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2 years ago

What is the speed of an object at rest? a. 0.0 m/s c. 9.8 m/s b. 1.0 m/s d. 9.81 m/s

Physics

1 answer:

pishuonlain [190]2 years ago

8 0

Answer:

A 0.0 m/s

An object at rear is not moving so it is morning 0m every second.

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What is the energy of moving electrical charges

Simora [160]

The energy of moving electrical charges is Electrical energy

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3 0

3 years ago

A modern compact fluorescent lamp contains 1.4 mg of mercury (Hg). If each mercury atom in the lamp were to emit a single photon

Reika [66]

Answer:

A. 1.64 J

Explanation:

First of all, we need to find how many moles correspond to 1.4 mg of mercury. We have:

$n=\frac{m}{M_m}$

where

n is the number of moles

m = 1.4 mg = 0.0014 g is the mass of mercury

Mm = 200.6 g/mol is the molar mass of mercury

Substituting, we find

$n=\frac{0.0014 g}{200.6 g/mol}=7.0\cdot 10^{-6} mol$

Now we have to find the number of atoms contained in this sample of mercury, which is given by:

$N=n N_A$

where

n is the number of moles

$N_A=6.022\cdot 10^{23} mol^{-1}$ is the Avogadro number

Substituting,

$N=(7.0\cdot 10^{-6} mol)(6.022\cdot 10^{23} mol^{-1})=4.22\cdot 10^{18}$ atoms

The energy emitted by each atom (the energy of one photon) is

$E_1 = \frac{hc}{\lambda}$

where

h is the Planck constant

c is the speed of light

$\lambda=508 nm=5.08\cdot 10^{-7}nm$ is the wavelength

Substituting,

$E_1 = \frac{(6.63\cdot 10^{-34} Js)(3\cdot 10^8 m/s)}{5.08\cdot 10^{-7} m}=3.92\cdot 10^{-19} J$

And so, the total energy emitted by the sample is

$E=nE_1 = (4.22\cdot 10^{18} )(3.92\cdot 10^{-19}J)=1.64 J$

4 0

3 years ago

Which of Newton's laws of motion best illustrates the principle of inertia?

andrew-mc [135]

Newton's first law of motion best illustrates the principle of inertia<span />

5 0

3 years ago

Define the term overload.

ryzh [129]

Answer:

Verb. Load with too great a burden or cargo.

"Both boats were overloaded and low in the water"

Noun. An excessive load or amount.

"an overload of stress"

Explanation:

Similar words are strain, excess, and overburden.

Have a good day and stay safe!

4 0

3 years ago

Read 2 more answers

A uniform ladder 5.0 m long rests against a frictionless, vertical wall with its lower end 3.0 m from the wall. The ladder weigh

DiKsa [7]

Answer:(a)360N,(b)171N,(c)2.702m

Explanation:

(a)Maximum Friction Force = $\mu \left ( N\right )=0.4\times \left ( 740+160\right )$

=360 N

$cos\theta =\frac{3}{5}$

$sin\theta =\frac{4}{5}$

(b)Moment about Ground Point

$740\times 1\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

$N_1tan\theta =1140$

$N_1=171 N$

$N_1=f=171 N$

(c)

$740\times x\times cos\theta +2.5\times 160\times cos\theta -N_15sin\theta$

Here maximum friction force can be 360 N

Therefore $N_1=360 N$

Where x is the maximum distance moved by man along the ladder

$360\times 5\times \frac{4}{3}=740x+160\times 2.5$

740x=2000

x=2.702m

5 0

3 years ago