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3 years ago

If 3.5 waves have a distance of 42M what is the wavelength of one wave?

Physics

1 answer:

docker41 [41]3 years ago

5 0

Answer:

The answer is 12

Explanation:

Just divide 42/3.5

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Determine the minimum applied force p required to move wedge a to the right. the spring is compressed a distance of 175 mm. negl

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<span>b. The coefficient of static friction for all contacting surfaces is Î¼s=0.35. neglect friction at the rollers.

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3 years ago

A solid disk has a mass of 162 kg and a radius of 1.30m. This

RideAnS [48]

Answer:

754.3 m

Explanation:

The moment of inertia of the solid disk:

$I = mR^2/2$

Where m is the disk mass and R is the radius of the disk.

$I = 162*1.3^2/2 = 136.89 kgm^2$

The angular kinetic energy of the disk is then:

$E_k = I\omega^2/2 = 136.89 * 18^2/2 = 22176.18 J$

By law of energy conservation, this energy is converted to potential energy to pick up the 3kg block

let g = 9.8 m/s2

$E_p = m_bgh = 22176.18 J$

where $m_b$ = 3 kg is the mass of block

$3*9.8h = 22176.18$

$h = \frac{22176.18}{3*9.8} = 754.3 m$

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3 years ago

O operate this generator, a human applies a force by turning the crank. This action generates an electric current, which lights

Shtirlitz [24]

Answer A I’m very positive about it

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3 years ago

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To determine if an object is in motion, you must use a __________. Question 1 options: Moving object Living thing Frame of Refer

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Answer: moving object :)

Explanation:

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3 years ago

A Foucault pendulum consists of a brass sphere with a diameter of 31.0 cm suspended from a steel cable 11.0 m long (both measure

kozerog [31]

Answer:

43.7 °C

Explanation:

$\alpha_b$ = Coefficient of linear expansion of brass = $18\times 10^{-6}\ ^{\circ}C$

$\alpha_s$ = Coefficient of linear expansion of steel = $11\times 10^{-6}\ ^{\circ}C$

$L_{0b}$ = Initial length of brass = 31 cm

$L_{0s}$ = Initial length of steel = 11 m

$\Delta L$ = Total change in length = 3 mm

Total change in length would be

$\Delta L=\Delta L_b+\Delta L_s\\\Rightarrow \Delta L=L_{0b}\alpha_b\Delta T+L_{0s}\alpha_b\Delta T\\\Rightarrow \Delta T=\frac{\Delta L}{L_{0b}\alpha_b+L_{0s}\alpha_b}\\\Rightarrow \Delta T=\frac{0.003}{0.31\times 18\times 10^{-6}+11\times 10^{-6}\times 11}\\\Rightarrow \Delta T=23.7\ ^{\circ}C$

$\Delta T=23.7\\\Rightarrow T_f-T_i=23.7\\\Rightarrow T_f=23.7+T_i\\\Rightarrow T_f=23.7+20\\\Rightarrow T_f=43.7\ ^{\circ}C$

The final temperature is 43.7 °C

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3 years ago