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tiny-mole [99]
3 years ago
9

How do scientist use information about an object or event

Physics
1 answer:
Arlecino [84]3 years ago
4 0
Many ways for example they look up related things and study them or they can test it and see what happens
You might be interested in
Resonance structures have _______ connectivity of atoms and _______ distribution of electrons.
Arte-miy333 [17]

Answer:

Resonance structures have <u> </u><u>same</u><u>  </u>  connectivity of atoms and <u> differ only in</u> distribution of electrons.

Explanation:

Atoms supply the electrons from their outer electron shells. Electrons are found free in nature and are grouped around the nucleus into shells. Electrons can be further explained as negatively charged subatomic particle. Electrons have properties of both particles and waves and they can be moved around.

Resonance structures are imaginary structures and not all of them are created equally. Resonance structures have two or more possible electron structures, and, the resonance structures for a particular substance sometimes have different energy and stability. When resonance structures are identical, they are important descriptions  of the molecule. The position of the atoms is the same in the various resonance structures of a compound, but the electrons are distributed differently around the structure.

3 0
3 years ago
Suppose an oven's radiation wavelength is 0.125 m. a container with 350.00 g of water was placed in the oven, and the temperatur
ivann1987 [24]
The heat (energy) needed to raise the temperature of the water is given by
Q=m C_S (T_f - T_i)=(350.0 g)(4.18 J/gC)(80C-20C)=87780 J

The wavelength of the radiation of the oven is \lambda=0.125 m, so the energy of a single photon of this radiation is
E=h \frac{c}{\lambda}=(6.6 \cdot 10^{-34}J) \frac{3\cdot 10^8 m/s}{0.125 m}=1.6 \cdot 10^{-24} J

So, the number of photons required to heat the water is the total energy absorbed by the water divided by the energy of a single photon:
N= \frac{Q}{E}= \frac{87780 J}{1.6\cdot 10^{-24}J}=  5.5 \cdot 10^{28} photons
8 0
2 years ago
Tommy runs around a track whose circumference is 400 meters. He runs a single lap in a time of 62 seconds. What is Tommy’s displ
bija089 [108]

Answer:

<h2>Angular Displacement 6.28 radians</h2>

Explanation:

for circular motion we are expected to solve for Angular Displacement it is measured in radian

Measurement of Angular Displacement.

we can measure it using the following relation

∅= s/r

where

s = the distance travelled by the body, and

r = radius of the circle along which it is moving.

given that

circumference c,  s= 400 m

r= ?

we have to solve for the radius

we know that circumference

c= 2\pi r

400= 2*3.142*r

400= 6.282*r

divide both sides by 6.284 we have

400/6.284

r= 63.63 m

Angular  displcament

∅= 400/63.63

∅= 6.28 radians

8 0
2 years ago
Suppose you are drinking root beer from a conical paper cup. The cup has a diameter of 10 centimeters and a depth of 13 centimet
sveta [45]

Answer:

The level of the root beer is dropping at a rate of 0.08603 cm/s.

Explanation:

The volume of the cone is :

V=\frac {1}{3}\times \pi\times r^2\times h

Where, V is the volume of the cone

r is the radius of the cone

h is the height of the cone

The ratio of the radius and the height remains constant in overall the cone.

Thus, given that, r = d / 2 = 10 / 2 cm = 5 cm

h = 13 cm

r / h = 5 / 13

r = {5 / 13} h

V=\frac {1}{3}\times \frac {22}{7}\times ({{{\frac {5}{13}\times h}}})^2\times h

V=\frac {550}{3549}\times h^3

Also differentiating the expression of volume w.r.t. time as:

\frac {dV}{dt}=\frac {550}{3549}\times 3\times h^2\times \frac {dh}{dt}

Given: \frac {dV}{dt} = -4 cm³/sec (negative sign to show leaving)

h = 10 cm

So,

-4=\frac{550}{3549}\times 3\times {10}^2\times \frac {dh}{dt}

\frac{55000}{1183}\times \frac {dh}{dt}=-4

\frac {dh}{dt}=-0.08603\ cm/s

<u>The level of the root beer is dropping at a rate of 0.08603 cm/s.</u>

3 0
2 years ago
Which statement accurately describes science?
slava [35]
Answer: We don’t really ever need to know how to dissect a frog but hey, one day maybe I’ll need to give a frog a discectomy
5 0
3 years ago
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