How to do this question

Answers

Answer 1

Answer:

(a) 10 m/s

(b) 22.4 m/s

Explanation:

(a) Draw a free body diagram of the car when it is at the top of the loop. There are two forces: weight force mg pulling down, and normal force N pushing down.

Sum of forces in the centripetal direction (towards the center):

∑F = ma

mg + N = mv²/r

At minimum speed, the normal force is 0.

mg = mv²/r

g = v²/r

v = √(gr)

v = √(10 m/s² × 10.0 m)

v = 10 m/s

(b) Energy is conserved.

Initial kinetic energy + initial potential energy = final kinetic energy

½ mv₀² + mgh = ½ mv²

v₀² + 2gh = v²

(10 m/s)² + 2 (10 m/s²) (20.0 m) = v²

v = 22.4 m/s

Related Questions

The momentum of an electron is 1.75 times larger than the value computed non-relativistically. What is the speed of the electron

Answers

Answer:

Speed of the electron is 2.46 x 10^8 m/s

Explanation:

momentum of the electron before relativistic effect = [tex]M_{0} V[/tex]

where [tex]M_{0}[/tex] is the rest mass of the electron

V is the velocity of the electron.

under relativistic effect, the mass increases.

under relativistic effect, the new mass M will be

M = [tex]M_{0}/ \sqrt{1 - \beta ^{2} }[/tex]

where

[tex]\beta = V/c[/tex]

c is the speed of light = 3 x 10^8 m/s

V is the speed with which the electron travels.

The new momentum will therefore be

==> [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]

It is stated that the relativistic momentum is 1.75 times the non-relativistic momentum. Equating, we have

1.75[tex]M_{0} V[/tex] = [tex]M_{0}V/ \sqrt{1 - \beta ^{2} }[/tex]

the equation reduces to

1.75 = [tex]1/ \sqrt{1 - \beta ^{2} }[/tex]

square both sides of the equation, we have

3.0625 = 1/[tex](1 - \beta ^{2} )[/tex]

3.0625 - 3.0625[tex]\beta ^{2}[/tex] = 1

2.0625 = 3.0625[tex]\beta ^{2}[/tex]

[tex]\beta ^{2}[/tex] = 0.67

β = 0.819

substitute for [tex]\beta = V/c[/tex]

V/c = 0.819

V = c x 0.819

V = 3 x 10^8 x 0.819 = 2.46 x 10^8 m/s

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+2z)j + (3z)k be a vector field (for example, the velocityfaild of a fluid flow). the solid object has five sides, S1:bottom(xy-plane), S2:left side(xz-plane), S3 rear side(yz-plane), S4:right side, and S5:cylindrical roof.

a. Sketch the solid object.
b. Evaluate the flux of F through each side of the object (S1,S2,S3,S4,S5).
c. Find the total flux through surface S.

Answers

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

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

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

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

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

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

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

[tex]\mathbf n_1=\dfrac{\partial\mathbf s_1}{\partial y}\times\dfrac{\partial\mathbf s_1}{\partial x}=-\mathbf k[/tex]

[tex]\mathbf n_2=\dfrac{\partial\mathbf s_2}{\partial u}\times\dfrac{\partial\mathbf s_2}{\partial v}=-u\,\mathbf j[/tex]

[tex]\mathbf n_3=\dfrac{\partial\mathbf s_3}{\partial z}\times\dfrac{\partial\mathbf s_3}{\partial y}=-\mathbf i[/tex]

[tex]\mathbf n_4=\dfrac{\partial\mathbf s_4}{\partial v}\times\dfrac{\partial\mathbf s_4}{\partial u}=u\,\mathbf i+u\,\mathbf j[/tex]

[tex]\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[/tex]

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

[tex]\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[/tex]

[tex]=\displaystyle\int_0^2\int_0^{6-x}0\,\mathrm dy\,\mathrm dx=0[/tex]

[tex]\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[/tex]

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

[tex]\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[/tex]

[tex]=\displaystyle\int_0^2\int_0^60\,\mathrm dy\,\mathrm dz=0[/tex]

[tex]\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[/tex]

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

[tex]\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[/tex]

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

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.

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

where R is the interior of S. We have

[tex]\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[/tex]

The integral is easily computed in cylindrical coordinates:

[tex]\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[/tex]

[tex]\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[/tex]

as expected.

The ability of a water strider to move along the surface of water without breaking the surface is due to:

Answers

Answer:

The ability of a water strider to move along the surface of water without breaking the surface is due to:

SURFACE TENSION

Explanation:

this is because Water molecules are more attracted to each other than they are to other materials, so they generate a force to stay together called surface tension. Which allows the strider to move without breaking the surface

Now the friends are ready to tackle a homework problem. A pulse is sent traveling along a rope under a tension of 29 N whose mass per unit length abruptly changes, from 19 kg/m to 45 kg/m. The length of the rope is 2.5 m for the first section and 2.8 m for the second, and the second rope is rigidly fixed to a wall. Two pulses will eventually be detected at the origin: the pulse that was reflected from the medium discontinuity and the pulse that was originally transmitted, which hits the wall and is reflected back and transmitted through the first rope. What is the time difference, Δt, between the two pulses detected at the origin? s

Answers

Answer:

The time difference is 2.97 sec.

Explanation:

Given that,

Tension = 29 N

Mass per unit length [tex]\mu_{1}=19\ kg/m[/tex]

Mass per unit length [tex]\mu_{2}=45\ kg/m[/tex]

Length of first section = 2.5 m

Length of second section = 2.8 m

We need to total distance of first pulse

Using formula for distance

[tex]d=2.5+2.5[/tex]

[tex]d_{1}=5.0\ m[/tex]

We need to total distance of second pulse

Using formula for distance

[tex]d=2.8+2.8[/tex]

[tex]d_{2}=5.6\ m[/tex]

We need to calculate the speed of pulse in the first string

Using formula of speed

[tex]v_{1}=\sqrt{\dfrac{T}{\mu_{1}}}[/tex]