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2 years ago

What is buoyant force

Physics

1 answer:

Xelga [282]2 years ago

7 0

Answer:

When an object is immersed in water. it is pulled downwards due to gravitational pull of earth. Water exerts upward force on the object. This makes object rise up. This upward force is called buoyancy or upthrust.

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Which actions are examples of conserving resources? Check all that apply.

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Answer:

1st, 2nd, and 4th

Explanation:

1st conserves gasoline/petroleum

2nd conserves electricity

4th conserves paper

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5) A girl with 40kg mass is riding an 8 kg bicycle. At the top of the hill her speed is 6.0 m/s, and her speed triples at the bo

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Answer:

Work = 6912 joules

Explanation:

Non-conservative forces are dissipative forces such as friction or air resistance. These forces take energy away from the system as the system progresses, energy that you can't get back. These forces are path dependent; therefore it matters where the object starts and stops.

Total mass = 40 + 8 = 48kg

Initial speed u= 6 m/s

Final speed v = 3*initial

Final speed v = 3* 6 = 18 m/s

Distance s = 15

Acceleration a is?

V² = U² + 2aS

18² = 6² + 2a*15

324 = 36 + 30a

324-36= 30a

288 = 30a

288/30 = a

9.6= a

Force = mass* acceleration

Force = 48*9.6

Force = 460.8N

Work = force*distance

Work = 460.8*15

Work = 6912 joules

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3 years ago

Scientists believe that Earth is becoming warmer. This idea will become unacceptable over time if

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Consider a semi-infinite (hollow) cylinder of radius R with uniform surface charge density. Find the electric field at a point o

VikaD [51]

Answer:

For the point inside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

For the point outside the cylinder: $E = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

where x0 is the position of the point on the x-axis and σ is the surface charge density.

Explanation:

Let us assume that the finite end of the cylinder is positioned at the origin. And the rest of the cylinder lies on the (-x) axis, which is the vertical axis in this question. In the first case (inside the cylinder) we will calculate the electric field at an arbitrary point -x0. In the second case (outside), the point will be +x0.

The cylinder is consist of the sum of the rings with the same radius.

First we will calculate the electric field at point -x0 created by the ring at an arbitrary point x.

We will also separate the ring into infinitesimal portions of length 'ds' where ds = Rdθ.

The charge of the portion 'ds' is 'dq' where dq = σds = σRdθ. σ is the surface charge density.

Now, the electric field created by the small portion is 'dE'.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2}$

The electric field is a vector, and it needs to be separated into its components in order us to integrate it. But, the sum of horizontal components is zero due to symmetry. Every dE has an equal but opposite counterpart which cancels it out. So, we only need to take the component with the sine term.

$dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rd\theta}{R^2+x^2} \frac{x}{\sqrt{x^2+R^2}} = dE = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rxd\theta}{(R^2+x^2)^{3/2}}$

We have to integrate it over the ring, which is an angular integration.

$E_{ring} = \int{dE} = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}\int\limits^{2\pi}_0 {} \, d\theta = \frac{1}{4\pi\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}2\pi = \frac{1}{2\epsilon_0}\frac{\sigma Rx}{(R^2+x^2)^{3/2}}$

This is the electric field created by a ring a distance x away from the point -x0. Now we can integrate this electric field over the semi-infinite cylinder to find the total E-field:

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{-2x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + 4x_0^2}}$

The reason we integrate over -2x0 to -inf is that the rings above -x0 and below to-2x0 cancel out each other. Electric field is created by the rings below -2x0 to -inf.

We will only change the boundaries of the last integration.

$E_{cylinder} = \int{E_{ring}} = \frac{\sigma R}{2\epsilon_0}\int\limits^{-\inf}_{x_0} \frac{x}{(R^2+x^2)^{3/2}}dx = \frac{\sigma R}{2\epsilon_0}\frac{1}{\sqrt{R^2 + x_0^2}}$

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Which subatomic particle has a negative charge? Please help i think its A but im not sure A. proton B. electron C. neutron D.all

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The answer is electron cause it head a negative power.

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What is buoyant force​

What is buoyant force