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3 years ago

Convertir 1200 ms a cs Convertir 0,3 mm a cm.

Physics

1 answer:

Vaselesa [24]3 years ago

3 0

Answer:

You can do the reverse unit conversion from cm/s to m/s, or enter any two units below: Metre per second (U.S. spelling: meter per second) is an SI derived unit of both speed (scalar) and velocity (vector quantity which specifies both magnitude and a specific direction), defined by distance in metres divided by time in seconds.

Explanation:

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A space station shaped like a giant wheel has a radius of a radius of 153 m and a moment of inertia of 4.16 × 10⁸ kg·m² (when it

Naya [18.7K]

Answer:

a = 1.709g

Explanation:

Given the absence of external forces being applied in the space station, it is possibly to use the Principle of Angular Momentum Conservation, which states that:

$I_{o} \cdot \omega_{o} = I_{f} \cdot \omega_{f}$

The required initial angular speed is obtained herein:

$g= \omega_{o}^{2}\cdot R_{ss}$

$\omega_{o}=\sqrt{\frac{g}{R_{ss}} }$

$\omega_{o}= \sqrt{\frac{9.807\,\frac{m}{s^{2}} }{153\,m} }$

$\omega_{o} \approx 0.253\,\frac{rad}{s}$

The initial moment of inertia is:

$I_{o} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{o} = 4.16\times 10^{8}\,kg\cdot m^{2}+(150)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{o} = 6.442\times 10^{8}\,kg\cdot m^{2}$

The final moment of inertia is:

$I_{f} =I_{ss}+n\cdot m_{person}\cdot R_{ss}^{2}$

$I_{f} = 4.16\times 10^{8}\,kg\cdot m^{2}+(50)\cdot (65\,kg)\cdot (153\,m)^{2}$

$I_{f} = 4.921\times 10^{8}\,kg\cdot m^{2}$

Now, the final angular speed is obtained:

$\omega_{f} = \frac{I_{o}}{I_{f}}\cdot \omega_{o}$

$\omega_{f} = \frac{6.442\times 10^{8}\,{kg\cdot m^{2}}}{4.921\times 10^{8}\,kg\cdot m^{2}} \cdot (0.253\,\frac{rad}{s} )$

$\omega_{f} = 0.331\,\frac{rad}{s^}$

The apparent acceleration is:

$a_{f} = \omega_{f}^{2}\cdot R_{ss}$

$a_{f} = (0.331\,\frac{rad}{s} )^{2}\cdot (153\,m)$

$a_{f} = 16.763\,\frac{m}{s^{2}}$

This is approximately 1.709g.

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4 years ago

Which agent of weathering breaks down limestone statues?

Fynjy0 [20]

Acid chemicals because acid chemicals break down limestone statues

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4 years ago

Read 2 more answers

A 73-kg Norwegian olympian ski champion is going down a hill sloped at 39 ◦ . The coefficient of kinetic friction between the sk

bazaltina [42]

Answer:

Explanation:

net force on the skier = mg sin 39 - μ mg cos39

mg ( sin39 - μ cos39 )

= 73 x 9.8 ( .629 - .116)

= 367 N

impulse = net force x time = change in momentum .

= 367 x 5 = 1835 kg m /s

velocity of the skier after 5 s = 1835 / 73

= 25.13 m /s

b )

net force becomes zero

mg ( sin39 - μ cos39 ) = 0

μ = tan39

= .81

c )

net force becomes zero , so he will continue to go ahead with constant speed of 25.13 m /s

so he will have speed of 25.13 m /s after 5 s .

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3 years ago

What is the complete back-and-forth motion of an object called?

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<span>Vibration is the Answer</span>

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3 years ago

A flat, 181 ‑turn, current‑carrying loop is immersed in a uniform magnetic field. The area of the loop is 4.97 cm2 and the angle

Sidana [21]

Answer:

B = 0.135T

Explanation:

To find the magnitude of the magnetic field you use the following formula, for the torque produced by a magnetic field B in a loop:

$\tau=NIABsin\theta$ (1)

τ: torque = 1.51*10^-5 Nm

I: current = 2.47mA = 2.47*10^-3 A

B: magnitude of the magnetic field

A: area of the loop = 4.97cm^2 = 4.97(10^-2m)^2=4.97*10^-4m^2

N: turns = 181

θ: angle between B and the magnetic dipole (same as the direction of the normal to the plane)

You replace the values of the parameters in (1). Furthermore you do B the subject of the formula:

$B=\frac{\tau}{NIAsin\tetha}=\frac{1.51*10^{-5}Nm}{(181)(2.47*10^{-3}A)(4.97*10^{-4}m^2)(sin30.1\°)}=0.135T$

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3 years ago