All categories

3 years ago

An example of an exothermic reaction is:

Physics

1 answer:

elena55 [62]3 years ago

6 0

Answer:

Explanation:

Exothermic refers to a RELEASE of energy through a reaction(usually chemical). When salt dissociates, polar water molecules pull apart ions, releasing energy.

You might be interested in

Three identical charges q form an equilateral triangle of side a with two charges on the x-axis and one on the positive y-axis.

shusha [124]

Answer:

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

Explanation:

Given:

- Three identical charges q.

- Two charges on x - axis separated by distance a about origin

- One on y-axis

- All three charges are vertices

Find:

- Find an expression for the electric field at points on the y-axis above the uppermost charge.

- Show that the working reduces to point charge when y >> a.

Solution

- Take a variable distance y above the top most charge.

- Then compute the distance from charges on the axis to the variable distance y:

$r = \sqrt{(\frac{\sqrt{3}*a }{2} + y)^2 + (a/2)^2 }$

- Then compute the angle that Force makes with the y axis:

cos(Q) = sqrt(3)*a / 2*r

- The net force due to two charges on x-axis, the vertical components from these two charges are same and directed above:

F_1,2 = 2*F_x*cos(Q)

- The total net force would be:

F_net = F_1,2 + kq / y^2

- Hence,

$F_n = k*q*(\frac{2*(y + \frac{\sqrt{3}*a }{2}) }{((y+ \frac{\sqrt{3}*a }{2})^2 + (a/2)^2)^1.5 } +\frac{1}{y^2} )$

- Now for the limit y >>a:

$F_n = k*q*(\frac{2*y(1 + \frac{\sqrt{3}*a }{2*y}) }{y^3((1+ \frac{\sqrt{3}*a }{2*y})^2 + (a/y*2)^2)^1.5 }) +\frac{1}{y^2} )$

- Insert limit i.e a/y = 0

$F_n = k*q*(\frac{2}{y^2} +\frac{1}{y^2}) \\\\F_n = 3*k*q/y^2$

Hence the Electric Field is off a point charge of magnitude 3q.

8 0

3 years ago

Need help finding the average speed.

Snowcat [4.5K]

Explanation:

To find the average of these numbers, we just have to add the three numbers together and divide by 3.

2.07 + 0. 74 + 1.33 = 4.14. 4.14 / 3 = 1.38
1.09 + 1.40 + 0.31 = 2.8. 2.8 / 3 ≈ 9.3333333/ 9 1/3
0.95 + 1.61 + 0.56 = 3.12 / 3 = 1.04
0.81 + 1.89 + 1.08 = 3.78 / 3 = 1.26

5 0

3 years ago

If Chris throws the baseball 60 meters forward in 4.2 sec, what is the velocity of the ball? I need answer immediately

vazorg [7]

Answer:

14.3 m/s

Explanation:

velocity equation

v= d/t

v= 60/4.2

v=14.28

round it to 1 decimal place

v= 14.3m/s

5 0

2 years ago

All objects at any temperature emit radiant energy true or false

kirill [66]

The answer is true.
All object at any temperature emit radiant energy.

4 0

3 years ago

Read 2 more answers

A 38.0 kg satellite has a circular orbit with a period of 1.30 h and a radius of 7.90 × 106 m around a planet of unknown mass. I

Alik [6]

Answer:

5.44×10⁶ m

Explanation:

For a satellite with period t and orbital radius r, the velocity is:

v = 2πr/t

So the centripetal acceleration is:

a = v² / r

a = (2πr/t)² / r

a = (2π/t)² r

This is equal to the acceleration due to gravity at that elevation:

g = MG / r²

(2π/t)² r = MG / r²

M = (2π/t)² r³ / G

At the surface of the planet, the acceleration due to gravity is:

g = MG / R²

Substituting our expression for the mass of the planet M:

g = [(2π/t)² r³ / G] G / R²

g = (2π/t)² r³ / R²

R² = (2π/t)² r³ / g

R = (2π/t) √(r³ / g)

Given that t = 1.30 h = 4680 s, r = 7.90×10⁶ m, and g = 30.0 m/s²:

R = (2π / 4680 s) √((7.90×10⁶ m)³ / 30.0 m/s²)

R = 5.44×10⁶ m

Notice we didn't need to know the mass of the satellite.

8 0

3 years ago