What will terminal velocity look like on a velocity v. time graph?

1) A thin ring made of uniformly charged insulating material has total charge Q and radius R. The ring is positioned along the x

allochka39001 [22]

Answer:

(A) considering the charge "q" evenly distributed, applying the technique of charge integration for finite charges, you obtain the expression for the potential along any point in the Z-axis:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$

With $(\epsilon_{0})$ been the vacuum permittivity

(B) The expression for the magnitude of the E(z) electric field along the Z-axis is:

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$

Explanation:

(A) Considering a uniform linear density $λ_{0}$ on the ring, then:

$dQ=\lambda dl$ (1)⇒ $Q=\lambda_{0} 2\pi R$ (2)⇒ $\lambda_{0}=\frac{Q}{2\pi R}$ (3)

Applying the technique of charge integration for finite charges:

$V(z)= 4\pi (ε_{0})\int\limits^a_b {\frac{1}{ r' }} \, dQ$ (4)

Been r' the distance between the charge and the observation point and a, b limits of integration of the charge. In this case a=2π and b=0.

Using cylindrical coordinates, the distance between a point of the Z-axis and a point of a ring with R radius is:

$r'=\sqrt{R^{2} +Z^{2}}$ (5)

Using the expressions (1),(4) and (5) you obtain:

$V(z)= 4\pi (\epsilon_{0})\int\limits^a_b {\frac{\lambda_{0}R}{ \sqrt{R^{2} +Z^{2}} }} \, d\phi$

Integrating results:

$V(z)=\frac{Q}{4\pi (\epsilon_{0}) \sqrt{R^{2} +z^{2}} }$ (S_a)

(B) For the expression of the magnitude of the field E(z), is important to remember:

$|E| =-\nabla V$ (6)

But in this case you only work in the z variable, soo the expression (6) can be rewritten as:

$|E| =-\frac{dV(z)}{dz}$ (7)

Using expression (7) and (S_a), you get the expression of the magnitude of the field E(z):

$E(z)=\frac{QZ}{4\pi (\epsilon_{0}) (R^{2} +z^{2})^{\frac{3}{2} } }$ (S_b)

4 0

3 years ago

A volumetric flask made of Pyrex is calibrated at 20.0°C. It is filled to the 285-mL mark with 40.5°C glycerin. After the flask

Charra [1.4K]

Answer:

V_f = 287.04 mL

Explanation:

We are given the initial/original volume of the glycerine as 285 mL.

Now, after it is finally cooled back to 20.0 °C , its volume is given by the formula;

V_f = V_i (1 + βΔT)

Where;

V_f is the final volume

V_i is the original volume = 285 mL

β is the coefficient of expansion of glycerine and from online tables, it has a value of 5.97 × 10^(-4) °C^(−1)

Δt is change in temperature = final temperature - initial temperature = 32 - 20 = 12 °C

Thus, plugging in relevant values;

V_f = 285(1 + (5.97 × 10^(-4) × 12))

V_f = 287.04 mL

7 0

3 years ago

Which among the states of matter has definite shape and volume?why?

Mashcka [7]

Answer:

Answer

Explanation:

A solid because each solids has various shapes and volume. scientist uses variety of tools such as, Cylinder,scaler, etc to differentiate the solids.

8 0

3 years ago

A 3.5 kg object moving at 8.1 m/s in the positive direction of an x axis has a one-dimensional elastic collision with an object

Tamiku [17]

Answer:

So the mass of the second object M will be 1.951 kg

Explanation:

We have given mass of the first object $m_1=3.5kg$ and its velocity $v_1=8.1m/sec$

Mass of the second object $m_2=M$ it is at rest so its velocity $v_2=0$

From conservation of momentum we know that

Initial momentum = final momentum

So $m_1v_1+m_2v_2=(m_1+m_2)v$

$3.5\times 8.1+M\times 0=(3.5+M)5.2$

$28.35=18.2+5.2M$

M = 1.951 kg

8 0

3 years ago

How far should pencil held from a convex mirror of radius 80 cm to form an image one-half the size of the pencil?

Cerrena [4.2K]

Answer:

40 centimeters

Explanation:

8 0

3 years ago