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bearhunter [10]
3 years ago
6

Mass (kg)

Physics
2 answers:
Tju [1.3M]3 years ago
8 0

Mass.

Because mass doesn't depend on weight but weight depends on mass.

blondinia [14]3 years ago
6 0
Mass is the answer to this
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You and your friends are having a discussion about weight. He/she claims that he/she weighs less on the 100th floor of a buildin
Viktor [21]

Answer:

if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

Explanation:

The weight of a person in the force with which the Earth attracts the person, therefore can be calculated using the law of universal attraction

          F = G m M / r²

Where m is the mass of the person, M the masses of the earth

Let's call the person's weight at ground level as Wo and suppose the distance to the center of the Earth is Re

            W₀ = G m M / Re²

In the calculation of the weight of the person on the 100th floor the only thing that changes is the distance

          r = Re + 100 r₀

Where r₀ is the distance between the floors, which is approximately 2.5 m, so the distance is

         r = Re + 250

We substitute

     W = G m M / r²

      W = G m M / (Re + 250)²

The value of Re is 6.37 10⁶ m, so we can take it out as a factor and perform a serial expansion of the remaining fraction

      W = G m M / Re² (1+ 250 / Re)²

      (1 + 250 / Re)⁻² = 1 + (-2) 250 / Re + (-2 (-2-1)) / 2 (250 / Re)² +….

The value of the expression is

      (1 + 250 / Re)⁻² = 1 -2 250 / 6.37 10⁶ -30 (250 / 6.37)² 10⁻¹² + ...

We can see that the quadratic term is very small, which is why we despise it, we substitute in the weight equation

      W = G m M / Re² (1 - 78.5 10⁻⁶)

Remains

     W = Wo (1 - 7.85  10⁻⁵)

We can see that if the weight theoretically decreases at this height, but in a fraction of 10⁻⁵, which is not appreciable in any scale, therefore, the reading of the scale in the two places is the same.

4 0
3 years ago
Read 2 more answers
Imagine an alternate universe where all of the quantum number rules were identical to ours except m_{s} had three allowed values
marishachu [46]

Answer:

so in a given orbital there can be 3 electrons.

Explanation:

The Pauli exclusion principle states that all the quantum numbers of an electron cannot be equal, if the spatial part of the wave function is the same, the spin part of the wave function determines how many electrons fit in each orbital.

In the case of having two values, two electrons change. In the case of three allowed values, one electron fits for each value, so in a given orbital there can be 3 electrons.

5 0
3 years ago
The moon has a diameter of 3.48 x 106 m and is a distance of 3.85 x 108 m from the earth. The sun has a diameter of 1.39 x 109 m
Mrrafil [7]

Answer:

0.00903 rad

0.00926 rad

6.268\times 10^{-6}

Explanation:

s = Diameter of the object

r = Distance between the Earth and the object

Angle subtended is given by

\theta=\frac{s}{r}

For the Moon

\theta_m=\dfrac{3.48\times 10^6}{3.85\times 10^8}\\\Rightarrow \theta_m=0.00903\ rad

The angle subtended by the Moon is 0.00903 rad

For the Sun

\theta_s=\dfrac{1.39\times 10^9}{1.5\times 10^{11}}\\\Rightarrow \theta_s=0.00926\ rad

The angle subtended by the Sun is 0.00926 rad

Area ratio is given by

\frac{A_m}{A_s}=\dfrac{\pi r_m^2}{\pi r_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{d_m^2}{d_s^2}\\\Rightarrow \frac{A_m}{A_s}=\dfrac{(3.48\times 10^{6})^2}{(1.39\times 10^9)^2}\\\Rightarrow \frac{A_m}{A_s}=6.268\times 10^{-6}

The area ratio is 6.268\times 10^{-6}

3 0
3 years ago
The picture shows a loop of wire rotating in a magnetic field in a generator. Determine the direction of the current through the
BaLLatris [955]
I believe the answer is b
3 0
2 years ago
Read 2 more answers
You attach a 2.90 kg mass to a horizontal spring that is fixed at one end. You pull the mass until the spring is stretched by 0.
Murljashka [212]

Answer:

Explanation:

The spring is stretched by .5 m and then released that means its amplitude of oscillation A is 0.5 m .

A = 0.5 m

After the release at one extreme point , the mass comes to rest again at another extreme point after half the time period ie

T / 2 = .3 s

T = 0.6 s

Angular velocity

ω = \frac{2\times \pi}{T}

ω = \frac{2\times \pi}{0.6}

ω = 10.45

Maximum velocity  = ω A

ω and  A are angular velocity and amplitude of oscillation.

Maximum velocity  = 10.45 x .5

= 5.23 m /s

7 0
3 years ago
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