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3 years ago

How do you think the temperature difference between the beakers relates to the rate of heat transfer?

Physics

1 answer:

Bogdan [553]3 years ago

8 0

Answer and Explanation:

The rate (in W) at which heat energy moves from the more blazing article to the colder item increments with the temperature distinction between the items.

The more noteworthy the temperature contrast, the more prominent the rate at which exchange of heat energy takes place.

Thus the temperature difference between the beakers relates in the same manner as described above.

The exchange of heat energy continues until a thermal equilibrium is attained.

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An automobile traveling on a straight, level road has an initial speed v when the brakes are applied. In coming to rest with a c

Molodets [167]

Answer:

Explanation:

Use $v^{2} = u^{2} +2as$ to do the question.

For first instance,

$0 = v^{2} +2ax$ -------------------( 1 )

for second instance,

$0 = (2v)^{2} +2as$ -----------------( 2 )

So by (1) and (2),

s = 4x

3 0

3 years ago

An 80-cm-long steel string with a linear density of 1.0 g/m is under 200 N tension. It is plucked and vibrates at its fundamenta

icang [17]

Answer:

Wavelength of the sound wave that reaches your ear is 1.15 m

Explanation:

The speed of the wave in string is

$v=\sqrt{\frac{T}{\mu} }$

where T= 200 N is tension in the string , $\mu$ =1.0 g/m is the linear mass density

$v=\sqrt{\frac{200}{1\times 10^{-3} }$

$v=447.2 m/s$

Wavelength of the wave in the string is

$\lambda =2L=2\times 0.8=1.6 m$

The frequency is

$f=\frac{v}{\lambda} \\f=\frac{447.2}{1.6}\\f=298.25 Hz$

The required wavelength pf the sound wave that reaches the ear is( take velocity of air v=344 m/s)

$\lambda=\frac{v_{air}}{f} \\\lambda=\frac{344}{298.25} \\\lambda=1.15 m$

8 0

2 years ago

he triceps muscle in the back of the upper arm extends the forearm. This muscle in a professional boxer exerts a force of 2.00 1

meriva

Answer:

Moment of inertia = 0.3862kg-m²

Explanation:

2.00x10³

2.80cm

145 rad

r = r⊥ x F

F is an applied force

r⊥ is the distance between the applied force and axis

Force exerted = 2.00x10³

r⊥ = 2.8cm = 0.028m

Alpha = 145rad/s²

r = 0.028m x 2.00x10³

r = 56.0N-m

To get the moment of inertia

56.0N-m² = (145rad/s²) x I

The I would be:

I = (56.0N-m²)/(145rad/s²)

I = 56/145

= 0.3862Kg-m²

This is the moment of inertia.

Thank you!

5 0

3 years ago

Your image appears 3.0 m from a plane mirror. How far IS YOUR IMAGE IN<br> RELATION TO YOU?*

Vsevolod [243]

Answer:

Being a plane mirror the Image is formed 3 metres beyond the mirror . So total distance is 3+3 = 6metres

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2 years ago

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Which of the following waves have the lowest energy?

borishaifa [10]

The radio waves have a very low frequency so they will always have a very low energy.

Good luck :)

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3 years ago

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