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lidiya [134]
3 years ago
8

A data chrt shows the height of a person on his birthday each year for 10 years what is the dependent variable

Physics
1 answer:
KiRa [710]3 years ago
8 0
The height of a person is the dependent variable because it depends on the years.
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Find the period of the leg of a man who is 1.83 m in height with a mass of 67 kg. The moment of inertia of a cylinder rotating a
In-s [12.5K]

Answer:

T = 1.108\,s

Explanation:

The period of a physical pendulum is:

T = \sqrt{\frac{I_{O}}{m\cdot g \cdot L} }

T=2\cdot \pi \sqrt{\frac{\frac{1}{3}\cdot m \cdot L^{2} }{m\cdot g\cdot L} }

T=2\cdot \pi \sqrt{\frac{L }{3\cdot g} }

The length of the leg is approximately the height of the person:

L = 0.915\,m

The period is:

T = 2\cdot \pi \sqrt{\frac{0.915\,m}{3\cdot (9.807\,\frac{m}{s^{2}} )} }

T = 1.108\,s

4 0
3 years ago
Unless otherwise authorized, the maximum indicated airspeed at which aircraft may be flown when at or below 2,500 feet AGL and w
andrew11 [14]

Answer: 200 knots

Explanation: the maximum indicated airspeed at which aircraft may be flown when at or below 2,500 feet AGL and within 4 nautical miles of the primary airport of Class C airspace is 200 KNOTS

3 0
3 years ago
Radio devices are used for communication in space. Why
Elena-2011 [213]
Because radio waves can travel in space but sound waves cannot.
6 0
3 years ago
The masses of the Moon and Earth are 7.35 x 1022 kg and 5.97 x 1024 kg, respectively. The strength of the gravitational force be
bixtya [17]

Answer:

Distance between centre of Earth and centre of Moon is 3.85 x 10⁸ m

Explanation:

The attractive force experienced by two mass objects is known as Gravitational force.

The gravitational force is determine by the relation:

F=\frac{Gm_{1} m_{2} }{d^{2} }      ....(1)

According to the problem,

Mass of Moon, m₁ = 7.35 x 10²² kg

Mass of Earth, m₂ = 5.97 x 10²⁴ kg

Gravitational force experienced by them, F = 1.98 x 10²⁰ N

Universal gravitational constant, G = 6.67 x 10⁻¹¹ Nm²kg⁻²

Substitute these values in equation (1).

1.98\times10^{20} =\frac{6.67\times10^{-11}\times7.35\times10^{22}\times5.97\times10^{24} }{d^{2} }

d^{2}=\frac{2.93\times10^{37}}{1.98\times10^{20}}

d=\sqrt{1.48\times10^{17}}

d = 3.85 x 10⁸ m

3 0
3 years ago
A carbon rod with a radius of 1.9 mm is used to make a resistor. What length of the carbon rod should be used to make a 3.7 Ω re
vladimir2022 [97]

Answer:

Length = 2.32 m

Explanation:

Let the length required be 'L'.

Given:

Resistance of the resistor (R) = 3.7 Ω

Radius of the rod (r) = 1.9 mm = 0.0019 m [1 mm = 0.001 m]

Resistivity of the material of rod (ρ) = 1.8\times 10^{-5}\ \Omega\cdot m

First, let us find the area of the circular rod.

Area is given as:

A=\pi r^2=3.14\times (0.0019)^2=1.13\times 10^{-5}\ m^2

Now, the resistance of the material is given by the formula:

R=\rho( \frac{L}{A})

Express this in terms of 'L'. This gives,

\rho\times L=R\times A\\\\L=\frac{R\times A}{\rho}

Now, plug in the given values and solve for length 'L'. This gives,

L=\frac{3.7\ \Omega\times 1.13\times 10^{-5}\ m^2}{1.8\times 10^{-5}\ \Omega\cdot m}\\\\L=\frac{4.181}{1.8}=2.32\ m

Therefore, the length of the material required to make a resistor of 3.7 Ω is 2.32 m.

5 0
3 years ago
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