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HACTEHA [7]
3 years ago
8

Explain the difference in polarity between co2 and h2o by referring to the polarity of the bonds and the shape of the molecules

Physics
2 answers:
Lina20 [59]3 years ago
8 0
<span>Both H2O and CO2 have polar covalent bondings (O is more electronegative than H or C) . The main molecular difference arises from their different geometry. </span>
erik [133]3 years ago
4 0
<span>CO2 is a linear molecule so despite of having to dipolar bonds they add up and exactly cancel each other rendering the molecule non-polar.</span>
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Which image shows the correct way of lining up vectors to add them
Marina CMI [18]

Answer:

it's A

Explanation:

wen aligning the vectors the head and the tail should meet

4 0
1 year ago
Hello i need help with this!!​
mr Goodwill [35]

Answer:

Yes

Explanation:

If lamp A burnt out there would still be a wire above it that connects lamp B and C to the power source

3 0
2 years ago
A football is kicked from the ground with a velocity of 38m/s at an angle of 40 degrees and eventually lands at the same height.
Anastasy [175]

Initially, the velocity vector is \langle 38cos(40^{\circ}),38sin(40^{\circ}) \rangle=\langle 29.110, 24.426 \rangle. At the same height, the x-value of the vector will be the same, and the y-value will be opposite (assuming no air resistance). Assuming perfect reflection off the ground, the velocity vector is the same. After 0.2 seconds at 9.8 seconds, the y-value has decreased by 4.9(0.2)^2, so the velocity is \langle 29.110, 24.426-0.196 \rangle = \langle 29.110, 24.23 \rangle.

Converting back to direction and magnitude, we get \langle r,\theta \rangle=\langle \sqrt{29.11^2+24.23^2},tan^{-1}(\frac{29.11}{24.23}) \rangle = \langle 37.87,50.2^{\circ}\rangle

4 0
3 years ago
a moving billiard ball collides with an identical stationary billiard ball in an elastic collision. after the collision, the sec
MArishka [77]

A billiard ball collides with a stationary identical billiard ball to make it move. If the collision is perfectly elastic, the first ball comes to rest after collision.

<h3>Why does the first ball comes to rest after collision ?</h3>

Let m be the mass of the two identical balls.  

u1 = velocity before the collision of ball 1

u2 = 0 = velocity of second ball that is at rest

v1 and v2 are the velocities of the balls after the collision.

From the conservation of momentum,

∴ mu1 + mu2 = mv1 + mv2

∴ mu1 = mv1 + mv2

∴ u1 = v1 + v2

In an elastic collision, the kinetic energy of the system before and after collision remains same.

\frac{1}{2}  mu_1^2+0=\frac{1}{2}  mv_1^2+\frac{1}{2}  mv_2^2

∴  \frac{1}{2}  m(v_1+v_2 )^2=\frac{1}{2} mv_1^2+\frac{1}{2}mv_2^2

∴ \frac{1}{2} mv_1^2+\frac{1}{2} mv_2^2+mv_1 v_2=\frac{1}{2}  mv_1^2+\frac{1}{2} mv_2^2

∴ mv₁v₂ = 0

  1. It is impossible for the mass to be zero.
  2. Because the second ball moves, velocity v2 cannot be zero.
  3. As a result, the velocity of the first ball, v1, is zero, indicating that it comes to rest after collision.
<h3>What is collision ?</h3>

An elastic collision is a collision between two bodies in which the total kinetic energy of the two bodies remains constant. There is no net transfer of kinetic energy into other forms such as heat, noise, or potential energy in an ideal, fully elastic collision.

Can learn more about elastic collision from brainly.com/question/12644900

#SPJ4

3 0
1 year ago
After landing on an unfamiliar planet, a space explorer constructs a simple pendulum of length 54.0 cm. The explorer finds that
Natasha2012 [34]

Answer:

g = 11.2 m/s²

Explanation:

First, we will calculate the time period of the pendulum:

T = \frac{t}{n}

where,

T = Time period = ?

t = time taken = 135 s

n = no. of swings in given time = 98

Therefore,

T = \frac{135\ s}{98}

T = 1.38 s

Now, we utilize the second formula for the time period of the simple pendulum, given as follows:

T = 2\pi \sqrt{\frac{l}{g}}

where,

l = length of pendulum = 54 cm = 0.54 m

g = acceleration due to gravity on the planet = ?

Therefore,

(1.38\ s)^2 = 4\pi^2(\frac{0.54\ m}{g} )\\\\g = \frac{4\pi^2(0.54\ m)}{(1.38\ s)^2}

<u>g = 11.2 m/s²</u>

3 0
3 years ago
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