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Vladimir79 [104]
3 years ago
8

A student pushes on a 5 kg box with a force of 20 N forward. The force of sliding friction

Physics
1 answer:
mart [117]3 years ago
4 0

Answer:

Explanation:

F=20N

mu=-10N

F=ma

20-10=10

10=(5)(a)

a=2 m/s^2

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Two loudspeakers emit sound waves of the same frequency along the x-axis. The amplitude of each wave is a. The sound intensity i
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Answer:

Explanation:

To find the amplitude of the sound, we must first determine the wavelength and the phase difference between the two speakers.

For the wavelength;

Recall that, the separation between two successive max. and min. intensity points are \dfrac{\lambda}{2}

Thus; for both speakers; the wavelength of the sound is:

\dfrac{\lambda}{2} = (10+30) cm

\dfrac{\lambda}{2} = (40) cm

λ = 80 cm

The relation between the path difference(Δx) and the phase difference(Δ∅) is:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

where;

Δx = 10 cm

λ = 80 cm

Δ∅ = π rad

∴

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{80 \ cm}(10 \ cm) + \Delta \phi_o

\pi \ rad  = \dfrac{2 \pi}{8}+ \Delta \phi_o

\pi \ rad  = \dfrac{ \pi}{4}+ \Delta \phi_o

\Delta \phi_o  =  \pi -\dfrac{ \pi}{4}

\Delta \phi_o  = \dfrac{ 4\pi - \pi}{4}

\Delta \phi_o  = \dfrac{ 3\pi}{4} \ rad

Suppose both speakers are placed side-by-side, then the path difference between the two speakers is: Δx = 0 cm

Thus, we have:

\Delta \phi = \dfrac{2 \pi}{\lambda}\Delta x + \Delta \phi_o

\Delta \phi = \dfrac{2 \pi}{\lambda}(0 \ cm ) + \dfrac{3 \pi}{4} \ rad

\Delta \phi = \dfrac{3 \pi}{4} \ rad

∴

The amplitude of the sound wave if the two speakers are placed side-by-side is:

A = 2a \ cos \bigg (\dfrac{\Delta \phi }{2} \bigg)

A = 2a \ cos \bigg (\dfrac{\dfrac{3 \pi}{4} }{2} \bigg)

A = 2a \ cos \bigg ({\dfrac{3 \pi}{8} } \bigg)

A = 0.765a

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Oh your from the other question you made I just saw it LOL.
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Answer:

This question is incomplete

Explanation:

This question is incomplete because the telescope's focal length was not provided. The formula to be used here is

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NOTE: Magnification can be described as the power of an instrument (in this case telescope) to enlarge an object. It has no unit and thus the two focal lengths mentioned in the formula above must be in the same unit (preferably meters since one of them is in meters already).

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