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Dimas [21]
2 years ago
10

A lamp is labelled '230 V, 100 W'. How many joules of electrical energy is changed to thermal energy and light if the lamp is sw

itched on for 2 hours?
Physics
2 answers:
vovikov84 [41]2 years ago
8 0

Answer:

Formula  E(J) = P(W) × t(s)

3600 Seconds in an hour

7200 seconds in two hours.

watts = 100

100 x 7200 = 720000

Explanation:

Watts to joules calculation

The energy E in joules (J) is equal to the power P in watts (W), times the time period t in seconds (s):

E(J) = P(W) × t(s)

Energy cannot be created or destroyed, only converted from one form to another. These conversions can be shown in Sankey diagrams. Efficiency is a measure of how much useful energy is converted. Attachment 1

Modern energy-saving lamps and LEDs (light-emitting diodes) work in a different way. They transfer a greater proportion of electrical energy as light energy. Attachment 2

From the diagram, you can see that much less electrical energy is transferred, or 'wasted', as heat energy from the energy-saving lamp. It's more efficient than the filament lamp.

Calculating efficiency

The efficiency of a device, such as a lamp, can be calculated:

efficiency = useful energy out ÷ total energy in (for a decimal efficiency)

or

efficiency = (useful energy out ÷ total energy in) × 100 (for a percentage efficiency)

The efficiency of the filament lamp is 10 ÷ 100 = 0.10 (or 10%). This means that 10 per cent of the electrical energy supplied is transferred as light energy (90 per cent is transferred as heat energy).

The efficiency of the energy-saving lamp is 75 ÷ 100 = 0.75 (or 75 per cent). This means that 75 per cent of the electrical energy supplied is transferred as light energy (25 per cent is transferred as heat energy).

Note that the efficiency of a device will always be less than 100 per cent. Occasionally the power is shown in W instead of the energy in J. The equations are the same – just substitute power for energy:

efficiency = useful power out ÷ total power in (for a decimal efficiency)

or

efficiency = (useful power out ÷ total power in) × 100 (for a percentage efficiency)

This is the Sankey diagram for a typical energy-saving lamp:

you can find the resistance using this formula

R = V ÷ I    

where; V is the potential difference in volts, V

I is the current in amperes (amps), A

R is the resistance in ohms, Ω

viva [34]2 years ago
4 0

Given;

V = 230V

Power, P = 100W

time, t = 2hrs = 7200s

from,

P = IV

and Energy, E= Pt

E = 100*7200

E = 720000 Joules

E = 720KJ

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6 0
3 years ago
Can somebody help me please? 10 point!
Vlada [557]
The correct answer is the last one:
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A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar
m_a_m_a [10]

Answer:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

#Where u is in meters and t in seconds.

Explanation:

Given that :m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s

From \omega=kL we have:

k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2

From F_d(t)=-\gamma u\prime(t) we have that:

\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m

Now,given that the initial value problem is given by:

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\

Hence,the position of u at time t is given by

13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\, u in meters,t in seconds.

4 0
3 years ago
An object 5.Ocm in the length is placed at a distance of 20cm in front of convex mirror
andrew-mc [135]

Answer:

Position = \frac{60}{7}\ cm behind the mirror

Nature = Virtual and Erect

Size = \frac{15}{7}\ cm : Diminished

Explanation:

Sign convention-Distance measured to the left of pole is negative and to the right of pole is positive.

Object distance = u = -20 cm

Focal length = f = Radius of curvature/2 = 30/2 = 15 cm

We have to use mirror formula to find image distance.

\frac{1}{u}+\frac{1}{v}=\frac{1}{f}\\ \frac{1}{-20}+\frac{1}{v}=\frac{1}{15}\\ \frac{1}{v}=\frac{7}{60}\\v=\frac{60}{7}\ cm

Since the image distance is positive, it is formed behind the mirror or a virtual image is formed.

Magnification = =\frac{h_{image}}{h_{object}}=-\frac{v}{u}=\frac{60}{7\times20}=\frac{3}{7}

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4 0
3 years ago
The rotor in a certain electric motor is a flat, rectangular coil with 84 turns of wire and dimensions 2.61 cm by 3.64 cm. The r
ira [324]

Given Information:

Number of turns = N = 84

Area of Rectangular coil = 2.61x3.64 cm = 0.0261x0.0364 m

Magnetic field = B = 0.80 T

Current = I = 10.5 mA = 0.0105 A

Angular speed = ω = 3.54x10³ rev/min

Required Information:

(a) Maximum torque = τmax = ?

(b) Peak output power = Ppeak = ?

(c) Work done = W = ?

(d) Average power = Pavg?

Answer:

(a) Maximum torque = 0.00067 N.m

(b) Peak output power = 0.248 W

(c) Work done = 0.00189 J

(d) Average power = 0.1115 W

Explanation:

(a) The toque τ acting on the rotor is given by,

τ = NIABsin(θ)

Where N is the number of turns, I is the current, A is the area of rectangular coil and B is the magnetic field

A = 0.0261x0.0364

A = 0.00095 m²

The maximum toque τ is achieved when θ = 90°

τmax = NIABsin(90°)

τmax = 84*0.0105*0.00095*0.80*1

τmax = 0.00067 N.m

(b) The peak output power of the motor is given by,

Pmax = τmax*ω

ω = 3.54x10³ x 2π/60

ω = 370.7 rad/sec

Pmax = 0.00067*370.7

Pmax = 0.248 W

(c) The amount of work done by the magnetic field on the rotor in every full revolution is given by

W = 2∫NIABωsin(ωt) dt

W = -2NIABcos(ωt)

Evaluating limits,

W = -2NIABcos(π) - (-2NIABcos(0))

W = 2NIAB + 2NIAB

W = 4NIAB

W = 4*84*0.0105*0.00067*0.80

W = 0.00189 J

(d) Average power of the motor is given by

Pavg = W/t

t = 2π/ω

t = 2π/370.7

t = 0.01694 sec

Pavg = W/t

Pavg = 0.00189/0.01694

Pavg = 0.1115 W

4 0
3 years ago
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