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3 years ago

Can someone answer these physics questions

Physics

1 answer:

USPshnik [31]3 years ago

6 0

1. The object moves -25 meters in a period of 3 seconds, so its average velocity is (-25 m)/(3 s) = -25/3 m/s ≈ -8.3 m/s

2. Ms. Patil's average velocity is the displacement over the period divided by the period length: (8 mi)/(0.25 h) = 32 mi/h

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A very good insulator has an R-value of 29. The material's heat transfer coefficient is

Fittoniya [83]

Answer:

b. 0.034

Explanation:

The heat transfer coefficient of a material (U-value) is equal to the reciprocal of its R-value, therefore:

$U = \frac{1}{R}$

where

R is the R-value of the material

For the insulator in this problem,

R = 29

Substituting into the equation, we find the heat transfer coefficient:

$U=\frac{1}{29}=0.034$

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3 years ago

Which instrument should, ideally, have zero resistance?

Temka [501]

The answer is C voltmeter

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2 years ago

Read 2 more answers

A piano wire with mass 2.95 g and length 79.0 cm is stretched with a tension of 29.0 N . A wave with frequency 105 Hz and amplit

Romashka-Z-Leto [24]

The concept needed to solve this problem is average power dissipated by a wave on a string. This expression ca be defined as

$P = \frac{1}{2} \mu \omega^2 A^2 v$

Here,

$\mu$ = Linear mass density of the string

$\omega =$ Angular frequency of the wave on the string

A = Amplitude of the wave

v = Speed of the wave

At the same time each of this terms have its own definition, i.e,

$v = \sqrt{\frac{T}{\mu}} \rightarrow$ Here T is the Period

For the linear mass density we have that

$\mu = \frac{m}{l}$

And the angular frequency can be written as

$\omega = 2\pi f$

Replacing this terms and the first equation we have that

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2(\sqrt{\frac{T}{\mu}})$

$P = \frac{1}{2} (\frac{m}{l})(2\pi f)^2 A^2 (\sqrt{\frac{T}{m/l}})$

$P = 2\pi^2 f^2A^2(\sqrt{T(m/l)})$

PART A ) Replacing our values here we have that

$P = 2\pi^2 (105)^2(1.8*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.2320W$

PART B) The new amplitude A' that is half ot the wavelength of the wave is

$A' = \frac{1.8*10^{-3}}{2}$

$A' = 0.9*10^{-3}$

Replacing at the equation of power we have that

$P = 2\pi^2 (105)^2(0.9*10^{-3})^2(\sqrt{(29.0)(2.95*10^{-3}/0.79)})$

$P = 0.058W$

8 0

2 years ago

A circular loop of wire with a radius of 20 cm lies in the xy plane and carries a current of 2 A, counterclockwise when viewed f

Zina [86]

To solve this problem we will apply the concept related to the magnetic dipole moment that is defined as the product between the current and the object area. In our case we have the radius so we will get the area, which would be

$A=\pi r^2$

$A =\pi (0.2)^2$

$A =0.1256 m^2$

Once the area is obtained, it is possible to calculate the magnetic dipole moment considering the previously given definition:

$\mu=IA$

$\mu=2(0.1256)$

$\mu=0.25 A\cdot m^2$

Therefore the magnetic dipole moment is $0.25A\cdot m^2$

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3 years ago

Example of an unbalenced force

lesya692 [45]

Answer:

Explanation:

unbalance force can be defined as the force that makes object slow up or down, in order words unbalanced force can be defined as anything that can alter the state of motion of an object.

an example of an unbalanced force is likened to a situation of someone inside a lift going for a meeting at the top floor, the force which propels the lift upward is unbalanced, because it is greater than the force acting on the lift.

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3 years ago