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Sidana [21]
3 years ago
6

In this also please

Physics
1 answer:
fgiga [73]3 years ago
8 0
Im not really sure but try Mendel genetics states laws are now part of theory of biological eveolotion

Also how silly would it be if Darwinist entire work was erased by another theory that doesn't even add information it is its entire work and Mendel just made another theroy
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When a student stands on a rotating table,the frictional force exerted on the student by the table is?
lakkis [162]

Answer:

Less

Explanation:

because static friction is more than rolling friction

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In a transformer a 120 volt dc Primary of 500 turns is connected to a secondary of 75 turns. What is the Induced voltage in the
oee [108]

Answer:

The induced voltage in the Secondary is 18 volt.

Explanation:

Given that,

Voltage = 120 volt

Number of turns in primary = 500

Number of turns in secondary = 75

We need to calculate the induced voltage in the Secondary

Using relation number of turns and voltage in primary and secondary

\dfrac{V_{p}}{V_{s}}=\dfrac{N_{p}}{N_{s}}

Where, N_{p} = Number of primary coil

N_{s} = Number of  secondary  coil

V_{p} = Voltage of primary coil

V_{p} = Voltage of primary coil

Put the value into the formula

\dfrac{120}{V_{s}}=\dfrac{500}{75}

V_{s}=\dfrac{120\times75}{500}

V_{s}=18\ Volt

Hence, The induced voltage in the Secondary is 18 volt.

7 0
3 years ago
An air-gap, parallel plate capacitor with area A and gap width d is connected to a battery that maintains the plates at potentia
sergejj [24]

Answer:

The new potential energy decreases by the factor of 2 to the old potential energy.

Explanation:

Capacitance of a parallel plate capacitor is given by the relation :

C = (ε₀A)/d

Here ε₀ is vacuum permittivity, A is area of the capacitor plate and d is the distance between them.

Potential energy of the capacitor, U = \frac{1}{2}CV^{2}

Here V is the potential difference between the plates.

According to the problem, the distance between the plates get double but area remains same. So,

d₁ = 2d

Here d₁ is new distance between the plates.

Hence, new capacitance is :

C₁ = (ε₀A)/d₁ = (ε₀A)/2d = C/2

The capacitor have same potential difference that is V. Hence, the new potential energy is :

U₁ = \frac{1}{2}C_{1} V^{2} = \frac{1}{2}\frac{C}{2} V^{2}

U₁ = U/2

\frac{U_{1} }{U} = \frac{1}{2}

7 0
3 years ago
A torus is formed when a circle of radius 3 centered at (5 comma 0 )is revolved about the​ y-axis. a. Use the shell method to wr
arlik [135]

Answer:

a) V=4\pi\int\limits^8_2 {x\sqrt{9-(x-5)^{2}}} \, dx

b) V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy

c) V=90\pi ^{2}

Explanation

In order to solve these problems, we must start by sketching a drawing of what the graph of the problem looks like, this will help us analyze the drawing better and take have a better understanding of the problem (see attached pictures).

a)

On part A we must build an integral for the volume of the torus by using the shell method. The shell method formula looks like this:

V=\int\limits^a_b {2\pi r y } \, dr

Where r is the radius of the shell, y is the height of the shell and dr is the width of the wall of the shell.

So in this case, r=x so dr=dx.

y is given by the equation of the circle of radius 3 centered at (5,0) which is:

(x-5)^{2}+y^{2}=9

when solving for y we get that:

y=\sqrt{9-(x-5)^{2}}

we can now plug all these values into the shell method formula, so we get:

V=\int\limits^8_2 {2\pi x \sqrt{9-(x-5)^{2}} } \, dx

now there is a twist to this problem since that will be the formula for half a torus.Luckily for us the circle is symmetric about the x-axis, so we can just multiply this integral by 2 to get the whole volume of the torus, so the whole integral is:

V=\int\limits^8_2 {4\pi x \sqrt{9-(x-5)^{2}} } \, dx

we can take the constants out of the integral sign so we get the final answer to be:

V=4\pi\int\limits^8_2 {x\sqrt{9-(x-5)^{2}}} \, dx

b)

Now we need to build an integral equation of the torus by using the washer method. In this case the formula for the washer method looks like this:

V=\int\limits^b_a{\pi(R^{2}-r^{2})} \, dy

where R is the outer radius of the washer and r is the inner radius of the washer and dy is the width of the washer.

In this case both R and r are given by the x-equation of the circle. We start with the equation of the circle:

(x-5)^{2}+y^{2}=9

when solving for x we get that:

x=\sqrt{9-y^{2}}+5

the same thing happens here, the square root can either give you a positive or a negative value, so that will determine the difference between R and r, so we get that:

R=\sqrt{9-y^{2}}+5

and

r=-\sqrt{9-y^{2}}+5

we can now plug these into the volume formula:

V=\pi \int\limits^3_{-3}{(5+\sqrt{9-y^{2}})^{2}-(5-\sqrt{9-y^{2}})^{2}} \, dy

This can be simplified by expanding the perfect squares and when eliminating like terms we end up with:

V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy

c) We are going to solve the integral we got by using the washer method for it to be easier for us to solve, so let's take the integral:

V=20\pi\int\limits^3_{-3} {\sqrt{9-y^{2}}} \, dy

This integral can be solved by using trigonometric substitution so first we set:

y=3 sin \theta

which means that:

dy=3 cos \theta d\theta

from this, we also know that:

\theta=sin^{-1}(\frac{y}{3})

so we can set the new limits of integration to be:

\theta_{1}=sin^{-1}(\frac{-3}{3})

\theta_{1}=-\frac{\pi}{2}

and

\theta_{2}=sin^{-1}(\frac{3}{3})

\theta_{2}=\frac{\pi}{2}

so we can rewrite our integral:

V=20\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {\sqrt{9-(3 sin \theta)^{2}}} \, 3 cos \theta d\theta

which simplifies to:

V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{9-(3 sin \theta)^{2}}} \, cos \theta d\theta

we can further simplify this integral like this:

V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{9-9 sin^{2} \theta}}} \, cos \theta d\theta

V=60\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {3(\sqrt{1- sin^{2} \theta})}} \, cos \theta d\theta

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{1- sin^{2} \theta})}} \, cos \theta d\theta

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(\sqrt{cos^{2} \theta})}} \, cos \theta d\theta

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {(cos \theta})} \, cos \theta d\theta

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {cos^{2} \theta}} \, d\theta

We can use trigonometric identities to simplify this so we get:

V=180\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {\frac{1+cos 2\theta}{2}}} \, d\theta

V=90\pi\int\limits^{\frac{\pi}{2}}_{-\frac{\pi}{2}} {1+cos 2\theta}}} \, d\theta

we can solve this by using u-substitution so we get:

u=2\theta

du=2d\theta

and:

u_{1}=2(-\frac{\pi}{2})=-\pi

u_{2}=2(\frac{\pi}{2})=\pi

so when substituting we get that:

V=45\pi\int\limits^{\pi}_{-\pi} {1+cos u}} \, du

when integrating we get that:

V=45\pi(u+sin u)\limit^{\pi}_{-\pi}

when evaluating we get that:

V=45\pi[(\pi+0)-(-\pi+0)]

which yields:

V=90\pi ^{2}

8 0
3 years ago
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