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15 days ago

You are at the park with your little brother, when you notice a small merry-go-round with a radius that looks to be about 1.5 m.

Your brother climbs on, and you give him a spin. From rest, you speed up smoothly, completing one full lap in 3.0 seconds. You wonder what sort of acceleration (magnitude and direction) your brother experiences at the very start of the motion, and at the very end of the first lap. He is quite young and hasn't studied any physics, so he does not know how to answer your question. You decide to give him your smartphone (which has a built-in accelerometer), and repeat the experiment. What will the smartphone record

Physics

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A metal, M, forms an oxide having the formula M2O3 containing 52.92% metal by mass. Determine the atomic weight in g/mole of the

ValentinkaMS [3465]

Answer:

The molar mass of the metal in grams per mole is calculated to be 8.87.

Explanation:

Initially, we can consider a sample of the compound weighing 100 g. This results in:

52.92% metal: 52.92 g M
47.80% oxygen: 47.80 g O

By utilizing the molar mass of oxygen, which is 16 g / mol, we can determine the quantity of moles of oxygen in the sample via the rule of three:

$moles of oxygen=\frac{47.8g*1mol}{16g}$

moles of oxygen=2.9875

The formula for the metal oxide indicates that:

2 M⁺³ + 3 O²⁻ ⇒ M₂O₃

From the previous equation, it is evident that 3 oxygen ions are necessary to react with 2 metal ions. Hence:

$2.9875 moles of oxygen*\frac{2 moles of metal M}{1 mol of oxygen} = 5.975 moles of metal M$

Given 52.92 g of metal in the sample, the molar mass of the metal is:

$molar mass=\frac{52.92 g}{5.975 mol}$

molar mass≅ 8.87 g/mol

The molar mass of the metal in grams per mole is 8.87.

The value that most closely corresponds to this is Beryllium (Be), which has an atomic mass of 9.0122 g / mol.

3 0

2 months ago

The magnitude of the net force versus time graph has a rectangular shape. Often in physics geometric properties of graphs have p

Maru [3345]

True. Explanation: In this instance, the area of the graph represents the impulse. Impulse is defined as the change in an object's momentum. Moreover, it is also expressed as the product of the force acting on an object and the duration of the impact. When we graph the force against time, if the force remains constant, the resultant graph will take on a rectangular shape, and the area under that graph will equal the impulse's definition.

8 0

2 months ago

When a downward force is applied at a point 0.60 m to the left of a fulcrum, equilibrium is obtained by placing a mass of 10–7 k

ValentinkaMS [3465]

Answer:

force = 6.53× $10^{-7}$ N

Explanation:

Provided data

downward force = 0.60 m

mass m = $10^{-7}$ kg

distance h = 0.40 m

to determine

magnitude of the downward force

solution

we know here mg is apply 0.4 m away from support and

thus applied force is d = 0.6 m from support

therefore

by balancing torque we can compute force

force = mass × g × h / d

substituting the values

force = mass × g × h / d

force = ( $10^{-7}$ × 9.81 × 0.4 ) / 0.6

force = 6.53× $10^{-7}$ N

4 0

1 month ago

Explain whether or not there is any difference between a light ray emitted by a candle flame and one reflected off the cover of

inna [3103]

The distinction lies in the fact that the candle emits an emission spectrum, while the book reflects an absorption spectrum. In the case of the book, light is observed from all directions, causing its reflection to be diffuse. Explanation: The light emitted by a candle originates from the heat of the flame, composed of a combination of emissions from a black body at that temperature along with emissions from the chemical elements within the candle. On the contrary, the light reflected off a book cover consists of the incident light spectrum minus the wavelengths that trigger electronic transitions in the cover's elements, resulting in dark areas on the spectrum. Consequently, the difference stems from the candle producing an emission spectrum, whereas the book showcases an absorption spectrum. For a book's cover to reflect light specularly, incident rays would need to reflect uniformly, creating dark areas. However, since light is observed from all directions when reflecting off a book, the result is diffuse reflection.

6 0

1 month ago

The two sides of the DNA double helix are connected by pairs of bases (adenine, thymine, cytosine, and guanine). Because of the

kicyunya [3294]

Answer:

Explanation:

An image of the bond resulting from the search is attached.

Consider the force directed towards thymine as negative.

For the O-H-N combination:

The resulting force from this combination is:

$F=-F_{OH}+F_{ON}\\\\=\frac{Ke^2}{r^2}+\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[(0.280-0.110)\times 10^{-9}m]^2}+\frac{1}{0.280\times 10^{-9}m)^2}]\\\\=-5.03354\times 10^{-9}N$

In the case of the N-H-N combination:

The total force acting from this combination is:

$F'=F_{NN}-F_{HN}\\\\=\frac{Ke^2}{r^2}-\frac{Ke^2}{r'^2}\\\\=Ke^2(\frac{-1}{r^2}+\frac{1}{r'^2})\\\\=(9.0\times 10^9Nm^2/kg^2)(1.6\times 10^{-19}C)^2[\frac{1}{[0.300\times 10^{-9}m]^2}-\frac{1}{((0300-0..110)\times 10^{-9})m)^2}]\\\\=-3.822\times 10^{-9}N$

The force that thymine applies on adenine is:

$F_{net}=F+F'\\\\=-5.03354\times 10^{-9}N-3.822\times 10^{-9}N\\\\=-8.8558\times 10^{-9}N$

When rounded to three significant figures, the net force is $8.86\times 10^{-9}N$

The negative value indicates that the force is attractive, as it is aimed towards thymi.

The force acting on the electron due to the proton is:

$F=\frac{Ke^2}{r^2}\\\\=\frac{(9.0\times 10^9Nm^2/C^2)(1.6\times 10^{-9}C)^2}{(5.29\times 10^{-11}m)^2}\\\\=8.233\times 10^{-8}N$

Since the electron and proton carry opposite charges, the force on the electron points towards the proton.

The ratio of the above forces is:

$\frac{F}{F_{net}}=\frac{8.233\times 10^{-8}N}{8.233\times 10^{-8}N}\\\\=9.3$

Therefore, the bonding strength of the electron in the hydrogen atom is 9.3 times greater than the bonding force between adenine and thymine molecules.

5 0

2 months ago