All categories

3 years ago

Ionic bond two ions what changes

1 answer:

7 0

Ionic bonds <span>are the type of bonds where there is </span>transfer<span> of electrons from one atom to another. The electrons are removed and from one atom and attached to another. A good example is salt which is composed of sodium and chlorine. Sodium readily loses one of its electrons and chlorine readily accepts it. Before losing the electron, sodium has a positive charge, but then becomes negatively charged after giving up the electron. Chlorine has a positive charge before gaining the electron but becomes negatively charged after gaining the electron. These opposite charges between sodium and chlorine attract the two elements together to form the ionic bond.</span>

You might be interested in

Question 3 A woman sitting in a stationary car notices a man cycling past her at 40 km/hr. Five seconds after he passes her car,

Naya [18.7K]

The shortest possible distance in which she catches up with the man is 129.7 m.

The given parameters:

speed of the man, v₁ = 40 km/hr

speed of the woman, v₂ = 70 km/hr

time when the woman started moving, t₁ = 5 seconds

To find:

the shortest distance when she catches up with the man

Convert the given speed in km/h to m/s

3.6 km/h = 1 m/s
40 km/h = 40/3.6 = 11.11 m/s
70 km/h = 70/3.6 = 19.44 m/s

Calculate the distance moved by the man before she started following him.

Distance moved by the man = speed x time

= 11.11 m/s x 5 s

= 55.55 m

Let the time the woman catches up with the man = t

Apply the following simple principle to calculate the time.

The man is moving forward and the woman is moving forward to close the gap between them, thus both speeds are working against each other.

$(v_2-v_1)t = 55.55 \ m$

the difference in the speed is because they are working against each other.

$(19.44 - 11.11)t = 55.55\\\\8.33t = 55.55\\\\t = \frac{55.55}{8.33} \\\\t = 6.67 \ s$

The time when the woman catches up with the man = 6.67 s

The shortest distance at which the woman catches up with the man is calculated as;

Shortest distance = woman's speed x time when she catches up

= 19.44 m/s x 6.67 s

= 129.7 m

Thus, the shortest possible distance in which she catches up with the man is 129.7 m.

Learn more here: https://brainly.in/question/9541933?tbs_match=3

5 0

3 years ago

A 0.86 kg rock is projected from the edge of the top of a building with an initial velocity of 8.65 m/s at an angle 46◦ above th

Georgia [21]

Answer:

Height of the building = 11.4 m

Explanation:

As we know that the stone is projected at an angle 46 degree with speed 8.65 m/s

so the two components of the speed is given as

$v_x = 8.65 cos46$

$v_x = 6 m/s$

vertical component of the speed is given as

$v_y = 8.65 sin46$

$v_y = 6.22 m/s$

now we know that the ball strike at horizontal distance of 13.7 m

so we will have

$x = v_x t$

$13.7 = 6 t$

$t = 2.28 s$

now we know that in vertical direction ball will move under uniform gravity so we can use kinematics

$y = v_y t + \frac{1}{2}at^2$

$y = 6.22(2.28) - \frac{1}{2}(9.81)(2.28^2)$

$y = -11.4 m$

Height of the building = 11.4 m

3 0

3 years ago

Read 2 more answers

In the Daytona 500 auto race, a Ford Thunderbird and a Mercedes Benz are moving side by side down a straightaway at 78.5 m/s. Th

Andrews [41]

Answer:

FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

Explanation:

First you have to consider that the Ford Thunderbird (FT) follows a rectilinear motion with varying acceleration, while Mercedez Benz (MB) has a constant velocity (no acceleration). So if you finde the time spent by FT in each section, and the distance, then you will find the distance for MB.

1) Vf² = Vi² + 2ad, where Vf: final velocity, Vi: ionitial velocity, a: acceleration and d: distance.

For the first portion (0 m/s)² = (78.5 m/s)² + 2a(250 m) ⇒

-(78.5 m/s)² / 2(250m) = a ⇒ a = -12.3 m/s².

Now, you can find the corresponding time for this section with the following formule: Vf = Vi + at ⇒ 0 m/s = 78.5 m/s + (-12.3 m/s²) t

⇒ t= (-78.5 m/s)/ (-12.3 m/s²) ⇒ t= 6.4 seconds.

2) Then FT spent 5 seconds in the pit.

3) The the FT accelerates until reach 78.5 m/s again in a distance of 370 m.

Vf² = Vi² + 2ad ⇒ (78.5 m/s)² = (0 m/s)² + 2a(370 m)

⇒ (78.5 m/s)²/ 2(370 m) = a ⇒ a = 8.3 m/s²

Then, Vf = Vi + at ⇒ 78.5 m/s = 0 m/2 + (8.3 m/s²) t

⇒ (78.5 m/s)/(8.3 m/s²) = t ⇒ t = 9.5 seconds.

4) Summarizing, the FT moves 620 meters (250 + 370 mts) in 20.9 seconds ( 6.4 s + 5 s + 9.5 s).

5) During this time, MB moves

Velocity = distance/ time ⇒ Velocity x time = Distance

⇒ Distance = (78.5 m/s) x (20.9 seconds) ⇒ Distance = 1640.6 meters

6) Finally, the FT is 1020.6 meters (1640.6 meters - 620 meters) far from MB

3 0

3 years ago

Lumbar nerves transmit signals that allow us to?

labwork [276]

Lumbar nerves transmit signals that allow us to walk. They are a pivotal part of the human body, and without them we would be immobile. Hope this helps.

7 0

4 years ago

Read 2 more answers

An electric motor spins at 1000 rpm and is slowing down at a rate of 10 t rad/s2 ; where t is measured in seconds. (a) If the mo

REY [17]

Answer:

a) The tangential component of acceleration at the edge of the motor at $t = 1.5\,s$ is -1.075 meters per square second.

b) The electric motor will take approximately 3.963 seconds to decrease its angular velocity by 75 %.

Explanation:

The angular aceleration of the electric motor ( $\alpha$ ), measured in radians per square second, as a function of time ( $t$ ), measured in seconds, is determined by the following formula:

$\alpha = -10\cdot t\,\left[\frac{rad}{s^{2}} \right]$ (1)

The function for the angular velocity of the electric motor ( $\omega$ ), measured in radians per second, is found by integration:

$\omega = \omega_{o} - 5\cdot t^{2}\,\left[\frac{rad}{s} \right]$ (2)

Where $\omega_{o}$ is the initial angular velocity, measured in radians per second.

a) The tangential component of aceleration ( $a_{t}$ ), measured in meters per square second, is defined by the following formula:

$a_{t} = R\cdot \alpha$ (3)

Where $R$ is the radius of the electric motor, measured in meters.

If we know that $R = 7.165\times 10^{-2}\,m$ , $\alpha = 10\cdot t$ and $t = 1.5\,s$ , then the tangential component of the acceleration at the edge of the motor is: