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2 years ago

11

If an object is an irregularly shaped solid and it is dropped into a graduated cylinder and it displaces 25 mL of water and has

a mass of 50 grams what is the density of this irregular shaped object.

1 answer:

zavuch27 [327]2 years ago

7 0

Answer:

1250

Explanation:

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An electron is pushed into an electric field where it acquires a 1-V electrical potential. Suppose instead that two electrons ar

sleet_krkn [62]

Answer:

0.5 V

Explanation:

The electric potential distance between different locations in an electric field area is unaffected by the charge that is transferred between them. It is solely dependent on the distance. Thus, for two electrons pushed together at the same distance into the same field, the electric potential will remain at 1 V. However, the electric potential of one of the two electrons will be half the value of the electric potential for the two electrons.

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3 years ago

What are the importance of regulare<br>health examination.

Xelga [282]

Answer:

The purpose of regular health examination is to evaluate health status, screen for risk factors and disease, and provide preventive counseling interventions. The major benefits of regular health examination is early detection of treatable disease.

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2 years ago

For installation with a 25-kVA, 3-phase transformer, a 440-volt primary, and a 120-volt secondary. Calculate the maximum overcur

andre [41]

Answer:

41.053 A

Explanation:

given,

three phase kVA = 25-kVA

voltage = 440 Volt

current = ?

To determine three phase kVA when volts and amperes are know

three phase kVA = 1.73 x V x I

$I = \dfrac{three\ phase\ kVA}{1.73 \times V}$

$I = \dfrac{25000}{1.73 \times 440}$

I = 32.84 A

the maximum overcurrent protection value is equal to

= 125 % of I

= 1.25 x 32.84

= 41.053 A

4 0

3 years ago

Two square air-filled parallel plates that are initially uncharged are separated by 1.2 mm, and each of them has an area of 190

julia-pushkina [17]

Answer:

$5.5\cdot 10^{-11} C$

Explanation:

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

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

where

$\epsilon_0 = 8.85\cdot 10^{-12} F/m$ is the vacuum permittivity

$A=190 mm^2 = 190 \cdot 10^{-6} m^2$ is the area of the plates

$d=1.2 mm = 0.0012 m$ is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}) \frac{190 \cdot 10^{-6}}{0.0012}=1.4\cdot 10^{-12}F$

The energy stored in the capacitor is given by

$U=\frac{Q^2}{2C}$

Since we know the energy

$U=1.1 nJ = 1.1 \cdot 10^{-9} J$

we can re-arrange the formula to find the charge, Q:

$Q=\sqrt{2UC}=\sqrt{2(1.1\cdot 10^{-9} J)(1.4\cdot 10^{-12}F )}=5.5\cdot 10^{-11} C$

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2 years ago

Ask Your Teacher In the air over a particular region at an altitude of 500 m above the ground, the electric field is 120 N/C dir

strojnjashka [21]

Answer:

$1.475\times 10^{-13}\ C/m^3$

Explanation:

$\epsilon_0$ = Permittivity of free space = $8.85\times 10^{-12}\ F/m$

A = Area

h = Altitude = 600 m

Electric flux through the top would be

$-110A$ (negative as the electric field is going into the volume)

At the bottom

$120A$

Total flux through the volume

$\phi=120-110\\\Rightarrow \phi=10A$

Electric flux is given by

$\phi=\dfrac{q}{\epsilon_0}\\\Rightarrow q=\phi\epsilon_0\\\Rightarrow q=10A\epsilon_0$

Charge per volume is given by

$\rho=\dfrac{q}{v}\\\Rightarrow \rho=\dfrac{10A\epsilon_0}{Ah}\\\Rightarrow \rho=dfrac{10\epsilon_0}{h}\\\Rightarrow \rho=\dfrac{10\times 8.85\times 10^{-12}}{600}\\\Rightarrow \rho=1.475\times 10^{-13}\ C/m^3$

The volume charge density is $1.475\times 10^{-13}\ C/m^3$

7 0

3 years ago