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timama [110]
3 years ago
11

One of the way atoms bond with each other would be through:

Physics
1 answer:
Galina-37 [17]3 years ago
8 0

Answer:

ehdgywi

Explanation:

djhcuowhciwurvgwyirgvy mm. ncmsmsmx. n. mssmsmiwvfiywrvvkjbwviverbladcnviwrgqecocqeboodqeugচঠচবি

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A new pest is found that only feeds on corn plant leaves. Which promoter would you choose if the modified gene produces toxins t
8090 [49]

It is actually B. Promoter 2.

Your welcome

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5 0
3 years ago
A uniform 2.50m ladder of mass 7.30kg is leaning against a vertical wall while making an angle of 51.0degree with the floor. A w
Zigmanuir [339]

Answer:

19.95 J

Explanation:

The center of mass of the ladder is initially at a height of:

h_1=\frac{L}{2}sin\theta

The center of mass of the ladder ends at a height of:

h_2= \frac{L}{2}sin90 =L/2

So, the work done is equal to the change in potential energy which is:

W = PE = mg(h_2-h_1)

now h_2-h_1= 1-sin\theta

therefore

W = [mgL/2]×[1 - sin(theta)]

W = [(7.30)(9.81)(2.50)/2]×[1-sin(51°)]

solving this we get

W = 19.95 J

8 0
3 years ago
Which of these is MOST LIKELY to cause natural pollution?
egoroff_w [7]
What are the options so that I can help.
8 0
3 years ago
Read 2 more answers
A 5-kg projectile is fired over level ground with a velocity of 200 m/s at an angle of 25°above the horizontal. Just before it h
Nutka1998 [239]

Answer:

the change in thermal energy of the projectile is 43.8 kJ

Explanation:

Given;

mass of the object, m = 5kg

initial velocity of the projectile, v₁ = 200 m/s

final  velocity of the projectile, v₂ = 150 m/s

To determine the the change in the thermal energy of the projectile and air, we consider change in potential and kinetic enrgy of the projectile. Since the projectile was fired over level ground, change in potential energy is zero.

Then, change in thermal energy of the projectile, KE = Δ¹/₂mv²

KE = Δ¹/₂mv² = ¹/₂m(v₁²-v₂²)

KE =  ¹/₂ × 5(200²-150²) = 2.5(17500) = 43750 J = 43.8 kJ

Therefore, the change in thermal energy of the projectile is 43.8 kJ

8 0
3 years ago
Read 2 more answers
A velocity selector has a magnetic field of magnitude 0.22 T perpendicular to an electric field of magnitude 0.51 MV/m.
ohaa [14]

Answer:

speed of a particle  = 2.31 × 10^{6} m/s

energy of proton required = 27.77 KeV

energy of electron required =  15.171 eV

Explanation:

given data

magnetic field of magnitude = 0.22 T

electric field of magnitude = 0.51 MV/m

to find out

speed of a particle and energy must protons have to pass through undeflected and  energy must electrons have to pass through undeflected

solution

we know that force due to magnetic and electric field is express as

force due to magnetic field B = qvB    ..............1

and force due to electric field E = qE   .....................2

so without deflection force due to magnetic field  = force due to electric field  

so here qvB = qE

and V = \frac{E}{B}    ...................3

put here value

V =  \frac{0.51*10^6}{0.22}

speed of a particle  = 2.31 × 10^{6} m/s

and

now energy of proton required will be here as

energy of proton required = mass of proton × \frac{V^2}{2}

put here value

energy of proton required = 1.67 × 10^{-27} × \frac{(2.31*10^6)^2}{2}

energy of proton required = 4.45 × 10^{-15} J

energy of proton required = 4.45 × 10^{-15} J ÷ (1.602 × 10^{-19}

energy of proton required = 27777.777 eV

energy of proton required = 27.77 KeV

and

now we get here energy of electron required that is

energy of electron required = mass of electron × \frac{V^2}{2}

put here value

energy of electron required = 9.11 × 10^{-31} × \frac{(2.31*10^6)^2}{2}

energy of electron required =  24.305× 10^{-19} J

energy of electron required =  24.305 × 10^{-19} J ÷ (1.602 × 10^{-19}  

energy of electron required =  15.171 eV

5 0
3 years ago
Read 2 more answers
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