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3 years ago

Which form of energy is equal to the sum of an object’s kinetic and potential energy?

Physics

1 answer:

Natali5045456 [20]3 years ago

4 0

Answer:

Mechanical Energy

Explanation:

The sum of kinetic energy and potential energy of an object is its total mechanical energy.

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Why does an object in motion stay in motion unless acted on by an unbalanced force?

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Answer: A) because forces are what stop and start motion

Explanation:

From Newton's first law, an object tends to stay in state of rest or motion unless acted upon by an unbalanced external force. This is also known law of inertia. This is because a force can stop or start a motion. A force cause body to accelerate to decelerate otherwise the body continues with constant speed.

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3 years ago

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Select whether the argument is an example of a deductive or an inductive argument:

viva [34]

The answer is a inductive

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3 years ago

FIGURE 1 shows part of a mass spectrometer. The whole arrangement is in a vacuum. Negative ions of mass 2.84 x 10-20 kg and char

yuradex [85]

Yes, the ions can exit slit P without being deflected, if the electric field strength is 170.6 N/C

Explanation:

When the ions are inside the container, they are subjected to two forces, with directions opposite to each other:

The force due to the electric field, whose magnitude is $F_E=qE$ , where q is the charge of the ion and E is the strength of the electric field
The force due to the magnetic field, whose magnitude is $F_B=qvB$ , where v is the speed of the ions and B is the strength of the magnetic field

The ions will move straight and undeflected if the two forces are equal and opposite. By using Fleming Left Hand rule, we notice that the magnetic force on the (negative) ions point upward: this means that the electric field must be also upward (so that the electric force on the ions is downward). Then, the two forces are balanced if

$F_E = F_B$

which translates into

$qE=qvB\\\rightarrow v = \frac{E}{B}$

Therefore, if the speed of the ions is equal to this ratio, the ions will go undeflected.

We can even calculate the value of E at which this occurs. In fact, we know that the ions are earlier accelerated by a potential difference $V=-3000 V$ , so we have that their kinetic energy is given by the change in electric potential energy:

$qV=\frac{1}{2}mv^2$

where

$q=-2.0\cdot 10^{-19}C\\m=2.84\cdot 10^{-20}kg$

Solving for v, the speed,

$v=\sqrt{\frac{2qV}{m}}=\sqrt{\frac{2(-2.0\cdot 10^{-19})(-3000)}{2.84\cdot 10^{-20}}}=205.6 m/s$

And since the magnetic field strength is

B = 0.83 T

The strength of the electric field must be

$E=vB=(205.6 m/s)(0.83 T)=170.6 N/C$

Learn more about electric and magnetic fields:

brainly.com/question/8960054

brainly.com/question/4273177

brainly.com/question/3874443

brainly.com/question/4240735

#LearnwithBrainly

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3 years ago

Let surface S be the boundary of the solid object enclosed by x^2+z^2=4, x+y=6, x=0, y=0, and z=0. and, let f(x,y,z)=(3x)i+(x+y+

babunello [35]

a. I've attached a plot of the surface. Each face is parameterized by

• $\mathbf s_1(x,y)=x\,\mathbf i+y\,\mathbf j$ with $0\le x\le2$ and $0\le y\le6-x$ ;

• $\mathbf s_2(u,v)=u\cos v\,\mathbf i+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ;

• $\mathbf s_3(y,z)=y\,\mathbf j+z\,\mathbf k$ with $0\le y\le 6$ and $0\le z\le2$ ;

• $\mathbf s_4(u,v)=u\cos v\,\mathbf i+(6-u\cos v)\,\mathbf j+u\sin v\,\mathbf k$ with $0\le u\le2$ and $0\le v\le\frac\pi2$ ; and

• $\mathbf s_5(u,y)=2\cos u\,\mathbf i+y\,\mathbf j+2\sin u\,\mathbf k$ with $0\le u\le\frac\pi2$ and $0\le y\le6-2\cos u$ .

b. Assuming you want outward flux, first compute the outward-facing normal vectors for each face.

$\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k$

$\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j$

$\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i$

$\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j$

$\mathbf n_5=\dfrac{\partial\mathbf s_5}{\partial y}\times\dfrac{\partial\mathbf s_5}{\partial u}=2\cos u\,\mathbf i+2\sin u\,\mathbf k$

Then integrate the dot product of f with each normal vector over the corresponding face.

$\displaystyle\iint_{S_1}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{6-x}f(x,y,0)\cdot\mathbf n_1\,\mathrm dy\,\mathrm dx$

$=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0$

$\displaystyle\iint_{S_2}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,0,u\sin v)\cdot\mathbf n_2\,\mathrm dv\,\mathrm du$

$\displaystyle=\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=-8$

$\displaystyle\iint_{S_3}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^6\mathbf f(0,y,z)\cdot\mathbf n_3\,\mathrm dy\,\mathrm dz$

$=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0$

$\displaystyle\iint_{S_4}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^2\int_0^{\frac\pi2}\mathbf f(u\cos v,6-u\cos v,u\sin v)\cdot\mathbf n_4\,\mathrm dv\,\mathrm du$

$=\displaystyle\int_0^2\int_0^{\frac\pi2}-u^2(2\sin v+\cos v)\,\mathrm dv\,\mathrm du=\frac{40}3+6\pi$

$\displaystyle\iint_{S_5}\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\int_0^{\frac\pi2}\int_0^{6-2\cos u}\mathbf f(2\cos u,y,2\sin u)\cdot\mathbf n_5\,\mathrm dy\,\mathrm du$

$=\displaystyle\int_0^{\frac\pi2}\int_0^{6-2\cos u}12\,\mathrm dy\,\mathrm du=36\pi-24$

c. You can get the total flux by summing all the fluxes found in part b; you end up with 42π - 56/3.

Alternatively, since S is closed, we can find the total flux by applying the divergence theorem.

$\displaystyle\iint_S\mathbf f(x,y,z)\cdot\mathrm d\mathbf S=\iiint_R\mathrm{div}\mathbf f(x,y,z)\,\mathrm dV$

where R is the interior of S. We have

$\mathrm{div}\mathbf f(x,y,z)=\dfrac{\partial(3x)}{\partial x}+\dfrac{\partial(x+y+2z)}{\partial y}+\dfrac{\partial(3z)}{\partial z}=7$

The integral is easily computed in cylindrical coordinates:

$\begin{cases}x(r,t)=r\cos t\\y(r,t)=6-r\cos t\\z(r,t)=r\sin t\end{cases},0\le r\le 2,0\le t\le\dfrac\pi2$

$\displaystyle\int_0^2\int_0^{\frac\pi2}\int_0^{6-r\cos t}7r\,\mathrm dy\,\mathrm dt\,\mathrm dr=42\pi-\frac{56}3$

as expected.

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