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SSSSS [86.1K]
3 years ago
6

Balance the equation- Al+Mn02 ———-> Mn + Al2O3

Physics
1 answer:
ivanzaharov [21]3 years ago
7 0

Answer:

4Al + 3Mno2 --> 2Al2o3 + 3Mn

Explanation:

Al = 1 x 4 = 4

Mn = 1 x 3 = 3

O = 2 x 3 = 6

----------

Al = 2 x 2 = 4

Mn  = 1 x 3 = 3

O = 2 x 3 = 6

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C

Explanation:

that's just what I learned in school

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2 years ago
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Hi gir_ls join nkd-mbja-nuj​
GREYUIT [131]

Answer:

never lol

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but why all are doing I don't know=_=

3 0
2 years ago
The brakes of a 125 kg sled are applied while it is moving at 8.1 m/s, which exerts a force of 261 N to slow the sled down. How
sammy [17]

Answer:

15.7 m

Explanation:

m = mass of the sled = 125 kg

v₀ = initial speed of the sled = 8.1 m/s

v = final speed of sled = 0 m/s

F = force applied by the brakes in opposite direction of motion = 261

d = stopping distance for the sled

Using work-change in kinetic energy theorem

- F d = (0.5) m (v² - v₀²)

- (261) d = (0.5) (125) (0² - 8.1²)

d = 15.7 m

6 0
3 years ago
Which of these would be an example of unbalanced forces?
mylen [45]

A snowball picks up speed as it rolls down the mountain.<em> (D)</em>

Since the description includes acceleration ("picks up speed"), we know that the forces on the snowball must be unbalanced.

3 0
3 years ago
Read 2 more answers
a rocket, initially at rest, is fired vertically with a net upward acceleration of 12 m/s2 . at an altitude of 0.50 km, the engi
kobusy [5.1K]

The rocket travelled a maximum height at 1.0102 km.

Given,

The acceleration of a rocket (a) = 12 m/s²

The altitude of the rocket (s) =  0.50 km = 0.5×10³m

The maximum height of the rocket (h) = ?

Solution,

A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.

The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²

(a) = Δv/Δt

Where , Δv is change in velocity and Δt is change in time.

The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.

(v)= Δx/Δt

Where,Δx is the change in position and Δt is change in time & v is velocity.

Therefore we know the equation of motion is written as,

v² = u² +2as

Where, v  is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.

Then putting the value ,

v² = 0 + ( 2× 10 × 0.5×10³)m/s

v² = \sqrt{10000} m/s

v = 100 m/s

Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,

v² = u² - 2(g)h

h = (v²- u² ) / 2g

h = 10,000/2×9.8

h = 510.2 m

So that the rocket travelled the maximum height ,

(h)= (0.5 km + 510.2m)

(h) = 1.0102 km

Hence, the rocket travelled at the maximum height h is 1.0102 km

To know more about acceleration

brainly.com/question/15135960

#SPJ4

4 0
1 year ago
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