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1 year ago

Which of the following best describes a capacitor?

Physics

2 answers:

noname [10]1 year ago

7 0

Answer:

APEX two conductors seperated by an insulator

Explanation:

galben [10]1 year ago

4 0

Answer:

Explanation:

The capacitor is a component which has the ability to store energy in the form of an electrical charge making a potential difference on those two metal plates

A capacitor consists of two or more parallel conductive (metal) plates. They are electrically seperated by an insulating material (ex: air, mica,ceramic etc.) which is called as Dielectric Layer

Due to this insulating layer, DC current can not flow through the capacitor.But it allows a voltage to be present across the plates in the form of an electrical charge.

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An object’s velocity is measured to be vx(t) = α - βt2, where α = 4.00 m/s and β = 2.00 m/s3. At t = 0 the object is at x = 0. (

Leona [35]

Answer:

Explanation:

Given

$v_x(t)=\alpha -\beta t^2$

$\alpha =4\ m/s$

$\beta =2\ m/s^3$

$v_x(t)=4-2t^2$

$v=\frac{\mathrm{d} x}{\mathrm{d} t}$

$\int dx=\int \left ( 4-2t^2\right )dt$

$x=4t-\frac{2}{3}t^3$

acceleration of object

$a=\frac{\mathrm{d} v}{\mathrm{d} t}$

$a=-4t$

(b)For maximum positive displacement velocity must be zero at that instant

i.e. $v=0$

$4-2t^2=0$

$t=\pm \sqrt{2}$

substitute the value of t

$x=4\times \sqrt{2}-\frac{2}{3}\times 2\sqrt{2}$

$x=3.77\ m$

7 0

1 year ago

Calculate the final temperature of a mixture of 0.350 kg of ice initially at 218°C and 237 g of water initially at 100.0°C.

kramer

Answer:

115 ⁰C

Explanation:

Step 1: The heat needed to melt the solid at its melting point will come from the warmer water sample. This implies

$q_{1} +q_{2} =-q_{3}$ -----eqution 1

where,

$q_{1}$ is the heat absorbed by the solid at 0⁰C

$q_{2}$ is the heat absorbed by the liquid at 0⁰C

$q_{3}$ the heat lost by the warmer water sample

Important equations to be used in solving this problem

$q=m *c*\delta {T}$ , where -----equation 2

q is heat absorbed/lost

m is mass of the sample

c is specific heat of water, = 4.18 J/0⁰C

$\delta {T}$ is change in temperature

Again,

$q=n*\delta {_f_u_s}$ -------equation 3

where,

q is heat absorbed

n is the number of moles of water

tex]\delta {_f_u_s}[/tex] is the molar heat of fusion of water, = 6.01 kJ/mol

Step 2: calculate how many moles of water you have in the 100.0-g sample

$=237g *\frac{1 mole H_{2} O}{18g} = 13.167 moles of H_{2}O$

Step 3: calculate how much heat is needed to allow the sample to go from solid at 218⁰C to liquid at 0⁰C

$q_{1} = 13.167 moles *6.01\frac{KJ}{mole} = 79.13KJ$

This means that equation (1) becomes

79.13 KJ + $q_{2} = -q_{3}$

Step 4: calculate the final temperature of the water

$79.13KJ+M_{sample} *C*\delta {T_{sample}} =-M_{water} *C*\delta {T_{water}$

Substitute in the values; we will have,

$79.13KJ + 237*4.18\frac{J}{g^{o}C}*(T_{f}-218}) = -350*4.18\frac{J}{g^{o}C}*(T_{f}-100})$

79.13 kJ + 990.66J* $(T_{f}-218})$ = -1463J* $(T_{f}-100})$

Convert the joules to kilo-joules to get

79.13 kJ + 0.99066KJ* $(T_{f}-218})$ = -1.463KJ* $(T_{f}-100})$

$79.13 + 0.99066T_{f} -215.96388= -1.463T_{f}+146.3$

collect like terms,

2.45366 $T_{f}$ = 283.133

∴ $T_{f} =$ = 115.4 ⁰C

Approximately the final temperature of the mixture is 115 ⁰C

6 0

2 years ago

A long thin uniform rod of length 1.50 m is to be suspended from a frictionless pivot located at some point along the rod so tha

Dvinal [7]

Answer:

0.087 m

Explanation:

Length of the rod, L = 1.5 m

Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.

time period, T = 3 s

the formula for the time period of the pendulum is given by

$T = 2\pi \sqrt{\frac{I}{mgd}}$ .... (1)

where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.

Moment of inertia of the rod about the centre of mass, Ic = mL²/12

By using the parallel axis theorem, the moment of inertia of the rod about the pivot is

I = Ic + md²

$I = \frac{mL^{2}}{12}+ md^{2}$

Substituting the values in equation (1)

$3 = 2 \pi \sqrt{\frac{\frac{mL^{2}}{12}+ md^{2}}{mgd}}$

$9=4\pi^{2}\times \left ( \frac{\frac{L^{2}}{12}+d^{2}}{gd} \right )$

12d² -26.84 d + 2.25 = 0

$d=\frac{26.84\pm \sqrt{26.84^{2}-4\times 12\times 2.25}}{24}$

$d=\frac{26.84\pm 24.75}{24}$

d = 2.15 m , 0.087 m

d cannot be more than L/2, so the value of d is 0.087 m.

Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.

3 0

1 year ago

A person travels distance πR along the circumference

lakkis [162]

Answer: 2R

Explanation:

Here the person travels пR distance. We know that the circumference of a circle is 2πR. So your imaginated person has traveled the distance which is half of the circumference of the circle. And this distance is equal to its diameter. We know that diameter of a circle is two times larger than the radius. So the person's displacement is two times of the radius, means 2R. [Here 'R' means the radius of the circle]

8 0

2 years ago

Read 2 more answers

Microwave ovens emit microwave energy with a wavelength of 12.6 cm. what is the energy of exactly one photon of this microwave r

Helga [31]

We need the frequency of the photon, it is v = c/ λ

Where c is 3 x 10^8 ms^-1 and λ is the wave length

We also need the expression of connecting frequency to energy of photon

which is E = hv where h is Planck’s constant

Combining the two equations will give us:

E = h x c/λ

Inserting the values, we will have:

E = 6.626 x 10^-34 x 3 x 10^8 / 0.126

E = 1.578 x 10^ -24 J

7 0

2 years ago

Read 2 more answers