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Evgen [1.6K]
3 years ago
14

When devising a model, scientists can only use the information available during their lifetime. This means that the current mode

l of the atom may be changed when new experiments are performed in the future. Does this mean that the currently accepted model of the atom is useless? Explain your answer.
Physics
2 answers:
11111nata11111 [884]3 years ago
6 0

no, it not useless. we still learn Bohr's model in HS n dats almost 200 yr old! while there may be new models, previous one is good for explaining the basics. it is also useful to learn previous model n see how our understanding improves over time.


Rom4ik [11]3 years ago
6 0

The currently accepted model, when replaced by a new one, has served the purpose being the best model we have at this time. So it is far from useless. It provides a basis for the new model. Combined with all previous replaced models, they show the history and process of our understanding of the atom.


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The message refers to which of the following?
Oliga [24]

The message is the information being communicated from one place to another.

It used to be called the "intelligence".  But as time went on, it became
harder to ignore the obvious fact that that was going too far, and the
label was changed to the more IQ-neutral "message".    

7 0
3 years ago
An artificial satellite is in a circular orbit around a planet of radius r= 2.05 x103 km at a distance d 310.0 km from the plane
lubasha [3.4K]

Answer:

\rho = 12580.7 kg/m^3

Explanation:

As we know that the satellite revolves around the planet then the centripetal force for the satellite is due to gravitational attraction force of the planet

So here we will have

F = \frac{GMm}{(r + h)^2}

here we have

F =\frac {mv^2}{(r+ h)}

\frac{mv^2}{r + h} = \frac{GMm}{(r + h)^2}

here we have

v = \sqrt{\frac{GM}{(r + h)}}

now we can find time period as

T = \frac{2\pi (r + h)}{v}

T = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{GM}{(r + h)}}}

1.15 \times 3600 = \frac{2\pi (2.05 \times 10^6 + 310 \times 10^3)}{\sqrt{\frac{(6.67 \times 10^{-11})(M)}{(2.05 \times 10^6 + 310 \times 10^3)}}}

M = 4.54 \times 10^{23} kg

Now the density is given as

\rho = \frac{M}{\frac{4}{3}\pi r^3}

\rho = \frac{4.54 \times 10^{23}}{\frac{4}[3}\pi(2.05 \times 10^6)^3}

\rho = 12580.7 kg/m^3

8 0
3 years ago
Match the block to its correct density. 1. Block A 0.64 Kg/L 2. Block B 0.917 kg/L 3. Block C 0.70 Kg/L 19.3 kg/L 4. Block D 3.5
Sedbober [7]

Answer:

quit school aghahahha

Explanation:

xD

4 0
3 years ago
A sound source producing 1.00 kHz waves moves toward a stationary listener at one-half the speed of sound.
SIZIF [17.4K]

Answer:

2000Hz and 1500Hz

Explanation:

Using

a) f = f0((c+vr)/(c+vs))

=>>> f0((c)/(c-0.5c))

=>>>1000/0.5 = 2000Hz

b) f = f0((c+vr)/(c+vs))

=>>>f0((c+0.5c)/(c))

=>>>>1000 x 1.5 = 1500Hz

5 0
3 years ago
if you push a box a distance of 2000 meters with a force of 1 newton, how many calories have you used
kap26 [50]
Note that
1 J = 0.239 cal

By definition,
Work = Force x Distance

Therefore work done is
W = (1 N)*(2000 m) = 2000 J

In calories,
W = (2000 J)*(0.239 cal/J) = 478 cal

Answer: 478 calories

4 0
3 years ago
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