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Dmitrij [34]
3 years ago
8

Balance the following equation Fe203+HCL fecl3+H20

Physics
1 answer:
Arturiano [62]3 years ago
3 0
Balanced equation:
6 HCl + Fe2O3 = 2 FeCl3 + 3 H2O
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IMPORTANT QUESTION ANSWER QUICK PLEASE!!!!!!
Flura [38]
Ⓘ ⒷⒺⓁⒾⒺⓋⒺ ⒾⓉ ⒾⓈ Ⓒ ⒽⒶⓋⒺ Ⓐ ⒼⓄⓄⒹ ⒹⒶⓎ
7 0
3 years ago
You are in charge of a cannon that exerts a force 11500 N on a cannon ball while the ball is in the barrel of the cannon. The le
IRISSAK [1]

Answer:

m = 7.48 kg

Explanation:

force (f) = 11,500 N

length of barrel (s) = 1.7 m

angle above the ground = 49.3 degrees

acceleration due to gravity (g) = 9.8 m/s^{2}

initial velocity (u) = 0 m/s

final velocity (v) = 72.3 m/s

mass (m) = ?

force = mass (m) x acceleration (a)

acceleration (a) = force / mass (m)

acceleration (a) = 11500 / m

applying the equation of motion v^{2} = u^{2} + 2as , we can get the mass

72.3^{2} = 0^{2} + (2 x \frac{11500}{m} x 1.7 )

5227.3 = 0 + \frac{39100}{m}

m =  \frac{39100}{5227.3}

m = 7.48 kg

5 0
3 years ago
Two cars are facing each other. Car A is at rest while car B is moving toward car A with a constant velocity of 20 m/s. When car
lapo4ka [179]

Answer:

Let's define t = 0s (the initial time) as the moment when Car A starts moving.

Let's find the movement equations of each car.

A:

We know that Car A accelerations with a constant acceleration of 5m/s^2

Then the acceleration equation is:

A_a(t)  = 5m/s^2

To get the velocity, we integrate over time:

V_a(t) = (5m/s^2)*t + V_0

Where V₀ is the initial velocity of Car A, we know that it starts at rest, so V₀ = 0m/s, the velocity equation is then:

V_a(t) = (5m/s^2)*t

To get the position equation we integrate again over time:

P_a(t) = 0.5*(5m/s^2)*t^2 + P_0

Where P₀ is the initial position of the Car A, we can define P₀ = 0m, then the position equation is:

P_a(t) = 0.5*(5m/s^2)*t^2

Now let's find the equations for car B.

We know that Car B does not accelerate, then it has a constant velocity given by:

V_b(t) =20m/s

To get the position equation, we can integrate:

P_b(t) = (20m/s)*t + P_0

This time P₀ is the initial position of Car B, we know that it starts 100m ahead from car A, then P₀ = 100m, the position equation is:

P_b(t) = (20m/s)*t + 100m

Now we can answer this:

1) The two cars will meet when their position equations are equal, so we must have:

P_a(t) = P_b(t)

We can solve this for t.

0.5*(5m/s^2)*t^2 = (20m/s)*t + 100m\\(2.5 m/s^2)*t^2 - (20m/s)*t - 100m = 0

This is a quadratic equation, the solutions are given by the Bhaskara's formula:

t = \frac{-(-20m/s) \pm \sqrt{(-20m/s)^2 - 4*(2.5m/s^2)*(-100m)}  }{2*2.5m/s^2} = \frac{20m/s \pm 37.42 m/s}{5m/s^2}

We only care for the positive solution, which is:

t = \frac{20m/s + 37.42 m/s}{5m/s^2} = 11.48 s

Car A reaches Car B after 11.48 seconds.

2) How far does car A travel before the two cars meet?

Here we only need to evaluate the position equation for Car A in t = 11.48s:

P_a(11.48s) = 0.5*(5m/s^2)*(11.48s)^2 = 329.48 m

3) What is the velocity of car B when the two cars meet?

Car B is not accelerating, so its velocity does not change, then the velocity of Car B when the two cars meet is 20m/s

4)  What is the velocity of car A when the two cars meet?

Here we need to evaluate the velocity equation for Car A at t = 11.48s

V_a(t) = (5m/s^2)*11.48s = 57.4 m/s

7 0
3 years ago
A ball has a mass of 0.046kg. Calculate the change in gravitational potential energy when the ball is lifted through a vertical
Nat2105 [25]

Answer:

Gravitational potential energy is 0.92414J

Kinetic energy is 0.28175J

5 0
3 years ago
The potential energy of a pear is 5.0 joules. The pear is 4 m high. What is the mass of the apple?​
gregori [183]

Answer:

rrteertewtagtewyewtyewtewy

Explanation:

8 0
4 years ago
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