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Brrunno [24]
3 years ago
8

Why are springs made of steel and not of copper.​

Physics
2 answers:
torisob [31]3 years ago
8 0

Answer:

Springs are made of a variety of materials including copper and various forms of steel. The most common is high carbon steel as it is cheap, easy to work and a couple of other important properties.

Copper springs exist as well, but copper is more expensive than steel. However, in an environment where corrosion resistance is important, copper springs are a good alternative.

Explanation:

just olya [345]3 years ago
4 0

Answer:

A good spring will be that in which a large restoring force is developed on being deformed

Explanation:

Since Young's modulus for steel is more than that of copper, therefore the springs made of steel and not of copper.

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Anestetic [448]
True would be the Correct answer
5 0
3 years ago
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Initially, a 2.00-kg mass is whirling at the end of a string (in a circular path of radius 0.750 m) on a horizontal frictionless
drek231 [11]

Answer:

v_f = 15 \frac{m}{s}

Explanation:

We can solve this problem using conservation of angular momentum.

The angular momentum \vec{L} is

\vec{L}  = \vec{r} \times \vec{p}

where \vec{r} is the position and \vec{p} the linear momentum.

We also know that the torque is

\vec{\tau} = \frac{d\vec{L}}{dt}  = \frac{d}{dt} ( \vec{r} \times \vec{p} )

\vec{\tau} =  \frac{d}{dt}  \vec{r} \times \vec{p} +   \vec{r} \times \frac{d}{dt} \vec{p}

\vec{\tau} =  \vec{v} \times \vec{p} +   \vec{r} \times \vec{F}

but, as the linear momentum is \vec{p} = m \vec{v} this means that is parallel to the velocity, and the first term must equal zero

\vec{v} \times \vec{p}=0

so

\vec{\tau} =   \vec{r} \times \vec{F}

But, as the only horizontal force is the tension of the string, the force must be parallel to the vector position measured from the vertical rod, so

\vec{\tau}_{rod} =   0

this means, for the angular momentum measure from the rod:

\frac{d\vec{L}_{rod}}{dt} =   0

that means :

\vec{L}_{rod} = constant

So, the magnitude of initial angular momentum is :

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i| cos(\theta)

but the angle is 90°, so:

| \vec{L}_{rod_i} | = |\vec{r}_i||\vec{p}_i|

| \vec{L}_{rod_i} | = r_i * m * v_i

We know that the distance to the rod is 0.750 m, the mass 2.00 kg and the speed 5 m/s, so:

| \vec{L}_{rod_i} | = 0.750 \ m \ 2.00 \ kg \ 5 \ \frac{m}{s}

| \vec{L}_{rod_i} | = 7.5 \frac{kg m^2}{s}

For our final angular momentum we have:

| \vec{L}_{rod_f} | = r_f * m * v_f

and the radius is 0.250 m and the mass is 2.00 kg

| \vec{L}_{rod_f} | = 0.250 m * 2.00 kg * v_f

but, as the angular momentum is constant, this must be equal to the initial angular momentum

7.5 \frac{kg m^2}{s} = 0.250 m * 2.00 kg * v_f

v_f = \frac{7.5 \frac{kg m^2}{s}}{ 0.250 m * 2.00 kg}

v_f = 15 \frac{m}{s}

8 0
3 years ago
Steam in a heating system flows through tubes whose outer diameter is 5 cm and whose walls are maintained at a temperature of 13
svet-max [94.6K]

Answer:

5945.27 W per meter of tube length.

Explanation:

Let's assume that:

  • Steady operations exist;
  • The heat transfer coefficient (h) is uniform over the entire fin surfaces;
  • Thermal conductivity (k) is constant;
  • Heat transfer by radiation is negligible.

First, let's calculate the heat transfer (Q) that occurs when there's no fin in the tubes. The heat will be transferred by convection, so let's use Newton's law of cooling:

Q = A*h*(Tb - T∞)

A is the area of the section of the tube,

A = π*D*L, where D is the diameter (5 cm = 0.05 m), and L is the length. The question wants the heat by length, thus, L= 1m.

A = π*0.05*1 = 0.1571 m²

Q = 0.1571*40*(130 - 25)

Q = 659.73 W

Now, when the fin is added, the heat will be transferred by the fin by convection, and between the fin and the tube by convection, thus:

Qfin = nf*Afin*h*(Tb - T∞)

Afin = 2π*(r2² - r1²) + 2π*r2*t

r2 is the outer radius of the fin (3 cm = 0.03 m), r1 is the radius difference of the fin and the tube ( 0.03 - 0.025 = 0.005 m), and t is the thickness ( 0.001 m).

Afin = 0.006 m²

Qfin = 0.97*0.006*40*(130 - 25)

Qfin = 24.44 W

The heat transferred at the space between the fin and the tube will be:

Qspace = Aspace*h*(Tb - T∞)

Aspace = π*D*S, where D is the tube diameter and S is the space between then,

Aspace = π*0.05*0.003 = 0.0005

Qspace = 0.0005*40*(130 - 25) = 1.98 W

The total heat is the sum of them multiplied by the total number of fins,

Qtotal = 250*(24.44 + 1.98) = 6605 W

So, the increase in heat is 6605 - 659.73 = 5945.27 W per meter of tube length.

5 0
3 years ago
What is the relationship between the normal force and weight
GalinKa [24]

Answer: they have the same magnitude.

Explanation:

normal force = mg

weight = mg

4 0
2 years ago
1 If you measured the distance travelled by a snail in
Fittoniya [83]
The answer is m/s hope it helps
3 0
2 years ago
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