The electrons that are gained or lost in an ionic bond are called...?

Por que no céu sem nuvens a lua nova não é visivel e a lua cheia aparece em grande destaque?

weeeeeb [17]

Porque la matematica es imposible

4 0

3 years ago

A dog is running at 15 m/s and then slows down to 10 m/s. It takes him two seconds to do this. What is the dog's acceleration

OLga [1]

Explanation:

a = (Vf-Vi)/t

= (10m/s - 15m/s)/2s

a = -5 / 2 m/s^2

• so the dog diaccelerate at the rate of 5/2 m/s^2

hope it helps

6 0

3 years ago

An airplane during departure has a constant acceleration of 3 m / s².

Rama09 [41]

Constant acceleration of plane = 3m/s²

a) Speed of the plane after 4s

Acceleration = speed/time

3m/s² = speed/4s

S = 12m/s

The speed of the plane after 4s is 12m/s.

b) Flight point will be termed as the point the plane got initial speed, u, 20m/s

Find speed after 8s, v

a = 3m/s²

from,

a = v - u

t

3 = v - 20

8

24 = v - 20

v = 44m/s

After 8s the plane would've 44m/s speed.

6 0

2 years ago

When the green arrow and solid red light is illuminated, __________?

otez555 [7]

When the green arrow and solid red light is illuminated, means you turn in the direction of the arrow.

5 0

3 years ago

The angle between the two force of magnitude 20N and 15N is 60 degrees (20N force being horizontal) determine the resultant in m

BARSIC [14]

A) The resultant force is 30.4 N at $25.3^{\circ}$

B) The resultant force is 18.7 N at $43.9^{\circ}$

Explanation:

A)

In order to find the resultant of the two forces, we must resolve each force along the x- and y- direction, and then add the components along each direction to find the components of the resultant.

The two forces are:

$F_1 = 20 N$ at $0^{\circ}$ above x-axis

$F_2 = 15 N$ at $60^{\circ}$ above y-axis

Resolving each force:

$F_{1x}=F_1 cos \theta = (20)(cos 0)=20 N\\F_{1y}=F_1 sin \theta =(20)(sin 0)=0 N$

$F_{2x}=F_2 cos \theta = (15)(cos 60)=7.5 N\\F_{2y}=F_2 sin \theta =(15)(sin 60)=13.0 N$

So, the components of the resultant are:

$F_x = F_{1x}+F_{2x}=20+7.5 = 27.5 N\\F_y = F_{1y}+F_{2y}=0+13.0=13.0 N$

And the magnitude of the resultant is:

$F=\sqrt{F_x^2+F_y^2}=\sqrt{27.5^2+13.0^2}=30.4 N$

And the direction is:

$\theta=tan^{-1}(\frac{F_y}{F_x})=tan^{-1}(\frac{13.0}{27.5})=25.3^{\circ}$

B)

In this case, the 15 N is applied in the opposite direction to the 20 N force. Therefore we need to re-calculate its components, keeping in mind that the angle of the 15 N force this time is

$\theta=180^{\circ}-60^{\circ}=120^{\circ}$