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Viefleur [7K]
2 years ago
14

What is a small portable device that counts every step taken throughout the day?

Physics
1 answer:
Valentin [98]2 years ago
7 0

Answer: B

Explanation: i learned it last year

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Would the frequency of the angular simple harmonic motion (SHM) of the balance wheel increase or decrease if the dimensions of t
storchak [24]

Answer:

Yes the frequency of the angular simple harmonic motion (SHM) of the balance wheel increases three times if the dimensions of the balance wheel reduced to one-third of original dimensions.

Explanation:

Considering the complete question attached in figure below.

Time period for balance wheel is:

T=2\pi\sqrt{\frac{I}{K}}

I=mR^{2}

m = mass of balance wheel

R = radius of balance wheel.

Angular frequency is related to Time period as:

\omega=\frac{2\pi}{T}\\\omega=\sqrt{\frac{K}{I}} \\\omega=\sqrt{\frac{K}{mR^{2}}

As dimensions of new balance wheel are one-third of their original values

R_{new}=\frac{R}{3}

\omega_{new}=\sqrt{\frac{K}{mR_{new}^{2}}}\\\\\omega_{new}=\sqrt{\frac{K}{m(\frac{R}{3})^{2}}}\\\\\omega_{new}={3}\sqrt{\frac{K}{mR^{2}}}\\\\\omega_{new}={3}\omega

5 0
3 years ago
the gravitational force between two objects is 1600 and what will be the gravitational force between the objects if the distance
Xelga [282]

I believe this is what you have to do:

The force between a mass M and a point mass m is represented by

F = G\frac{Mm}{r^{2} }

So lets compare it to the original force before it doubles, it would just be the exact formula so lets call that F₁

So F₁ = G(Mm/r^2)

Now the distance has doubled so lets account for this in F₂:

F₂ = G(Mm/(2r)^2)

Now square the 2 that gives you four and we can pull that out in front to give

F₂ = \frac{1}{4} G(Mm/r^2)

Now we can replace G(Mm/r^2) with F₁ as that is the value of the force before alterations

now we see that:

F₂ = \frac{1}{4} F₁

So the second force will be 0.25 (1/4) x 1600 or 400 N.



6 0
3 years ago
A cylindrical can 150 mm in diameter is filled to a depth of 100 mm with a fuel oil. The oil has a mass of 1.56 kg. Calculate it
Kazeer [188]

Answer:

Density (φ) = 0,8827 Kg/L

Specific weight (Ws) = 8,65 N/L

Specific gravity (Gs) = 0,8827 (without unit)

Explanation:

The density formula: φ = \frac{m}{V}

I know the mass "m", I need to find out the volume of the cylinder (V)

V = π* r²*h

The radius "r" is equal to half the diameter (150mm) = 75mm

Now I can find out the density (φ)

φ = \frac{1,56Kg}{1,767145L} = 0,8827 Kg/L

The specific weight (Ws) is the relationship between the weight of substance (oil) and its volume. We apply the following formula:

Ws = φ*g

(g = gravity = 9,8 m/s²)

Finally, specific gravity (Gs) is the ratio between the density of a substance (oil) "φ(o)" and the density of water "φ(w)" :

Gs = φ(o) /  φ(w)

(φ(w) = 1 Kg/L

Hope this can help you !!

3 0
3 years ago
Which of the following is Not true about the noble gases
Alexandra [31]
Noble gases are not highly reactive
3 0
4 years ago
A power plant running at 39 % efficiency generates 330 MW of electric power. Part A At what rate (in MW) is heat energy exhauste
marusya05 [52]

516.154 megawatts of heat are <em>exhausted</em> to the river that cools the plant.

By definition of energy efficiency, we derive an expression for the energy rate exhausted to the river (Q_{out}), in megawatts:

Q_{out} = Q_{in} - W

Q_{out} = \left(\frac{1}{\eta}-1 \right)\cdot W(1)

Where:

  • \eta - Efficiency.
  • W - Electric power, in megawatts.

If we know that \eta = 0.39 and W = 330\,MW, then the energy rate exhausted to the river is:

Q_{out} = \left(\frac{1}{0.39}-1 \right)\cdot (330\,MW)

Q_{out} = 516.154\,MW

516.154 megawatts of heat are <em>exhausted</em> to the river that cools the plant.

We kindly to check this question on first law of thermodynamics: brainly.com/question/3808473

7 0
3 years ago
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