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Ksju [112]
3 years ago
14

A bonding molecular orbital is of lower energy (more stable) than the atomic orbitals from which it was formed.

Physics
1 answer:
ioda3 years ago
7 0
I think the correct answer would be true. A bonding molecular orbital is of lower energy than the atomic orbitals from which it was formed. Having a lower energy, it makes them more stable than that of the atomic orbitals. They are attracted by nuclei at the same time.
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What is the part of the racquet directly below the head that connects the head to the handle?
4vir4ik [10]
It's called Throat. You can refer the figure attached.

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4 years ago
A ball, initially at rest, reaches a speed of 20 m/s at the bottom of the apparatus. If it takes the ball 10 seconds to reach th
MariettaO [177]

Answer:

2 m/s²

Explanation:

Given:

v₀ = 0 m/s

v = 20 m/s

t = 10 s

Find: a

a = (v − v₀) / t

a = (20 m/s − 0 m/s) / 10 s

a = 2 m/s²

3 0
3 years ago
A ufo was detected on the radar flying 7400 miles in 3 min, what is the estimated speed per hour?
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8 0
4 years ago
A rock is thrown upward from level ground in such a way that the maximumheight of its flight is equal to its horizontal rangeR.
Ne4ueva [31]

The range of a projectile motion is given by:

\frac{u_o^2 sin2\theta}{g}

where, u is the initial speed of the projectile, \theta is the angle of the projectile and g is the acceleration due to gravity.

The maximum height reached is given by:

\frac{u_o^2 sin^2\theta}{2g}

Part a

It is given that the maximum height reached is equal to the horizontal range. we need to find the angle of the projectile.

Equating the two:

\frac{u_o^2 sin2\theta}{g}=\frac{u_o^2 sin^2\theta}{2g}\\ \Rightarrow 2 sin2\theta =sin^2\theta\\ \Rightarrow 2\times 2 sin\theta cos\theta=sin^2 \theta\\ \Rightarrow tan\theta =4\\ \Rightarrow \theta=tan^{-1}4=75.96^o

Hence, the projectile was thrown at an initial angle of \theta=75.96^o.

Part b

we need to find the angle for which range would be maximum and then write this maximum range in terms of original range.

So, we know that range is given by:

R= \frac{u_o^2sin2\theta}{g}

It would be maximum when sin2\theta=1\Rightarrow 2\theta=90^o\Rightarrow \theta=45^o

Hence, R_{max}=\frac{u_o^2}{g}

Original range,

R=\frac{u_o^2sin2\times75.96^o}{g}\\ \Rightarrow R=\frac{u_o^2}{g}\times 0.47\\ \Rightarrow R=R_{max}0.47\\ \Rightarrow R_{max}=2.125R

Part c:

In the part a, we know that the angle of the projectile is independent of the g i.e. the acceleration due to gravity and this is the only factor that varies with the different planets. Hence, the answer would remain same.

7 0
3 years ago
A temperature unit is known as a _____. <br> kilogram<br> calorie<br> joule <br> kelvin
svlad2 [7]
A temperature unit is known as a Kelvin
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3 years ago
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