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Naddika [18.5K]

3 years ago

9

Instead of using his book to hold down fluttering papers, Sam looks for a paperweight. What type of problem-solving obstacle is

this?

1 answer:

lesya [120]3 years ago

7 0

I think it is which is better in other words which one would hold better

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A 1.0 kg cart moving right at 5.0 m/s on a frictionless track collides with a second cart moving at 2.0m/s. The 1.0 kg cart has

____ [38]

Answer:

m₂ = 3kg

Explanation:

The question wasn't clear about what direction the initial velocity of the second cart was, so I'll assume it was going left at 2.0m/s.

Anyway, this is a conservation of momentum problem. The equation you need to use is the one written in blue. They want you to solve for the mass of the second cart, so do some algebra and rearrange that blue equation in term of m₂.

Now that you have the equation for m₂, plug in all the values given from the question and you'll get 3kg.

3 0

4 years ago

What are the three rules for writing the electron configuration of elements

astraxan [27]

<em></em>

Answer:

<u><em>The aufbau principle</em></u>

<u />

<u><em>The Pauli exclusion principle</em></u>

<u><em></em></u>

<u><em>Hund's rule of maximum multiplicity</em></u>

Explanation:

<u><em>The aufbau principle:</em></u>

<em></em>

The fundamental electronic configuration is achieved by placing the electrons one by one in the different orbitals available for the atom, which are arranged in increasing order of energy.

<u><em>The Pauli exclusion principle:</em></u>

<em></em>

Two electrons of the same atom cannot have their four equal quantum numbers. Because each orbital is defined by the quantum numbers n, l, and m, there are only two possibilities ms = -1/2 and ms = +1/2, which physically reflects that each orbital can contain a maximum of two electrons, having opposite spins

<u><em>Hund's rule of maximum multiplicity:</em></u>

This rule says that when there are several electrons occupying degenerate orbitals, of equal energy, they will do so in different orbitals and with parallel spins, whenever this is possible. Because electrons repel each other, the minimum energy configuration is one that has electrons as far away as possible from each other, and that is why they are distributed separately before two electrons occupy the same orbital.

4 0

3 years ago

I'm completely stumped as to how to do this.

worty [1.4K]

Explanation:

You have already determined the components of the known forces so I won't repeat your work here. Since the resultant force $\vec{\textbf{R}}$ and F1 are completely along the x-axis, we can conclude that

$F_{2y} = F_{3y} \Rightarrow F_{3y} = F_3\cos{\theta} = 250\:\text{lb}$

We can now solve for the magnitude of $F_3:$

$F_3 = \sqrt{F_{3x}^2 + F_{3y}^2} = \sqrt{(467)^2 + (250)^2}$

$\:\:\:\:=529.7\:\text{lb}$

The angle $\theta$ is then

$\tan{\theta} = \dfrac{F_{3y}}{F_{3x}} = \dfrac{250}{467}$

or

$\theta = 49.2°$

6 0

3 years ago

f the mass of the block is 2 kg, the radius of the circle is 0.8 m, and the speed of the block is 3 m/s, what is the tension in

alisha [4.7K]

Answer:

the size are components relative to the whole.

Explanation:

they are particularly good at showing percentage or proportional data

6 0

3 years ago

A car has a mass of 2000 kg. While it is traveling along a perfectly flat road, it goes around an unbanked turn that has a radiu

Papessa [141]

Answer:

2.5 m/s²

Explanation:

The given parameters are;

The mass of the car, m = 2,000 kg

The radius of the car, r = 40.0 m

The coefficient of friction between the car tires and the road, μ = 0.500

The constant speed with which the car moves, v = 10.0 m/s

The normal reaction of the road on the car, N = The weight of the car;

∴ N = m × g

Where;

g = The acceleration due to gravity ≈ 9.81 m/s²

N ≈ 2,000 kg × 9.81 m/s² = 19,620 N

The frictional force, $F_f$ = μ × N

The centripetal force, $F_c$ = m·v²/r

The car moves without slipping when $F_f$ = $F_c$

Therefore, $F_f$ = 0.500 × 19,620 N = 2,000 kg × $v_{max}$ ²/40.0 m

∴ $v_{max}$ = √(0.500 × 19,620 N × 40.0 m/2,000 kg) ≈ 14.007 m/s

Therefore, the velocity with which the car moves, v < $v_{max}$

The cars centripetal acceleration, $a_c$ = v²/r

∴ $a_c$ = (10.0 m/s)²/40.0 m = 2.5 m/s²

The cars centripetal acceleration as it goes round the turn, $a_c$ = 2.5 m/s².

7 0

3 years ago