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kifflom [539]
3 years ago
11

Which feature of a heating curve indicates a change of state?

Physics
2 answers:
Andreas93 [3]3 years ago
7 0
A "heating curve" is a graph that shows the temperature of the substance
against the amount of heat you put into it.

For most of the graph, as you'd expect, the temperature goes up as you
add heat, and it goes down as you take heat away.  BUT ... While the
substance is changing state, its temperature doesn't change even though
you're putting heat in or taking heat out.

So that part of the graph is a horizontal line. 
BaLLatris [955]3 years ago
5 0

Answer: rizontal

Explanation:

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Two long parallel wires carry currents of 3.35 A and 6.99 A . The magnitude of the force per unit length acting on each wire is
Nady [450]

Answer:

244mm

Explanation:

I₁ = 3.35A

I₂ = 6.99A

μ₀ = 4π*10^-7

force per unit length (F/L) = 6.03*10⁻⁵N/m

B = (μ₀ I₁ I₂ )/ 2πr .........equation i

B = F / L ..........equation ii

equating equation i & ii,

F / L = (μ₀ I₁ I₂ )/ 2πr

Note F/L = B = F

F = (μ₀ I₁ I₂ ) / 2πr

2πr*F = (μ₀ I₁ I₂ )

r = (μ₀ I₁ I₂ ) / 2πF

r = (4π*10⁻⁷ * 3.35 * 6.99) / 2π * 6.03*10⁻⁵

r = 1.4713*10⁻⁵ / 6.03*10⁻⁵

r = 0.244m = 244mm

The distance between the wires is 244m

7 0
3 years ago
Read 2 more answers
How can accuracy be limited?
lozanna [386]

Answer:

accuracy refers to the deviation of a measurement from a standard or true value of the quantity being measured

5 0
3 years ago
Consider a uniformly charged sphere of radius Rand total charge Q. The electric field Eout outsidethe sphere (r≥R) is simply tha
AlexFokin [52]

1) Electric potential inside the sphere: \frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2) Ratio Vcenter/Vsurface: 3/2

3) Find graph in attachment

Explanation:

1)

The electric field inside the sphere is given by

E=\frac{1}{4\pi \epsilon_0}\frac{Qr}{R^3}

where

\epsilon_0=8.85\cdot 10^{-12}F/m is the vacuum permittivity

Q is the charge on the sphere

R is the radius of the sphere

r is the distance from the centre at which we compute the field

For a radial field,

E(r)=-\frac{dV(r)}{dr}

Therefore, we can find the potential at distance r by integrating the expression for the electric field. Calculating the difference between the potential at r and the potential at R,

V(R)-V(r)=-\int\limits^R_r  E(r)dr=-\frac{Q}{4\pi \epsilon_0 R^3}\int r dr = \frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)

The potential at the surface, V(R), is that of a point charge, so

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore we can find the potential inside the sphere, V(r):

V(r)=V(R)+\Delta V=\frac{Q}{4\pi \epsilon_0 R}+\frac{-Q}{8\pi \epsilon_0 R^3}(R^2-r^2)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})

2)

At the center,

r = 0

Therefore the potential at the center of the sphere is:

V(r)=\frac{Q}{8\pi \epsilon_0 R}(3-\frac{r^2}{R^2})\\V(0)=\frac{3Q}{8\pi \epsilon_0 R}

On the other hand, the potential at the surface is

V(R)=\frac{Q}{4\pi \epsilon_0 R}

Therefore, the ratio V(center)/V(surface) is:

\frac{V(0)}{V(R)}=\frac{\frac{3Q}{8\pi \epsilon_0 R}}{\frac{Q}{4\pi \epsilon_0 R}}=\frac{3}{2}

3)

The graph of V versus r can be found in attachment.

We observe the following:

- At r = 0, the value of the potential is \frac{3}{2}V(R), as found in part b) (where V(R)=\frac{Q}{4\pi \epsilon_0 R})

- Between r and R, the potential decreases as -\frac{r^2}{R^2}

- Then at r = R, the potential is V(R)

- Between r = R and r = 3R, the potential decreases as \frac{1}{R}, therefore when the distance is tripled (r=3R), the potential as decreased to 1/3 (\frac{1}{3}V(R))

Learn more about electric fields and potential:

brainly.com/question/8960054

brainly.com/question/4273177

#LearnwithBrainly

7 0
3 years ago
An object moves along the x axis. The graph shows it’s position x as a function of time t. Find the average velocity of the obje
nadezda [96]
Question is from B to C

Answer: (b) 1.5m/s

x1=3m, x2=9m
t1=1s, t2=5s
Displacement, ∆x=(9-3)m=6m
Time elapsed, ∆t=(5-1)s=4s

So average velocity v =∆x/∆t=6/4=1.5m/s
6 0
3 years ago
1. According to paragraph 3 in the text, MOST of the electromagnetic waves from
Fed [463]

Answer:

Most of the EM waves from the sun that reach Earth are infrared waves, visible light, and UV radiation.

Explanation:

I hope this helps! Have a good day!

5 0
3 years ago
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