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3 years ago

1. An electric iron has a rating of 750W, 220V. Calculate:

1 answer:

6 0

The equation for electrical power isP=VIwhere V is the voltage and I is the current. This can be rearranged to solve for I in 6(a). 6(b) can be solved with Ohm's LawV=IRor if you'd like, from power, after substituting Ohm's law in for IP=V2R For 7, realize that because they are in parallel, their voltages are the same. We can find the resistance of each lamp fromP=V2RThen the equivalent resistance as1R∗=1R1+1R2Then the total power asPt=V2R∗However, this will reveal that (with a bit of algebra)Pt=P1+P2 For 8, again the resistance can be found asP=V2RThe energy usage is simplyE=P⋅t

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A plane designed for vertical takeoff has a mass of 8.0 × 10³ kg. Find the net work done by all forces on the plane as it accele

artcher [175]

Answer:

the net work done after starting from rest is = 2.4 × 10⁵ J

Explanation:

Work: Work can be defined as the product of force and distance. The fundamental unit of work is Joules (J), The unit of Energy is Joules (J), as such Energy and work are interchangeable during calculation, This is illustrated below

E = W = 1/2mv².......................... Equation 1

Where m = mass of the plane, v = velocity of the plane, E = Energy, W = work done.

v² = u² + 2as ................................. Equation 2.

Where v = final velocity of the plane, u = initial velocity of the plane, a = acceleration of the plane, distance of the plane.

Given: a = 1.0 m/s², s = 30 m, u = 0 m/s (at rest)

Substituting these values into equation 2

v² = 0² +2×1×30

v² = 60

v = √60

v = 7.75 m/s

Also given: m = 8.0 × 10³ kg, and v = 7.75 m/s

Substituting these values into equation 1,

W = 1/2(8.0×10³)(7.75)²

W = (4.0×10³)(60)

W = 240 × 10³ J

W = 2.4 × 10⁵ J

Therefore the net work done after starting from rest is = 2.4 × 10⁵ J

4 0

3 years ago

Waves that move the particles of the medium parallel to the direction in which the waves are traveling are called

Nikolay [14]

1. a. longitudinal waves.

There are two types of waves:

- Transverse waves: in transverse waves, the oscillations of the wave occur in a direction perpendicular to the direction of propagation of the wave

- Longitudinal waves: in longitudinal waves, the oscillations of the waves occur parallel to the direction in which the waves are travelling.

So, these types of waves are called longitudinal waves.

2. d. a medium

There are two types of waves:

- Electromagnetic waves: these waves are produced by the oscillations of electric and magnetic field, and they can travel both in a medium and also in a vacuum (they do not need a medium to propagate)

- Mechanical waves: these waves are produced by the oscillations of the particles in a medium, so they need a medium to propagate - therefore, the correct choice is d. a medium

3. a. AM/FM radio

Analogue signals consist of continuous signals, which vary in a continuous range of values. On the contrary, digital signals consist of discrete signals, which can assume only some discrete values. For AM and FM radios, signals are transmitted by using analogue signals.

5 0

2 years ago

3. A cat pushes a 0.25-kg toy with a net force of 8 N. According to Newton's second

jek_recluse [69]

Mass=0.25kg
Force=8N

$\\ \sf{:}\!\implies F=ma$

$\\ \sf{:}\!\implies Acceleration=\dfrac{F}{m}$

$\\ \sf{:}\!\implies Acceleration=\dfrac{8}{0.25}$

$\\ \sf{:}\!\implies Acceleration=32m/s^2$

5 0

3 years ago

What is the importance of star ranting labelilling of an electrical appliance?

Misha Larkins [42]

Answer:

The standard of efficiency varies by product category

these are called BEE star label and they show how much electricity the appliance consumes in year each appliance gets btw one and five stars with five stars meaning that it's extremely efficienct andis likely to keep your electricity bills in check

 hope it's helpful

5 0

2 years ago

How many electrons flow through a point in a wire in 7.00 s if there is a constant current of I = 4.35 A?

OLEGan [10]

Answer:

1.90×10²⁰ Electrons

Explanation:

From the question,

Q = It.................... Equation 1

Where Q = charge flowing through the wire, I = current, t = time

Given: I = 4.35 A, t = 7.00 s

Substitute these values into equation 1

Q = 4.35(7.00)

Q = 30.45 C.

But,

1 electron contains 1.6×10⁻¹⁹ C

therefore,

30.45 C = 30.45/1.6×10⁻¹⁹ electrons

= 1.90×10²⁰ Electrons

8 0

3 years ago