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3 years ago

Higher energy efficiency is desirable because

Physics

1 answer:

natita [175]3 years ago

6 0

Because less energy is being wasted. non-renewable resources will last longer.
the cost is kept lower. hope it helps :)

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suppose that during any period of 1/4 second there is one instant at which the crests or troughs of component waves are exactly

ivanzaharov [21]

The correct answer for the question that is being presented above is this one: "(a)4." Suppose that during any period of 1/4 second there is one instant at which the crests or troughs of component waves are exactly in phase and maximum <span>reinforcement occurs, in 1 second, there will be 4 beats.</span>

8 0

2 years ago

Read 2 more answers

When light of wavelength 240 nm falls on a cobalt surface, electrons having a maximum kinetic energy of 0.17 eV are emitted. Fin

dusya [7]

Answer:

(a) 5.04 eV (B) 248.14 nm (c) $1.21\times 10^{15}Hz$

Explanation:

We have given Wavelength of the light \lambda = 240 nm

According to plank's rule ,energy of light

$E = h\nu = \frac{hc}{}\lambda$

$E = h\nu = \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{ 240\times 10^{-9} m\times 1.6\times 10^{-19}J/eV}= 5.21 eV$

Maximum KE of emitted electron i= 0.17 eV

Part( A) Using Einstien's equation

$E = KE_{max}+\Phi _{0}$ , here $\Phi _0$ is work function.

$\Phi _{0}=E - KE_{max}$ = 5.21 eV-0.17 eV = 5.04 eV

Part( B) We have to find cutoff wavelength

$\Phi _{0} = \frac{hc}{\lambda_{cuttoff}}$

$\lambda_{cuttoff}= \frac{hc}{\Phi _{0} }$

$\lambda_{cuttoff}= \frac{6.67\times 10^{-34} J.s\times 3\times 10^{8}m/s}{5.04 eV\times 1.6\times 10^{-19}J/eV }=248.14 nm$

Part (C) In this part we have to find the cutoff frequency

$\nu = \frac{c}{\lambda_{cuttoff}}= \frac{3\times 10^{8}m/s}{248.14 \times 10^{-19} m }= 1.21\times 10^{15} Hz$

5 0

3 years ago

A dielectric-filled parallel-plate capacitor has plate area A = 30.0 cm2 , plate separation d = 9.00 mm and dielectric constant

PIT_PIT [208]

Answer:

$9.96\cdot 10^{-10}J$

Explanation:

The capacitance of the parallel-plate capacitor is given by

$C=\epsilon_0 k \frac{A}{d}$

where

ϵ0 = 8.85x10-12 C2/N.m2 is the vacuum permittivity

k = 3.00 is the dielectric constant

$A=30.0 cm^2 = 30.0\cdot 10^{-4}m^2$ is the area of the plates

d = 9.00 mm = 0.009 m is the separation between the plates

Substituting,

$C=(8.85\cdot 10^{-12}F/m)(3.00 ) \frac{30.0\cdot 10^{-4} m^2}{0.009 m}=8.85\cdot 10^{-12} F$

Now we can calculate the energy of the capacitor, given by:

$U=\frac{1}{2}CV^2$

where

C is the capacitance

V = 15.0 V is the potential difference

Substituting,

$U=\frac{1}{2}(8.85\cdot 10^{-12}F)(15.0 V)^2=9.96\cdot 10^{-10}J$

4 0

2 years ago

How much work is done on a satellite in a circular orbit about earth?

alexgriva [62]

Answer:

0 J

Explanation:

$W$ = Work done on the satellite in circular orbit about earth by earth

$F$ = Force on gravity on satellite by earth

$d$ = displacement of the satellite

$\theta$ = Angle between the force on gravity and displacement = 90

We know that, Work done is given as

$W = F d Cos\theta\\W = F d Cos90\\W = F d (0)\\W = 0 J$

3 0

2 years ago

Regular exercise is positively related to wellness

Harman [31]

Answer:

yes ( true)

Explanation:

positive effects on all the body systems.

7 0

2 years ago