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2 years ago

Why can you just change the code on a number lock?

Physics

2 answers:

jasenka [17]2 years ago

7 0

Answer:

because it has fingers

Explanation:

dimulka [17.4K]2 years ago

5 0

Answer:

Not that I know of. You might be able to but I don't think so.

Explanation:

Did you lose the code? If so, just go to your local locksmith.

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A rock is dropped from a sea cliff, and the sound of it striking the ocean is heard 3.2 s later. If the speed of sound is 340 m/

Anastaziya [24]

Speed of sound = 340 m/s
Time taken for sound to reach observer = 3.2 seconds
Height = speed x time
= 340 x 3.2
= 1,088 meters

5 0

3 years ago

P14.003A spherical gas-storage tank with an inside diameter of 8.1 m is being constructed to store gas under an internal pressur

Artist 52 [7]

Answer:

The minimum wall thickness required for the spherical tank is 0.0189 m

Explanation:

Given data:

d = inside diameter = 8.1 m

P = internal pressure = 1.26 MPa

σ = 270 MPa

factor of safety = 2

Question: Determine the minimum wall thickness required for the spherical tank, tmin = ?

The allow factor of safety:

$\sigma _{a} =\frac{\sigma }{factor-of-safety} =\frac{270}{2} =135MPa$

The minimun wall thickness:

$t=\frac{Pd}{4\sigma _{a} } =\frac{1.26*8.1}{4*135} =0.0189m$

3 0

3 years ago

What is the gravitational force between two identical 5000 kg asteroids whose centers of mass are separated by 100 m?

Alenkinab [10]

Answer: 1.67 x 10^-7N

Explanation:

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3 years ago

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Mariana [72]

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3 years ago

Read 2 more answers

A 2.0-cm-diameter parallel-plate capacitor with a spacing of 0.50 mm is charged to 200 V?What is the total energy stores in the

Rama09 [41]

1) $1.11\cdot 10^{-7} J$

The capacitance of a parallel-plate capacitor is given by:

$C=\frac{\epsilon_0 A}{d}$

where

$\epsilon_0$ is the vacuum permittivity

A is the area of each plate

d is the distance between the plates

Here, the radius of each plate is

$r=\frac{2.0 cm}{2}=1.0 cm=0.01 m$

so the area is

$A=\pi r^2 = \pi (0.01 m)^2=3.14\cdot 10^{-4} m^2$

While the separation between the plates is

$d=0.50 mm=5\cdot 10^{-4} m$

So the capacitance is

$C=\frac{(8.85\cdot 10^{-12} F/m)(3.14\cdot 10^{-4} m^2)}{5\cdot 10^{-4} m}=5.56\cdot 10^{-12} F$

And now we can find the energy stored,which is given by:

$U=\frac{1}{2}CV^2=\frac{1}{2}(5.56\cdot 10^{-12} F/m)(200 V)^2=1.11\cdot 10^{-7} J$

2) 0.71 J/m^3

The magnitude of the electric field is given by

$E=\frac{V}{d}=\frac{200 V}{5\cdot 10^{-4} m}=4\cdot 10^5 V/m$

and the energy density of the electric field is given by

$u=\frac{1}{2}\epsilon_0 E^2$

and using

$E=4\cdot 10^5 V/m$ , we find

$u=\frac{1}{2}(8.85\cdot 10^{-12} F/m)(4\cdot 10^5 V/m)^2=0.71 J/m^3$

7 0

4 years ago