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JulsSmile [24]
3 years ago
11

18. When is an atom unlikely to react and form chemical bonds?

Physics
1 answer:
katrin2010 [14]3 years ago
5 0
Fission and fusion are two processes that release very little amounts of energy.
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Two mirrors are touching so they have an angle of 35.4 degrees with one another. A light ray is incident on the first at an angl
alexandr1967 [171]

Answer:

54.6°

Explanation:

From law of reflection i=r.

So, construct the reflected ray at 55.7°degrees from the normal and let it fall on the other mirror.  

Now draw the second normal at the point of incidence and again measure the angle of incidence, and draw the angle of reflection.

If you consider triangle AOB, one angle is ∠AOB=90°

 and ∠OAB is 54.6°

 

From angle sum property third angle ie ∠ABO=180°-90°-54.6°=35.4°

 

So, the second incident angle will be 54.6°

Hence, the second reflected angle will be 54.6 degrees.

8 0
3 years ago
What specific structures are in electroreceptors?
Slav-nsk [51]
Electroreception is limited to aquatic environments because on here is the resistivity of the medium is low enough for electric currents to be generated as the result of electric fields of biological origin.  In air, the resistivity of the environment is so high that electric fields from biological sources do not generate a significant electric current.  Electroreceptor are found in a number of species of fish,  and in at least one species of mammal,  the Duck-Billed platypus.
7 0
3 years ago
A child on a tricycle is moving at a speed of 1.40 m/s at the start of a 2.25 m high and 12.4 m long incline. The total mass is
goblinko [34]

Answer:

The work done by the child as the tricycle travels down the incline is 416.96 J

Explanation:

Given;

initial velocity of the child, v_i = 1.4 m/s

final velocity of the child, v_f = 6.5 m/s

initial height of the inclined plane, h = 2.25 m

length of the inclined plane, L = 12.4 m

total mass, m = 48 kg

frictional force, f_k = 41 N

The work done by the child is calculated as;

\Delta E_{mech} = W - f_{k} \Delta L\\\\W = \Delta E_{mech}  + f_{k} \Delta L\\\\W = (K.E_f - K.E_i) + (P.E_f - P.E_i) + f_{k} \Delta L\\\\W = \frac{1}{2} m(v_f^2 - v_i^2) + mg(h_f - h_i) + f_{k} \Delta L\\\\W = \frac{1}{2} \times 48(6.5^2 - 1.4^2) + 48\times 9.8(0-2.25) + (41\times 12.4)\\\\W = 966.96  \ - \ 1058.4 \ + \ 508.4\\\\W = 416.96 \ J

Therefore, the work done by the child as the tricycle travels down the incline is 416.96 J

5 0
3 years ago
The earth rotates every 86,160 seconds. What is the tangential speed (in m/s) at Livermore (Latitude 37.6819° measured up from e
Lena [83]

Answer:

The tangential speed at Livermore is approximately 284.001 meters per second.

Explanation:

Let suppose that the Earth rotates at constant speed, the tangential speed (v), measured in meters per second, at Livermore (37.6819º N, 121º W) is determined by the following expression:

v = \left(\frac{2\pi}{\Delta t}\right)\cdot R \cdot \sin \phi (1)

Where:

\Delta t - Rotation time, measured in seconds.

R - Radius of the Earth, measured in meters.

\phi - Latitude of the city above the Equator, measured in sexagesimal degrees.

If we know that \Delta t = 86160\,s, R = 6.371\times 10^{6}\,m and \phi = 37.6819^{\circ}, then the tangential speed at Livermore is:

v = \left(\frac{2\pi}{86160\,s} \right)\cdot (6.371\times 10^{6}\,m)\cdot \sin 37.6819^{\circ}

v\approx 284.001\,\frac{m}{s}

The tangential speed at Livermore is approximately 284.001 meters per second.

4 0
3 years ago
Sam, whose mass is 75 kg, straps on his skis and starts down a 50-m-high, 20 frictionless slope. A strong headwind exerts a hori
LenaWriter [7]

Answer:

Explanation:

Sam mass=75kg

Height is 50m

20° frictionless slope

Horizontal force on Sam is 200N

According to the work energy theorem, the net work done on Sam will be equal to his change in kinetic energy.

Therefore

Wg - Ww =∆K.E

Note initial the body was at rest at top of the slope.

Then, ∆K.E is K.E(final) - K.E(initial)

K.E Is given as ½mv²

Since initial velocity is zero then, K.E(initial ) is zero

Therefore, ∆K.E=½mVf²

Wg is work done by gravity and it is given by using P.E formulas

Wg=mgh

Wg=75×9.8×50

Wg=36750J

Ww is work done by wind and it's is given by using formulae for work

Work=force × distance

Ww=horizontal force × horizontal distance

Using Trig.

TanX=opposite/adjacent

Tan20=h/x

x=h/tan20

x=50/tan20

x=137.37m

Then,

Ww=F×x

Ww=200×137.37

We=27474J

Now applying the formula

Wg - Ww =∆K.E

36750 - 27474 =½×75×Vf²

9276=37.5Vf²

Vf²=9275/37.5

Vf²= 247.36

Vf=√247.36

Vf=15.73m/s

3 0
3 years ago
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