Explain why magnesium chloride cannot be made from one atom of magnesium and one atom of chlorine

True or False? Sometimes, substances can be added to a reaction to cause the reaction to start or speed up. These are not consid

Sphinxa [80]

The correct anwser is A true

8 0

4 years ago

Read 2 more answers

A horse gallops for 3.5 minutes around the raceing track. During this time it changes velocity from 543 min to 628 min. Calculat

enot [183]

Answer:

The acceleration of the horse during this time interval is 24.286 m/min²

Explanation:

Given;

initial velocity of the horse during the gallop, u = 543 m /min

final velocity of the horse during the gallop, v = 628 m /min

time of motion, t = 3.5 minutes

The acceleration of the horse is given by the change in velocity per change in time;

$a = \frac{v-u}{t}\\\\a = \frac{628-543}{3.5}\\\\a = 24.286 \ m/min^2$

Therefore, the acceleration of the horse during this time interval is 24.286 m/min²

4 0

3 years ago

Please help!!

pantera1 [17]

Answer:

(a) The time the ball stays in the air is approximately 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is approximately 83.18 meters

(c) The maximum altitude of the ball is approximately 26.62 m

Explanation:

The given parameters of the question are;

The initial velocity at which a baseball cannon fires balls, u = 29 m/s

The angle at which the cannon is tilted, θ = 52°

(a) The time duration the ball stays in the air is given by the time of flight of the projected ball as follows;

$2 \cdot t = \dfrac{2 \cdot u \cdot sin (\theta)}{g}$

Where;

t = The time it takes the baseball to reach maximum height

2·t = The total time of flight = The time the ball stays in the air

θ = The angle at which the ball is tilted = 52°

g = The acceleration due to gravity ≈ 9.81 m/s²

u = The initial velocity = 29 m/s

Therefore, we have;

$2 \cdot t = \dfrac{2 \times 29 \ m/s\times sin (52^{\circ})}{9.81 \ m/s^2} \approx 4.66 \ s$

The time the ball stays in the air, 2·t ≈ 4.66 seconds

(b) The distance from the cannon at which the ball will hit the ground is given by the horizontal range, 'R', of the projectile as follows;

$Horizontal \ range, R = \dfrac{u^2 \cdot sin(2 \cdot \theta) }{g}$

∴ The distance from the cannon at which the ball will hit the ground = R

$Horizontal \ range, R = \dfrac{(29 \ (m/s))^2 \cdot sin(2 \times 52^{\circ}) }{9.81 \ m/s^2} \approx 83.18 \, m$

The distance from the cannon at which the ball will hit the ground = R ≈ 83.18 meters

(c) The maximum altitude of the ball is equal to the maximum height reached by the ball, H, which is given as follows;

$H = \dfrac{u^2 \cdot sin^2 (\theta)}{2 \cdot g}$

Therefore, we have;

$H = \dfrac{(29 \ m/s)^2 \times sin^2 (52^{\circ})}{2 \times 9.81 \ m/s^2} \approx 26.62 \ m$

The maximum altitude of the ball ≈ 26.62 m

3 0

3 years ago

What type of boundries are the collision zones and subduction zones ?

IrinaK [193]

In a subduction zone<span>, the </span>subducting<span> plate, which is normally a plate with oceanic crust, moves beneath the other plate,</span>

3 0

4 years ago

Given that two objects interact via some force, the accelerations of these two objects have the same magnitude but opposite dire

frosja888 [35]

Answer:

False.

Explanation:

Newton's 3rd Law tells us that if two objects interact via some force, then the value one exerts on the other have the same magnitude but opposite directions. Newton's 2nd Law tells us that the value of this force is F=ma. So, for two objects with different masses, if the forces they are exerting on each other are equal (as they must be by Newton's 3rd Law), the accelerations they are experimenting must necessarily be different.

7 0

3 years ago