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2 years ago

12

I understand what makes a kettle so efficient, but what are some improvements you can make to a kettle to increase it’s energy e

fficiency?

1 answer:

shusha [124]2 years ago

3 0

Answer:

Reheat the cold cup of tea or coffee in the microwave. ...

Make iced tea or coffee with what's left. ...

Use a thermos for either tea or coffee.

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A bullet is shot horizontally from shoulder height (1.5 m) with an initial speed 200 m/s. (a) How much time elapses before the b

daser333 [38]

Answer:

<h2>a) Time elapsed before the bullet hits the ground is 0.553 seconds.</h2><h2>b) The bullet travels horizontally 110.6 m</h2>

Explanation:

a) Consider the vertical motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 0 m/s

Acceleration, a = 9.81 m/s²

Displacement, s = 1.5 m

Substituting

s = ut + 0.5 at²

1.5 = 0 x t + 0.5 x 9.81 xt²

t = 0.553 s

Time elapsed before the bullet hits the ground is 0.553 seconds.

b) Consider the horizontal motion of bullet

We have equation of motion s = ut + 0.5 at²

Initial velocity, u = 200 m/s

Acceleration, a = 0 m/s²

Time, t = 0.553 s

Substituting

s = ut + 0.5 at²

s = 200 x 0.553 + 0.5 x 0 x 0.553²

s = 110.6 m

The bullet travels horizontally 110.6 m

6 0

2 years ago

A particle moves along a straight path through a displacement d = 2.5i + cj while a force F = 8.5i + -8.5j acts on it. The displ

Zanzabum

Answer:

Explanation:

Work is defined as the scalar product of force and distance

W=F•d

Given that

F = 8.5i + -8.5j. +×-=-

F=8.5i-8.5j

d = 2.5i + cj

If the work in the practice is zero, then W=0

therefore,

W=F•ds

0=F•ds

0=(8.5i -8.5j)•(2.5i + cj)

Note that

i.i=j.j=k.k=1

i.j=j.i=k.i=i.k=j.k=k.j=0

So applying this

0=(8.5i -8.5j)•(2.5i + cj)

0= (8.5×2.5i.i + 8.5×ci.j -8.5×2.5j.i-8.5×cj.j)

0=21.25-8.5c

Therefore,

8.5c=21.25

c=21.25/8.5

c=2.5

7 0

2 years ago

A 53.0 kg sled is sliding on snow with μk=0.110. how much friction force does it feel?

Anon25 [30]

Answer:

57N

Explanation:

$F_F = \mu F_N = \mu mg = 0.11 \times 53 kg \times 9,8 \frac{m}{s^{2} }$

6 0

2 years ago

Read 2 more answers

Calculate the energy, wavelength, and frequency of the emitted photon when an electron moves from an energy level of -3.40 eV to

jasenka [17]

Answer:

(a) The energy of the photon is 1.632 x $10^{-8}$ J.

(b) The wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency of the photon is 2.47 x $10^{25}$ Hz.

Explanation:

Let;

$E_{1}$ = -13.60 ev

$E_{2}$ = -3.40 ev

(a) Energy of the emitted photon can be determined as;

$E_{2}$ - $E_{1}$ = -3.40 - (-13.60)

= -3.40 + 13.60

= 10.20 eV

= 10.20(1.6 x $10^{-9}$ )

$E_{2}$ - $E_{1}$ = 1.632 x $10^{-8}$ Joules

The energy of the emitted photon is 10.20 eV (or 1.632 x $10^{-8}$ Joules).

(b) The wavelength, λ, can be determined as;

E = (hc)/ λ

where: E is the energy of the photon, h is the Planck's constant (6.6 x $10^{-34}$ Js), c is the speed of light (3 x $10^{8}$ m/s) and λ is the wavelength.

10.20(1.6 x $10^{-9}$ ) = (6.6 x $10^{-34}$ * 3 x $10^{8}$ )/ λ

λ = $\frac{1.98*10^{-25} }{1.632*10^{-8} }$

= 1.213 x $10^{-17}$

Wavelength of the photon is 1.2 x $10^{-17}$ m.

(c) The frequency can be determined by;

E = hf

where f is the frequency of the photon.

1.632 x $10^{-8}$ = 6.6 x $10^{-34}$ x f

f = $\frac{1.632*10^{-8} }{6.6*10^{-34} }$

= 2.47 x $10^{25}$ Hz

Frequency of the emitted photon is 2.47 x $10^{25}$ Hz.

6 0

2 years ago

5. Which of these is necessary for electricity to flow?

CaHeK987 [17]

Answer:

the circuit must be closed

Explanation:

4 0

2 years ago

Read 2 more answers