3 years ago

What is an example of gravitational potential to kinetic to electrical current?

Alecsey [184]

Chemical to thermal to electrical current: Burning of coal or natural gases. Gravitational potential to kinetic to electrical current.

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2 years ago

What body parts were scientists wanting to image that prompted the development of the CT scanner

zalisa [80]

Answer:

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Explanation:

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2 years ago

A 5.67-kg block of wood is attached to a spring with a spring constant of 150 N/m. The block is free to slide on a horizontal fr

nikklg [1K]

Answer:

v=0.94 m/s

Explanation:

Given that

M= 5.67 kg

k= 150 N/m

m=1 kg

μ = 0.45

The maximum acceleration of upper block can be μ g.

a= μ g ( g = 10 m/s²)

The maximum acceleration of system will ω²X.

ω = natural frequency

X=maximum displacement

For top stop slipping

μ g =ω²X

We know for spring mass system natural frequency given as

$\omega=\sqrt{\dfrac{k}{M+m}}$

By putting the values

$\omega=\sqrt{\dfrac{150}{5.67+1}}$

ω = 4.47 rad/s

μ g =ω²X

By putting the values

0.45 x 10 = 4.47² X

X = 0.2 m

From energy conservation

$\dfrac{1}{2}kX^2=\dfrac{1}{2}(m+M)v^2$

$kX^2=(m+M)v^2$

150 x 0.2²=6.67 v²

v=0.94 m/s

This is the maximum speed of system.

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2 years ago

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