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3 years ago

What is the mass of a plastic spoon

Physics

2 answers:

mafiozo [28]3 years ago

8 0

Answer:

i would think 1.3 g

Explanation:

m_a_m_a [10]3 years ago

3 0

The mass of a plastic spoon

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Which of the following statements concerning moral development is not true?

Goshia [24]

Answer:

C- what is determined as right or wrong is universally defined and agreed upon.

Explanation:

Correct on edge

6 0

3 years ago

A cylinder with a moving piston expands from an initial volume of 0.250 L against an external pressure of 2.00 atm. The expansio

Step2247 [10]

The final volume of the gas is 144.25 L

Explanation:

For an ideal gas kept at constant pressure, the work done by the gas on the surroundings is given by

$W=p\Delta V = p(V_f - V_i)$

where

p is the pressure of the gas

$V_i$ is the initial volume

$V_f$ is the final volume

For the gas in the cylinder in this problem,

p = 2.00 atm

$V_i = 0.250 L$

And we also know the work done,

W = 288 J

So we can solve the equation for $V_f$ , the final volume:

$V_f = V_i + \frac{W}{p}=0.250 + \frac{288}{2.00}=144.25 L$

Learn more about ideal gases:

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7 0

3 years ago

On a clear day at a certain location, a 119-V/m vertical electric field exists near the Earth's surface. At the same place, the

IrinaVladis [17]

Answer:

(a) 62.69 nJ/m^3

(b) 1015.22 μJ/m^3

Explanation:

Electric field, E = 119 V/m

Magnetic field, B = 5.050 x 10^-5 T

(a) Energy density of electric field = $\frac{1}{2}\varepsilon _{0}E^{2}$

$=\frac{1}{2}\times 8.854\times 10^{-12}\times 119\times 119$

= 6.269 x 10^-8 J/m^3 = 62.69 nJ/m^3

(b) energy density of magnetic field = $\frac{B^{2}}{2\mu _{0}}$

$=\frac{\left ( 5.05\times 10^{-5} \right )^{2}}{2\times 4\times 3.14\times 10^{-7}}$

= 1.01522 x 10^-3 J/m^3 = 1015.22 μJ/m^3

8 0

3 years ago

A capacitor C is fully charged by connecting it to battery of V Volt. Then it is disconnected from battery. If the separation be

vodomira [7]

Answer:

Explanation:

i )

When it is disconnected with the battery , the charge stored in it becomes fixed . When the plate distance becomes half , its capacitance becomes twice from C to 2C . Let charge stored in it at the time of disconnection from battery be Q . Let plate separation reduces from d to d / 2

So charged stored in it will remain unchanged .

ii )

Potential difference = charge / capacitance

in the first case potential difference = Q / C

in the second case potential difference = Q / 2C

So potential difference becomes half .

iii ) electric field = potential diff / plate separation

in the first case electric field = Q / (d x C )

in the second case electric field = 2 Q / (d x 2C)

= Q / (d x C )

So electric field remains unchanged .

iv)