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2 years ago

Which statement best describes the electric field?

Physics

1 answer:

OleMash [197]2 years ago

6 0

Answer:

Explanation:

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As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s. The circle is parallel to the xy pl

Lynna [10]

It's 60.... i have used the formula
L=r^2mw
and didn't use the value 0.75...

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3 years ago

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explain why when someone uses his thumb to push a pin into a block of wood the pressure on the wood is greater than the pressure

Charra [1.4K]

Answer:

the smaller the area the bigger the pressure

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3 years ago

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In 2000, NASA placed a satellite in orbit around an asteroid. Consider a spherical asteroid with a mass of 5.00�1015kg and a rad

irakobra [83]

Answer:

a). $v_{1}=13.49 \frac{m}{s}$

b). $v_{2}=17.54\frac{m}{s}$

Explanation:

Using the conservation of energy and the kinetic energy of the satellite around the asteroid can model the motion in

$\frac{1}{2}*m*v^2=\frac{G*M*m}{a_{o}}$

$v^2=\frac{2*G*M*m}{a_{o}}$

$G=6.673x10^{-11}\frac{m^3}{kg*s^2}$

$M=1.20x10^{16}kg$

a).

$a_{o}=8.8km*\frac{1000m}{1km}=8800m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{8800m}}$

$v_{1}=13.49 \frac{m}{s}$

b).

$a_{f}=5.2km*\frac{1000m}{1km}=5200m$

$v=\sqrt{\frac{2*6.673x10^{-11}\frac{m^3}{kg*s^2}*1.2x10^{10}kg}{5200m}}$

$v_{2}=17.54\frac{m}{s}$

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3 years ago

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Problem 12: A 20.0-m tall hollow aluminum flagpole is equivalent in strength to a solid cylinder 4.00 cm in diameter. A strong w

avanturin [10]

Answer:

The deformation in the pole due to force is 0.70 mm.

Explanation:

Given that,

Height = 20.0 m

Diameter = 4.00 cm

Force = 1100 N

We need to calculate the area

Using formula of area

$A=\pi\times r^2$

$A=\pi\times(2.00\times10^{-2})^2$

$A=0.00125\ m^2$

$A=1.25\times10^{-3}\ m^2$

We need to calculate the deformation

Using formula of deformation

$\Delta x=\dfrac{1}{s}(\dfrac{F}{A}\times L)$

Where, s = shear modulus

F = force

l = length

A = area

Put the value into the formula

$\Delta x=\dfrac{1}{2.5\times10^{10}}\times(\dfrac{1100}{1.25\times10^{-3}}\times 20.0)$

$\Delta x=0.000704\ m$

$\Delta x=7.04\times10^{-4}\ m$

$\DElta x=0.70\ mm$

Hence, The deformation in the pole due to force is 0.70 mm.

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3 years ago

How does a PH probe work?

ziro4ka [17]

Answer:

It works by testing the ph levels in those areas it is placed in.

Explanation:

5 0

3 years ago