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3 years ago

10

Shakina and Juliette set the car's initial velocity to zero and set the acceleration to +1.2 m/s2, then clicked "start." Answer

the following questions.
What is the car's displacement between 0 and 10.0 s?
What was the total distance traveled by the car during this time interval?

1 answer:

Tcecarenko [31]3 years ago

4 0

Given:
u = 0, initial velocity
a = 1.2 m/s², acceleration
t = 10.0 , time of travel.

The distance traveled is
s = ut + (1/2)at²
= (1/2)*(1.2 m/s²)*(10 s)²
= 60 m

Answer: 60 m

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Newton’s law of gravity says the gravitational force between two objects is proportional to:

Nat2105 [25]

Answer:

Mass of the object or earth

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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 <em>f</em> 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 <em>S</em> 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 <em>R</em> is the interior of <em>S</em>. 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.

4 0

2 years ago

An athlete kicks a soccer ball that starts at rest so that it leaves their foot with a speed of 10m/s from the top o f a rectang

kirza4 [7]

Answer:

$a=500m/s^2$

Explanation:

We need only to apply the definition of acceleration, which is:

$a=\frac{v_f-v_i}{t_f-t_i}$

In our case the final velocity is $v_f=10m/s$ , the initial velocity is $v_i=0m/s$ since it departs from rest, the final time is $t_f=0.02s$ and the initial time we are considering is $t_i=0s$

So for our values we have:

$a=\frac{10m/s-0m/s}{0.02s-0s}=500m/s^2$

3 0

3 years ago

Two 0.20-kg balls, moving at 4 m/s east, strike a wall. Ball A bounces backwards at the same speed. Ball B stops. Which statemen

muminat

Answer:

Option A

Explanation:

From the question we are told that:

Mass $m=0.20kg$

Velocity $v=4m/s$

Generally the equation for momentum for Ball A is mathematically given by

Initial Momentum

$M_{a1}=mV$

$M_{a1}=0.2*4$

$M_{a1}=0.8$

Final Momentum

$M_{a2}=-0.8kgm/s$

Therefore

$\triangle M_a=-1.6kgm/s$

Generally the equation for momentum for Ball B is mathematically given by

Initial Momentum

$M_{b1}=mV$

$M_{b1}=0.2*4$

$M_{b1}=0.8$

Final Momentum

$M_{b2}=-0 kgm/s$

Therefore

$|\triangle M_a|>|\triangle Mb|$

Option A

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3 years ago

El tiempo que habria que esperar para que el dia fuera 1 hora mas largo que es hoy

myrzilka [38]

Answer: el tiempo que habria que esperar para que el dia fuera 1 hora mas largo que es hoy

Explanation:

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3 years ago