All categories

3 years ago

Which of the following electrical components is a temporary electrical energy storage device?

Physics

1 answer:

7nadin3 [17]3 years ago

6 0

Answer:

A capacitor

Explanation:

Because it can store electric energy when disconnected from its charging circuit. Commonly used in electronic devices to maintain power supply while batteries change.

Hope this helps! :)

You might be interested in

Which of the following solutions will have the lowest freezing point?

gregori [183]

C. 1.0 mol/kg sodium phosphide (Na3P)

This is because it will produce 4 ions, and thus has the highest molalitiy. (more particles in solution lowers the freezing point)

remember if you ever need anything and you dont want to waste points you can just post it on my pfp or message me the question

7 0

4 years ago

A wire of resistance 7 ohms and length 2.8 m is bent into a circle and is concentric with a solenoid in which the magnetic flux

Serggg [28]

Answer:

The emf, electric field and the current in the wire are 10 V, 3.57 V/m and 1.43 A.

Explanation:

Given that,

Resistance = 7 ohms

Length = 2.8 m

Time t =0.2

We need to calculate the change in magnetic flux

Using formula of induced emf

$\epsilon=\dfrac{d\phi}{dt}$

Put the value into the formula

$\epsilon=\dfrac{6-4}{0.2}$

$\epsilon=10\ V$

We need to calculate the electric field in the wire

Using formula of electric field

$E=\dfrac{V}{l}$

$E=\dfrac{10}{2.8}$

$E=3.57\ V/m$

We need to calculate the current in the wire,

Using formula of ohm's law

$V=IR$

$I=\dfrac{V}{R}$

Put the value into the formula

$I=\dfrac{10}{7}$

$I=1.43\ A$

Hence, The emf, electric field and the current in the wire are 10 V, 3.57 V/m and 1.43 A.

7 0

3 years ago

A horizontal clothesline is tied between 2 poles, 16 meters apart. When a mass of 3 kilograms is tied to the middle of the cloth

Alla [95]

Answer:

The magnitude of the tension on the ends of the clothesline is 41.85 N.

Explanation:

Given that,

Poles = 2

Distance = 16 m

Mass = 3 kg

Sags distance = 3 m

We need to calculate the angle made with vertical by mass

Using formula of angle

$\tan\theta=\dfrac{8}{3}$

$\thta=\tan^{-1}\dfrac{8}{3}$

$\theta=69.44^{\circ}$

We need to calculate the magnitude of the tension on the ends of the clothesline

Using formula of tension

$mg=2T\cos\theta$

Put the value into the formula

$3\times9.8=2T\times\cos69.44$

$T=\dfrac{3\times9.8}{2\times\cos69.44}$

$T=41.85\ N$

Hence, The magnitude of the tension on the ends of the clothesline is 41.85 N.

4 0

3 years ago

Read 2 more answers

engineers are using computer models to study train collisions design safer train cars. They start by modeling an elastic collisi

Archy [21]

Answer:

The final velocity of the second car is 57 m/s south.

Explanation:

This is an elastic collision between two train cars. In this case, the total kinetic energy between the two bodies will remain the same.

The formula to apply is :

$m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f$

where ;

$m_1=mass of object 1\\v_1i=initial velocity object1\\m_2=mass of object2\\v_2i=initial velocity object 2\\v_1f=final velocity object 1\\v_2f=final velocity object 2$

Given in the question that;

$m_1=14650kg\\v_1i=18m/s\\m_2=3825kg\\v_2i=11m/s\\v_1f=6m/s\\v_2f=?$

Apply the formula as;

$m_1v_1i +m_2v_2i=m_1v_1f+m_2v_2f$

{14650*18}+{3825*11} = {14650 *6} + {3825 * v₂f}

263700+42075=87900 + 3825v₂f

305775 =87900 + 3825v₂f

305775-87900 = 3825v₂f

217875=3825v₂f

217875/3825 =v₂f

56.96 = v₂f

57 m/s = v₂f { nearest whole number}

4 0

3 years ago

Read 2 more answers

A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the

Gennadij [26K]

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

Goodluck

Explanation:

4 0

2 years ago