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PtichkaEL [24]
3 years ago
5

a 4.0 kg ball is attached to 0.70 m string and spun at 2.0 m/s. what is the centripetal acceleration ?

Physics
2 answers:
konstantin123 [22]3 years ago
4 0
A(cp) = v^2/R

a(cp) = (2.0)^2/0.70

a(cp) = 4.00/0.70

a(cp) = 5.7 m/s^2 (approximately)
sergejj [24]3 years ago
3 0
Hope this helps, have a great day ahead!

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Elena-2011 [213]
The answer is a, series circuit.
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At a depth of 1030 m in Lake Baikal (a fresh water lake in Siberia), the pressure has increased by 100 atmospheres (to about 107
dangina [55]

Answer:

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

Explanation:

The bulk modulus is represented by the following differential equation:

K = - V\cdot \frac{dP}{dV}

Where:

K - Bulk module, measured in pascals.

V - Sample volume, measured in cubic meters.

P - Local pressure, measured in pascals.

Now, let suppose that bulk remains constant, so that differential equation can be reduced into a first-order linear non-homogeneous differential equation with separable variables:

-\frac{K \,dV}{V} = dP

This resultant expression is solved by definite integration and algebraic handling:

-K\int\limits^{V_{f}}_{V_{o}} {\frac{dV}{V} } = \int\limits^{P_{f}}_{P_{o}}\, dP

-K\cdot \ln \left |\frac{V_{f}}{V_{o}} \right| = P_{f} - P_{o}

\ln \left| \frac{V_{f}}{V_{o}} \right| = \frac{P_{o}-P_{f}}{K}

\frac{V_{f}}{V_{o}} = e^{\frac{P_{o}-P_{f}}{K} }

The final volume is predicted by:

V_{f} = V_{o}\cdot e^{\frac{P_{o}-P_{f}}{K} }

If V_{o} = 1\,m^{3}, P_{o} - P_{f} = -10132500\,Pa and K = 2.3\times 10^{9}\,Pa, then:

V_{f} = (1\,m^{3}) \cdot e^{\frac{-10.1325\times 10^{6}\,Pa}{2.3 \times 10^{9}\,Pa} }

V_{f} \approx 0.996\,m^{3}

Change in volume due to increasure on pressure is:

\Delta V = V_{o} - V_{f}

\Delta V = 1\,m^{3} - 0.996\,m^{3}

\Delta V = 0.004\,m^{3}

A volume of a cubic meter of water from the surface of the lake has been compressed in 0.004 cubic meters.

8 0
3 years ago
infrared radiation from young stars can pass through the heavy dust clouds surrounding them, allowing astronomers here on earth
Mashcka [7]

The wavelength of the infrared radiation is  λ = 3.174×10^{-5}m.

<h3>What is infrared radiation?</h3>

An infrared telescope is tuned to detect infrared radiation with a frequency of 9.45 THz.

We know that,

1 THz = 10¹² Hz

So,

f = 9.45 × 10¹² Hz

We need to find the wavelength of the infrared radiation.

λ=c/f

λ = 3×10^{8}/9.45×10^{12}

λ = 3.174 ×10^{-5} m

The term "infrared radiation" (IR) refers to a part of the electromagnetic radiation spectrum with wavelengths between about 700 nanometers (nm) and one millimeter (mm). Longer than visible light waves but shorter than radio waves are infrared waves.

Electromagnetic radiation with wavelengths longer than those of visible light is known as infrared, also known as infrared light. Since it is undetectable to the human eye, The typical range of wavelengths considered to be infrared (IR) is from about 1 millimeter to the nominal red edge of the visible spectrum, or about 700 nanometers.

To learn more about infrared radiation from the given link:

brainly.com/question/13163856

#SPJ4

5 0
1 year ago
The center of mass is
PolarNik [594]
D is the best answer. In many physics problems we treat an extended object as if it were a point with the same mass located at the center of mass.
5 0
3 years ago
A truck covers 47.0 m in 8.60 s while smoothly slowing down to final speed of 2.30 m/s. (a) Find its original speed.
Kruka [31]

Explanation:

Given that,

Distance, s = 47 m

Time taken, t = 8.6 s

Final speed of the truck, v = 2.3 m/s

Let u is the initial speed of the truck and a is its acceleration such that :

a=\dfrac{v-u}{t}.............(1)

Now, the second equation of motion is :

s=ut+\dfrac{1}{2}at^2

Put the value of a in above equation as :

s=ut+\dfrac{1}{2}\times \dfrac{v-u}{t}\times t^2

s=\dfrac{t(u+v)}{2}

u=\dfrac{2s}{t}-v

u=\dfrac{2\times 47}{8.6}-2.3

u = 8.63 m/s

So, the original speed of the truck is 8.63 m/s. Hence, this is the required solution.

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