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2 years ago

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A man stands still on a moving walkway that is going the speed of 0.4 m a second to the east what is the velocity of the man acc

ording to a stationary observer

1 answer:

MaRussiya [10]2 years ago

5 0

Answer:

gtfcddcghjhgtthghyy

Explanation:

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A mass of 13kg stretches a spring 14cm. The mass is acted on by an external force of 6sin(t/2)N and moves in a medium that impar

m_a_m_a [10]

Answer:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

#Where $u$ is in meters and $t$ in seconds.

Explanation:

Given that : $m=13.0kg, \ L=0.14m, \ F(t)=6sin\frac{t}{2}N, F_d(t*)=-4N^{-1}, u\prime(t*)=0.12m/s\\u(0)=0,u\prime(0)=0.04m/s$

From $\omega=kL$ we have:

$k=\frac{\omega}{L}=\frac{mg}{0.14m}=\frac{13.0\times 9.8m/s}{0.14m}\\\\k=910kg/s^2$

From $F_d(t)=-\gamma u\prime(t)$ we have that:

$\gamma=-\frac{F_d(t*)}{u\prime(t*)}=\frac{4N}{0.12m/s}\\=33.33Ns/m$

Now,given that the initial value problem is given by:

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$

Hence,the position of u at time t is given by

$13u\prime\prime+33.33u\prime+910u=6sin\frac{t}{2}, \ u(0)=0,u\prime(0)=0.04\\$ , $u$ in meters, $t$ in seconds.

4 0

3 years ago

Why a drop of spirit on the hand feels colder than a drop of water at the same temperature

brilliants [131]

Answer:

This is because spirit has a lower boiling point when compared to water

Explanation:

spirit has a lower boiling point when compared to water which means it has the capacity to pull more heat from your hand and also it can do this very fast. This is why our hand feels colder.

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2 years ago

What is the final velocity?

Reil [10]

The final velocity is a vector quantity that measures the speed and direction of a moving body after it has reached its maximum acceleration

5 0

3 years ago

Read 2 more answers

You want to determine your speed running across the room. what is one piece of data you need

Reika [66]

The distance you run divided by how long it takes

hope this is what you waned :)

3 0

3 years ago

Two charges are located in the x x – y y plane. If q 1 = − 2.75 nC q1=−2.75 nC and is located at ( x = 0.00 m , y = 0.600 m ) (x

ludmilkaskok [199]

Answer:

$\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

Explanation:

In this case we have to work with vectors. Firs of all we have to compute the angles between x axis and the r vector (which points the charges):

$\theta_1=tan^{-1}(\frac{0.6m}{0m})=90\° \\\\\theta_2=tan^{-1}(\frac{0.4m}{1.3m})=17.10\°$

the electric field has two components Ex and Ey. By considering the sign of the charges we obtain that:

$\vec{E} = E_x \hat{i}+E_y\hat{j}\\\\\vec{E}=(E_1cos\theta_1-E_2cos\theta_2)\hat{i}+(E_1sin\theta_1-E_2sin\theta_2)\hat{j}\\\\E_1=k\frac{q}{r_1^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{2.75*10^{-9}C}{(0.6m)^2}=68.67\frac{N}{C}\\\\E_2=k\frac{q}{r_2^2}=(8.99*10^{9}\frac{N}{m^2C^2})\frac{3.40*10^{-9}C}{((1.3m)^2+(0.4m)^2)}=16.52\frac{N}{C}$

Hence, by replacing E1 and E2 we obtain:

$\vec{E}=[(68.67N/C)cos(90\°)-(16.52N/C)cos(17.10\°)]\hat{i}+[(68.67N/C)sin(90\°)-(16.52N/C)sin(17.10\°)]\hat{j}\\\\\vec{E}=(-15.78\hat{i}+63.81\hat{j})\frac{N}{C}$

hope this helps!!

3 0

3 years ago