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3 years ago

9. ¿Por qué no es lo mismo distancia que desplazamiento?

Physics

1 answer:

masya89 [10]3 years ago

5 0

Answer:

Distance is a scalar quantity that refers to "how much ground an object has covered" during its motion. Displacement is a vector quantity that refers to "how far out of place an object is"; it is the object's overall change in position.

Explanation:

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What type of energy transformation occurs when a light bulb is turned on?

Vikentia [17]

Electical energy is transformed into heat and light energy

4 0

2 years ago

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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

In still water, a boat averages 18 18 miles per hour. it takes the same amount of time to travel 16 miles 16 miles downstream,

Vladimir79 [104]

<span>The current is 6 miles per hour.
Let's create a few equations:
Traveling with the current:
(18 + c)*t = 16

Traveling against the current:
(18 - c)*t = 8

Let's multiply the 2nd equation by 2
(18 - c)*t*2 = 16

Now subtract the 1st equation from the equation we just doubled.
(18 - c)*t*2 = 16
(18 + c)*t = 16

(18 - c)*t*2 - (18 + c)*t = 0
Divide both sides by t
(18 - c)*2 - (18 + c) = 0

Now solve for c
(18 - c)*2 - (18 + c) = 0
36 - 2c - 18 - c = 0
36 - 2c - 18 - c = 0
18 - 3c = 0
18 = 3c
6 = c

So the current is 6 mph.
Let's verify that.
(18 + 6)*t = 16
24*t = 16
t = 16/24 = 2/3

(18 - 6)*t = 8
12*t = 8
t = 8/12 = 2/3

And it's verified.</span>

4 0

3 years ago

Four particles are in a 2-d plane with masses, x- and y- positions, and x- and y- velocities as given in the table below: what i

Arte-miy333 [17]

I attached the picture of the missing table.
Center of mass is the point such that if you apply force to it, the system would move without rotating.
We can use following formula to calculate the center of mass:
$R=\frac{1}{M}\sum_{i=1}^{n=i}m_ir_i$
Where M is the sum of the masses of all particles.
Part 1
To calculate the x coordinate of the center of mass we will use this formula:
$R_x=\frac{1}{M}\sum_{i=1}^{n=i}m_ix_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_x=0.96m$
Part 2
To calculate the y coordinate of the center of mass we will use this formula:
$R_y=\frac{1}{M}\sum_{i=1}^{n=i}m_iy_i$
I will do all the calculations in the google sheet that I will share with you.
For the x coordinate of the center of mass we get:
$R_y=-0.84m$
Part 3
We will calculate speed along x and y-axis separately and then will add them together.
$v_x=\frac{\sum_{i=1}^{n=i}m_iv_x_i}{M}$
$v_y=\frac{\sum_{i=1}^{n=i}m_iv_y_i}{M}$
Total velocity is:
$v=\sqrt{v_x^2+v_y^2}$
Once we calculate velocities we get:
$v_x=-1.08\frac{m}{s}\\ v_y=-0.03\frac{m}{s}\\ v=\sqrt{(-1.08)^2+(-0.03)^2}=1.08\frac{m}{s}$
Part 4
Because origin is left to our center of mass(please see the attached picture) placing fifth mass in the origin would move the center of mass to the left along the x-axis.
Part 5
If you place fifth mass in the center of the mass nothing would change. The center of mass would stay in the same place.
Here is the link to the spreadsheet:
https://docs.google.com/spreadsheets/d/1SkQHbI1BxiJnwpWbLmP0XWgcNPrGquH1K2MfN6cznVo/edit?usp=sharing

3 0

2 years ago

Suzette had prepared the graph below to add to her lab

Charra [1.4K]

Answer:

A title

Explanation:

Because this is middle school.

4 0

3 years ago

Read 2 more answers