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svlad2 [7]
3 years ago
15

In order to generate electricity from mechanical energy, one essential element needed to transform the energy into electrical en

ergy is?
A) a steam turbine.
B) a mechanical linkage.
C) a large piece of iron.
D) a magnet or magnetic field.
Physics
2 answers:
Delvig [45]3 years ago
6 0
Hey there!

In order to generate electricity from mechanical energy, one essential element needed to transform the energy into electrical energy is a magnet or magnetic field

Hope this helps! :)
frutty [35]3 years ago
6 0

Answer:

D) a magnet or magnetic field.

Explanation:

Electricity can be generated from mechanical energy due to a phenomenon called electromagnetic induction.

This phenomenon occurs when the magnetic flux (due to the presence of a magnetic field) through an enclosed circuit is changed: when this happens, an electromotive force is induced in the circuit, and as a consequence, an electrical current is induced in the circuit.

This phenomenon can be produced in different ways; for example:

1- By rotating a coil in a fixed magnetic field --> by rotating the coil, the magnetic flux through the circuit changes over time, so a current is induced in the circuit

2- By varying the intensity of the magnetic field over time --> this way, the magnetic flux through the circuit changes as well

Number #1 is used in electrical generators to produce electrical energy from mechanical energy (kinetic energy): in fact, a coil is made rotating (so, kinetic energy) in a magnetic field, producing an electrical current (electrical energy). It is clear that an essential element of this system is the presence of a magnet or a magnetic field, so D) is the correct answer.

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galina1969 [7]

Answer:

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Explanation:

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3 years ago
An iron ball and an aluminum ball of mass 100 g each are heated to the same temperature and then cooled to a temperature of 20°
MrRa [10]

Answer:

The specific heat of aluminum is greater.

Explanation:

It lost the most heat.

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3 years ago
Read 2 more answers
Consider two identical objects of mass m = 0.250 kg and charge q = 4.00 μC. The first charge is held in place at the origin of a
Gnom [1K]

Answer:

a = 640 m/s²

Explanation:

From work-kinetic energy principles,

The net force acting on the second object is the gravitational force and the electric force due to the first object.

So, the gravitational force on the mass is F₁ = Gm₁m₂/r² since m₁ = m₂ = m, U = -Gm²/r²

Also, the electric force on the charge is F₂ = kq₁q₂/r² since q₁ = q₂ = q, U = kq²/r²

The net Force F = ma

So, -F₁ + F₂ = F     (F₁ is negative since it is an attractive force in the negative x -direction and F₂ is positive since it is a repulsive force in the positive x- direction)

-Gm²/r² + kq²/r² = ma

ma = -Gm²/r² + kq²/r²

a = (-Gm²/r² + kq²/r²)/m

a = (-G + kq²/m²)m/r²

Since m = 0.250 kg, q = 4.00 μC = 4.00 × 10⁻⁶ C, r = 3.00 cm = 3.00 × 10⁻² m, G = 6.67 × 10⁻¹¹ Nm²/kg², k = 9 × 10⁹ Nm²/C² and a = acceleration of second mass.

Substituting the variables into the equation, we have

a = (m/r²)(-G + k(q/m)²)]

a = (0.250 kg/{3.00 × 10⁻² m}²)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(4.00 × 10⁻⁶ C/0.250 kg)²)

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(16 × 10⁻⁶ C/kg)²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 9 × 10⁹ Nm²/C²(256 × 10⁻¹² C²/kg²)]

a = (0.250 kg/9.00 × 10⁻⁴ m)(-6.67 × 10⁻¹¹ Nm²/kg² + 2304 × 10⁻³ Nm²/kg²  ]

a = (0.250 kg/9.00 × 10⁻⁴ m)(2.304 Nm²/kg²)

a = 0.576 Nm²/kg /9.00 × 10⁻⁴ m²

a = 0.064 × 10⁴N/kg

a = 64 × 10 N/kg)

a = 640 m/s²

8 0
3 years ago
The robot arm is elevating and extending simultaneously. At a given instant, θ = 30°, ˙ θ = 10 deg / s = constant θ˙=10 deg/s=co
motikmotik

Explanation:

The position vector r:

\overrightarrow{r(t)}=lcos\theta\hat{i}+lsin\theta\hat{j}

The velocity vector v:

\overrightarrow{v(t)}=\overrightarrow{\frac{dr}{dt}}=\dot{l}cos\theta-lsin\theta\dot{\theta}\hat{i}+\dot{l}sin\theta+lcos\theta\dot{\theta}\hat{j}

The acceleration vector a:

\overrightarrow{a(t)}}=cos\theta(\ddot{l}-l\dot{\theta}^2)-sin\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})\hat{i}+cos\theta(2\dot{l}\dot{\theta}+l\ddot{\theta})+sin\theta(\ddot{l}-l\dot{\theta}^2)\hat{j}

\overrightarrow{v(t)}=0.13\hat{i}+0.18\hat{j}

\overrightarrow{a(t)}}=-0.3\hat{i}-0.1\hat{j}

5 0
3 years ago
A gold wire that is 1.8 mm in diameter and 15 cm long carries a current of 260 mA. How many electrons per second pass a given cr
Musya8 [376]

Answer:

162500000.  

Explanation:

Given that

Diameter of the wire , d= 1.8 mm

The length of the wire ,L = 15 cm

Current ,I = 260 m A

The charge on the electron ,e= 1.6 x 10⁻¹⁹ C

We know that Current I is given as

I=\dfrac{q}{t}

I=Current

q=Charge

t=time

q= I t

q= 260 m t

The total number of electron = n

q= n e

n=\dfrac{260\times 10^{-3}\ t}{1.6\times 10^{-9}}

n=162500000 t

\dfrac{n}{t}=16250000

The number of electron passe per second will be 162500000.

4 0
3 years ago
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