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1 year ago

Please Help! The thing is in the photo.

Physics

1 answer:

ivolga24 [154]1 year ago

5 0

The reaction 4Al + 3O₂ → 2Al₂O₃ is a synthesis or also called combination reaction because reactants combine to produce a new compound (Option 2).

<h3>What is a synthesis combination reaction?</h3>

In chemistry, a synthesis or combination reaction is any type of chemical reaction in which two or more molecules called reactants combine to produce one or more molecules called products of such reaction.

Therefore, with this data, we can see that a synthesis or combination reaction is a chemical process where two or more molecules combine to produce a new molecule known as a product.

Learn more about synthesis combination reactions here:

brainly.com/question/14703806

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Three children are riding on the edge of a merry-go-round that is a solid disk with a mass of 102 kg and a radius of 1.53 m. The

Mnenie [13.5K]

Three children of masses and their position on the merry go round

M1 = 22kg

M2 = 28kg

M3 = 33kg

They are all initially riding at the edge of the merry go round

Then, R1 = R2 = R3 = R = 1.7m

Mass of Merry go round is

M =105kg

Radius of Merry go round.

R = 1.7m

Angular velocity of Merry go round

ωi = 22 rpm

If M2 = 28 is moves to center of the merry go round then R2 = 0, what is the new angular velocity ωf

Using conservation of angular momentum

Initial angular momentum when all the children are at the edge of the merry go round is equal to the final angular momentum when the second child moves to the center of the merry go round Then,

L(initial) = L(final)

Ii•ωi = If•ωf

So we need to find the initial and final moment of inertia

NOTE: merry go round is treated as a solid disk then I= ½MR²

I(initial)=½MR²+M1•R²+M2•R²+M3•R²

I(initial) = ½MR² + R²(M1 + M2 + M3)

I(initial) = ½ × 105 × 1.7² + 1.7²(22 + 28 + 33)

I(initial) = 151.725 + 1.7²(83)

I(initial) = 391.595 kgm²

Final moment of inertial when R2 =0

I(final)=½MR²+M1•R²+M2•R2²+M3•R²

Since R2 = 0

I(final) = ½MR²+ M1•R² + M3•R²

I(final) = ½MR² + (M1 + M3)• R²

I(final)=½ × 105 × 1.7² + ( 22 +33)•1.7²

I(final) = 151.725 + 158.95

I(final) = 310.675 kgm²

Now, applying the conservation of angular momentum

L(initial) = L(final)

Ii•ωi = If•ωf

391.595 × 22 = 310.675 × ωf

Then,

ωf = 391.595 × 22 / 310.675

ωf = 27.73 rpm

Answer: So, the final angular momentum is 27.73 revolution per minute

7 0

3 years ago

The four wheels of a car are connected to the car's body by spring assemblies that let the wheels move up and down over bumps an

Marrrta [24]

Answer: $k=55.590 KN/m$

Explanation:

Given

mass of person $\left ( m\right )$ =68 kg

car dips about 1.2 cm

We know

F=kx

Where k=combined spring constant

mg=kx

$k=\frac{mg}{x}$

$k=\frac{68\times 9.81}{1.2\time 10^{-2}}$

$k=55.590 KN/m$

7 0

3 years ago

Fill in the correct answers. A big wheel with twice the circumference of a small wheel will rotate with ____ the force and ___ t

Lina20 [59]

Answer:

Explanation:

""""""" Twice,Half """"""

3 0

3 years ago

Read 2 more answers

A toroid with a square cross section 3.0 cm ✕ 3.0 cm has an inner radius of 25.1 cm. It is wound with 600 turns of wire, and it

coldgirl [10]

Answer:

B = 1.353 x 10⁻³ T

Explanation:

The Magnetic field within a toroid is given by

B = μ₀ NI/2πr, where N is the number of turns of the wire, μ₀ is the permeability of free space, I is the current in each turn and r is the distance at which the magnetic field is to be determined from the center of the toroid.

To find r we need to add the inner radius and outer radius and divide the value by 2. Hence,

r = (a + b)/2, where a is the inner radius and b is the outer radius which can be found by adding the length of a square section to the inner radius.

b = 25.1 + 3 = 28.1 cm

a = 25.1 cm

r = (25.1 + 28.1)/2 = 26.6 cm = 0.266m

B = 4π x 10⁻⁷ x 600 x 3/2π x 0.266

B = 1.353 x 10⁻³ T

The strength of the magnetic field at the center of the square cross section is 1.3 x 10⁻³ T

5 0

3 years ago

The flywheel of a steam engine begins to rotate from rest with a constant angular acceleration of 1.33 rad/s2. It accelerates fo

Svetradugi [14.3K]

Answer:

We first to know that if the wheel rotates from rest means that at t=0 the velocity and the angle rotated is 0.

Then, we know:

$\alpha = 1.33 = \frac{dw}{dt}$

Integrating 2 times, we have:

$w = 1.33t\\angle =0.665t^{2}$

For the first 27.9 s, we have:

w = 37.107 rad/s

angle = 517.6426 rad

For the next seconds, according to the text, the angular velocity is constant so

w = 37.107 rad/s and hence, integrating:

$angle =37.107t$

Then, the time remaining is:

53.5 - 27.9 = 25.6

So for the next 25.6 seconds we have:

$angle = 37.107*25.6=949.9392 rad$

Finally, we add the 2 angles and we have as a result:

$angle = 517.6426+949.9392=1467.5818$

3 0

3 years ago