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3 years ago

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A textbook is dropped from a second floor window and falls to the ground. What happens to the acceleration as the book falls to

the ground
-the acceleration increases
-the acceleration decreases
-the acceleration remains constant *

1 answer:

Sav [38]3 years ago

6 0

Answer:

The correct option is;

The acceleration remains constant

Explanation:

The acceleration is due to the force of gravitational attraction between the text book and the Earth

According to Newton's law of gravitation, there is an attractive force between all objects given by the following relation;

F = G×M₁×m₂/r²

Where;

G, M₁, m₂, and r are constant such that we have;

G×M₁/r² = Constant = The acceleraton due to gravity, g

F = g×m₂

So the acceleration of the textbook as it is being attracted by the force of gravity towards the ground (Earth) is remains constant.

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A helium-neon laser emits light of wavelength 632.8 nm and has a power output of 16 mw. how many photons are emitted per second

NemiM [27]

The power of the laser is
$P=16 mW=0.016 W$
by using the relationship between power, energy and time, we can find the energy delivered by the laser in 1 second:
$E=Pt=(0.016 W)(1 s)=0.016 J$

The frequency of the photons of this light is given by
$f= \frac{c}{\lambda}= \frac{3\cdot 10^8 m/s}{632.8 \cdot 10^{-9} m} =4.74 \cdot 10^{14} Hz$
and the energy of a single photon is
$E_1=hf=(6.6\cdot 10^{-34} Js)(4.74 \cdot 10^{14} Hz)=3.13 \cdot 10^{-19} J$

In order to find the number of photons emitted per second, we must divide the total energy emitted by the laser by the energy of a single photon, and we get:
$N= \frac{E}{E_1}= \frac{0.016 J}{3.13 \cdot 10^{-19}J} =5.11 \cdot 10^{16} photons$

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3 years ago

A wire is oriented along the x-axis. It is connected to two batteries, and a conventional current of 2.4 A runs through the wire

Deffense [45]

Answer:

$\vec{F}=0.40176 N \hat{k}$

Explanation:

To calculate the force we need to use this equation

$\vec{F}=\int\limits^L_0 {i(\vec{dl}\times\vec{B})}$

where L is the total length of the wire

So in this case the small element of current is

$\vec{dl} = dx \hat{i}$

Because x is the direction of the current flow.

As is said in the problem B is such that

$\vec{B} = B \hat{j} = 0.62\hat{j} [ T]$

so to use the equation above we first calculate the following cross product:

$\vec{dl}\times\vec{B}=dx \hat{i}\times B \hat{j} = Bdx\hat{k}$

so the force:

$F = \int\limits^L_0 {i(\vec{dl}\times\vec{B})}=\int\limits^L_0{iBdx\hat{k}}$

So here we use the fact that B=0 in any point of the x axis that is not $x^{'}=0.27 [m]$ , that means that we only need to do the integration between a very short distant behind the point $x^{'}=0.27 [m]$ and a very short distant after that point, meaning:

$\vec{F}= \lim_{h \to 0}{\int\limits^{x^{'}+h}_{x^{'}-h}{iBdx\hat{k}} }$

so is the same as evaluating $iBx$ at $x=x^{'}$

that is:

$2,4 A * 0,62 T * 0,27 m \hat{k}$

$2,4 A * 0,62 (\frac{Kg}{A s^{2}}) * 0,27 m \hat{k}$

$2,4*0,62*2,7 ( \frac{ kgm }{ s^{2} } ) \hat{k}$

$\vec{F}=0.40176 N \hat{k}$

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nordsb [41]

Answer:

I belive its THAT HELPS US BE A BETTER PERSON

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3 years ago

A large, 68.0-kg cubical block of wood with uniform density is floating in a freshwater lake with 20.0% of its volume above the

LenaWriter [7]

Answer:

a) V = 0.085 m^3

b) m = 17 kg

Explanation:

1) Data given

mb = 68 kg (mass for the block)

20% of the block volume is floating

100-20= 80% of the block volume is submerged

2) Notation

mb= mass of the block

Vw= volume submerged

mw = mass water displaced

V= total volume for the block

3) Forces involved (part a)

For this case we have two forces the buoyant force (B), defined as the weight of water displaced acting upward and the weight acting downward (W)

Since we have an equilibrium system we can set the forces equal. By definition the buoyant force is given by :

B = (mass water displaced) g = (mw) g (1)

The definition of density is :

$\rho_w = \frac{m_w}{V_w}$

If we solve for mw we got $m_w = \rho_w V_w$ (2)

Replacing equation (2) into equation (1) we got:

$B = \rho_w V_w g$ (3)

On this case Vw represent the volume of water displaced = 0.8 V

If we replace the values into equation (3) we have

0.8 ρ_w V g = mg (4)

And solving for V we have

V = (mg)/(0.8 ρ_w g )

We cancel the g in the numerator and the denominator we got

V = (m)/(0.8 ρ_w)

V = 68kg /(0.8 x 1000 kg/m^3) = 0.085 m^3

4) Forces involved (part b)

For this case we have bricks above the block, and we want the maximum mass for the bricks without causing it to sink below the water surface.

We can begin finding the weight of the water displaced when the block is just about to sink (W1)

W1 = ρ_w V g

W1 = 1000 kg/m^3 x 0.085 m^3 x 9.8 m/s^2 = 833 N

After this we can calculate the weight of water displaced before putting the bricks above (W2)

W2 = 0.8 x 833 N = 666.4 N

So the difference between W1 and W2 would represent the weight that can be added with the bricks (W3)

W3 = W1 -W2 = 833-666.4 N = 166.6 N

And finding the mass fro the definition of weight we have

m3 = (166.6 N)/(9.8 m/s^2) = 17 Kg

8 0

3 years ago