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laiz [17]
3 years ago
12

a 2kg mass of copper falls from a height of 3m to an insulated floor. what is the maximum possible temperature increase of the c

opper?​
Physics
1 answer:
Rainbow [258]3 years ago
6 0

The maximum possible temperature increase of the copper is 0.076^{\circ}C

Explanation:

The energy gained by the mass of copper is equal to its change in gravitational potential energy, which is:

E=mg \Delta h

where

m = 2 kg is the mass of copper

g=9.8 m/s^2 is the acceleration of gravity

\Delta h= 3m is the change in height of the copper

Substituting,

E=(2)(9.8)(3)=58.8 J

The energy gained by the copper is related to its change in temperature by

E=mC_s \Delta T

where

m = 2 kg is the mass

C_s = 385 J/kg^{\circ} is the copper specific heat capacity

\Delta T is the change in temperature

And solving for \Delta T,

\Delta T = \frac{E}{mC_s}=\frac{58.8}{(2)(385)}=0.076^{\circ}C

Learn more about specific heat capacity:

brainly.com/question/3032746

brainly.com/question/4759369

#LearnwithBrainly

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Hii! help asap. i’ll give brainliest thanks!
o-na [289]

I believe it’s the mass of the box but I don’t no if I’m right

Hope this helped
5 0
2 years ago
A wire, 1.0 m long, with a mass of 90 g, is under tension. A transverse wave is propagated on the wire, for which the frequency
Mice21 [21]

Answer:

T = 712.9 N

Explanation:

First, we will find the speed of the wave:

v = fλ

where,

v = speed of the wave = ?

f = frequency = 890 Hz

λ = wavelength = 0.1 m

Therefore,

v = (890 Hz)(0.1 m)

v = 89 m/s

Now, we will find the linear mass density of the wire:

\mu = \frac{m}{L}

where,

μ = linear mass density of wie = ?

m = mass of wire = 90 g = 0.09 kg

L = length of wire = 1 m

Therefore,

\mu = \frac{0.09\ kg}{1\ m}

μ = 0.09 kg/m

Now, the tension in wire (T) will be:

T = μv² = (0.09 kg/m)(89 m/s)²

<u>T = 712.9 N</u>

7 0
2 years ago
URGENT!!!
madreJ [45]

Answer:

f=6.97\times 10^{14}\ Hz

Explanation:

Given that,

Wavelength, \lambda=430\ nm=430\times 10^{-9}\ m

We need to find the frequency of the violet light.

We know that the relation between frequency and wavelength is given by :

f=\dfrac{c}{\lambda}\\\\f=\dfrac{3\times 10^8}{430\times 10^{-9}}\\\\f=6.97\times 10^{14}\ Hz

So, the frequency of violet light is 6.97\times 10^{14}\ Hz.

7 0
3 years ago
Is physical science part of physics?
telo118 [61]
Yes, it is. Physical science<span>, the systematic study of the inorganic world</span>
6 0
3 years ago
Read 2 more answers
There are four charges, each with a magnitude of 4.25 C. Two are positive and two are negative. The charges are fixed to the cor
VMariaS [17]

Answer:

 F = 7.68 10¹¹ N,  θ = 45º

Explanation:

In this exercise we ask for the net electric force. Let's start by writing the configuration of the charges, the charges of the same sign must be on the diagonal of the cube so that the net force is directed towards the interior of the cube, see in the attached numbering and sign of the charges

The net force is

          F_ {net} = F₂₁ + F₂₃ + F₂₄

bold letters indicate vectors. The easiest method to solve this exercise is by using the components of each force.

let's use trigonometry

          cos 45 = F₂₄ₓ / F₂₄

          sin 45 = F_{24y) / F₂₄

          F₂₄ₓ = F₂₄ cos 45

          F_{24y} = F₂₄ sin 45

let's do the sum on each axis

X axis

          Fₓ = -F₂₁ + F₂₄ₓ

          Fₓ = -F₂₁₁ + F₂₄ cos 45

Y axis  

         F_y = - F₂₃ + F_{24y}

         F_y = -F₂₃ + F₂₄ sin 45

They indicate that the magnitude of all charges is the same, therefore

         F₂₁ = F₂₃

Let's use Coulomb's law

         F₂₁ = k q₁ q₂ / r₁₂²

       

the distance between the two charges is

         r = a

         F₂₁ = k q² / a²

we calculate F₂₄

           F₂₄ = k q₂ q₄ / r₂₄²

the distance is

           r² = a² + a²

           r² = 2 a²

         

we substitute

           F₂₄ = k  q² / 2 a²

we substitute in the components of the forces

          Fx = - k \frac{q^2}{a^2} +  k \frac{q^2}{2 a^2}  \ cos 45

          Fx = k \frac{q^2}{a^2}  ( -1 + ½ cos 45)

          F_y = k \frac{q^2}{a^2} ( -1 +  ½ sin 45)    

         

We calculate

            F₀ = 9 10⁹ 4.25² / 0.440²

            F₀ = 8.40 10¹¹ N

       

            Fₓ = 8.40 10¹¹ (½ 0.707 - 1)

            Fₓ = -5.43 10¹¹ N

         

remember cos 45 = sin 45

             F_y = - 5.43 10¹¹  N

We can give the resultant force in two ways

a) F = Fₓ î + F_y ^j

          F = -5.43 10¹¹ (i + j)   N

b) In the form of module and angle.

For the module we use the Pythagorean theorem

          F = \sqrt{F_x^2 + F_y^2}

          F = 5.43 10¹¹  √2

          F = 7.68 10¹¹ N

in angle is

           θ = 45º

7 0
2 years ago
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