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Bad White [126]

4 years ago

14

A research-level Van de Graaff generator has a 2.15 m diameter metal sphere with a charge of 5.05 mC on it. What is the potentia

l near its surface in MV? (Assume the potential is equal to zero far away from the surface.)

1 answer:

Llana [10]4 years ago

5 0

Answer:

42.3 MV

Explanation:

d = diameter of the metal sphere = 2.15 m

r = radius of the metal sphere

diameter of the metal sphere is given as

d = 2r

2.15 = 2 r

r = 1.075 m

Q = charge on sphere = 5.05 mC = 5.05 x 10⁻³ C

Potential near the surface is given as

$V = \frac{kQ}{r}$

$V = \frac{(9\times 10^{9})(5.05\times 10^{-3})}{1.075}$

V = 4.23 x 10⁷ volts

V = 42.3 MV

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Water moves through a constricted pipe in steady, ideal flow. At the

Irina-Kira [14]

A) Speed in the lower section: 0.638 m/s

B) Speed in the higher section: 2.55 m/s

C) Volume flow rate: $1.8\cdot 10^{-3} m^3/s$

Explanation:

A)

To solve the problem, we can use Bernoulli's equation, which states that

$p_1 + \rho g h_1 + \frac{1}{2}\rho v_1^2 = p_2 + \rho g h_2 + \frac{1}{2}\rho v_2^2$

where

$p_1=1.75\cdot 10^4 Pa$ is the pressure in the lower section of the tube

$h_1 = 0$ is the heigth of the lower section

$\rho=1000 kg/m^3$ is the density of water

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

$v_1$ is the speed of the water in the lower pipe

$p_2$ is the pressure in the higher section

$h_2 = 0.250 m$ is the height in the higher pipe

$v_2$ is hte speed in the higher section

We can re-write the equation as

$v_1^2-v_2^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}$ (1)

Also we can use the continuity equation, which state that the volume flow rate is constant:

$A_1 v_1 = A_2 v_2$

where

$A_1 = \pi r_1^2$ is the cross-section of the lower pipe, with

$r_1 = 3.00 cm =0.03 m$ is the radius of the lower pipe (half the diameter)

$A_2 = \pi r_2^2$ is the cross-section of the higher pipe, with

$r_2 = 1.50 cm = 0.015 m$ (radius of the higher pipe)

So we get

$r_1^2 v_1 = r_2^2 v_2$

And so

$v_2 = \frac{r_1^2}{r_2^2}v_1$ (2)

Substituting into (1), we find the speed in the lower section:

$v_1^2-(\frac{r_1^2}{r_2^2})^2v_1^2=\frac{2(p_2-p_1)+\rho g h_2}{\rho}\\v_1=\sqrt{\frac{2(p_2-p_1+\rho g h_2)}{\rho(1-\frac{r_1^4}{r_2^4})}}=0.638 m/s$

B)

Now we can use equation (2) to find the speed in the lower section:

$v_2 = \frac{r_1^2}{r_2^2}v_1$

Substituting

v1 = 0.775 m/s

And the values of the radii, we find:

$v_2=\frac{0.03^2}{0.015^2}(0.638)=2.55 m/s$

C)

The volume flow rate of the water passing through the pipe is given by

$V=Av$

where

A is the cross-sectional area

v is the speed of the water

We can take any point along the pipe since the volume flow rate is constant, so

$r_1=0.03 cm$

$v_1=0.638 m/s$

Therefore, the volume flow rate is

$V=\pi r_1^2 v_1 = \pi (0.03)^2 (0.638)=1.8\cdot 10^{-3} m^3/s$

Learn more about pressure in a liquid:

brainly.com/question/9805263

#LearnwithBrainly

0 0

3 years ago

Which of ONE of the following four elements has the most metallic properties?

olchik [2.2K]

Answer:

12 (Magnesium- Mg)

Explanation:

Looking at the four numbers, we have:

Magnesium, Silicon, Sulfur, and Chlorine.

We can eliminate two of the answers immediately just by looking at the periodic table.

Sulfur and Chlorine are on the nonmetal side of the periodic table. So that's <em>definitely</em> not it. That leaves Magnesium and Silicon.

Silicon is a Metalloid. Magnesium is an Alkaline earth Metal.

Metaloids are elements that have a mix of both<em> metal</em> and<em> nonmetal </em>properties (luster, how it feels, etc.). Since it's a MIX and Magnesium is just straight METAL-

We can say Magnesium has the most metallic properties.

hope this helps!!

7 0

3 years ago

Analyze at the image below and answer the question that follows.

Rudiy27

the correct answer is c.

8 0

3 years ago

Read 2 more answers

A flying squirrel has a mass of 0. 765 kg. He jumps off the tree and hits the ground with 125 joules of energy. Determine how hi

uranmaximum [27]

Answer:

Height, h = 16.67 m

Explanation:

We have,

Mass of a squirrel is 0.765 kg.

He jumps off the tree and hits the ground with 125 joules of energy.

It is required to find the height up on the tree the squirrel was when it jumped.

The energy possessed by the squirrel is called its gravitational potential energy. It can be given by :

$E=mgh$

h is height up on the tree the squirrel was when it jumped

$h=\dfrac{E}{mg}\\\\h=\dfrac{125\ J}{0.765\ kg\times 9.8\ m/s^2}\\\\h=16.67\ m$

So, the squirrel will go to a height of 16.67 m.

6 0

3 years ago

The graph above shows the position and time calculate the velocity of the particle from T=0s to T=4s?

zzz [600]

Answer:

The velocity of the particle from T = 0 s to T = 4 s is;

0.5 m/s

Explanation:

The given parameters from the graph are;

The initial displacement (covered) at time, t₁ = 0 s is x₁ = 1 m

The displacement covered at time, t₂ = 4 s is x₂ = 3 m

The graph of distance to time, from time t = 0 to time t = 4 is a straight line graph, with the velocity given by the rate of change of the displacement to the time which is dx/dt which is also the slope of the graph given as follows;

$The \ slope \ of \ the \ displacement \ time \ graph, \ m =velocity, \ v= \dfrac{x_2 - x_1}{t_2 - t_1}$

$Therefore, \ the \ velocity \ of \ the \ particle \ v = \dfrac{3 \ m - 1 \ m}{4 \ s - 0 \ s} = \dfrac{2 \ m}{4 \ s} = \dfrac{1}{2} \ m/s$

The velocity of the particle from t = 0 s to t = 4 s = 1/2 m/s = 0.5 m/s.

3 0

3 years ago