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klemol [59]
3 years ago
14

Anyone know what the answer is?..

Physics
1 answer:
Licemer1 [7]3 years ago
8 0

Answer:

sometimes harmful and sometimes beneficial

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A particle initially located at the origin has an acceleration of vector a = 2.00ĵ m/s2 and an initial velocity of vector v i =
natali 33 [55]

With acceleration

\mathbf a=\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j

and initial velocity

\mathbf v(0)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i

the velocity at time <em>t</em> (b) is given by

\mathbf v(t)=\mathbf v(0)+\displaystyle\int_0^t\mathbf a\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\displaystyle\int_0^t\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)\,\mathbf j\,\mathrm du

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\bigg|_{u=0}^{u=t}

\mathbf v(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j

We can get the position at time <em>t</em> (a) by integrating the velocity:

\mathbf x(t)=\mathbf x(0)+\displaystyle\int_0^t\mathbf v(u)\,\mathrm du

The particle starts at the origin, so \mathbf x(0)=\mathbf0.

\mathbf x(t)=\displaystyle\int_0^t\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\,\mathrm du

\mathbf x(t)=\left(\left(8.00\dfrac{\rm m}{\rm s}\right)u\,\mathbf i+\dfrac12\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=t}

\mathbf x(t)=\left(8.00\dfrac{\rm m}{\rm s}\right)t\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)t^2\,\mathbf j

Get the coordinates at <em>t</em> = 8.00 s by evaluating \mathbf x(t) at this time:

\mathbf x(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)(8.00\,\mathrm s)\,\mathbf i+\left(1.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)^2\,\mathbf j

\mathbf x(8.00\,\mathrm s)=(64.0\,\mathrm m)\,\mathbf i+(64.0\,\mathrm m)\,\mathbf j

so the particle is located at (<em>x</em>, <em>y</em>) = (64.0, 64.0).

Get the speed at <em>t</em> = 8.00 s by evaluating \mathbf v(t) at the same time:

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(2.00\dfrac{\rm m}{\mathrm s^2}\right)(8.00\,\mathrm s)\,\mathbf j

\mathbf v(8.00\,\mathrm s)=\left(8.00\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(16.0\dfrac{\rm m}{\rm s}\right)\,\mathbf j

This is the <em>velocity</em> at <em>t</em> = 8.00 s. Get the <em>speed</em> by computing the magnitude of this vector:

\|\mathbf v(8.00\,\mathrm s)\|=\sqrt{\left(8.00\dfrac{\rm m}{\rm s}\right)^2+\left(16.0\dfrac{\rm m}{\rm s}\right)^2}=8\sqrt5\dfrac{\rm m}{\rm s}\approx17.9\dfrac{\rm m}{\rm s}

5 0
2 years ago
Can someone help me?​
Leviafan [203]

Car X traveled 3d distance in t time.  Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t,  the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.

In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.

V_x .t' = 3d\\ \frac{3d}{t} .t' = 3d\\ t'=t

Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.

V_y.t = \frac{2d}{t} .t = 2d

We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.

4 0
2 years ago
PART 2 OF ENERGY AND FORCES UNIT TEST
katrin2010 [14]

Answer:

1. at least two charged interacting parts

2. from the electric fields of charged subatomic particles

3 an arrow released from the bow

4Electrical fields of charged particles interact, bonding those with opposite charges.

5 the interaction of the electric fields of protons and electrons

6 The energy stored in the system increases.

7 Kinetic energy increases because the magnets move in the direction of the field.

8 Iron pieces accelerate toward the magnet, and the energy stored in the system decreases.

9  

The energy stored in the field decreases because the magnet moves in the direction of the field.

10 The energy stored increases and then decreases.

11 The wire was not connected to the source.

12  The electromagnet will become more powerful.

the rest are written, hope this helps (:

4 0
3 years ago
Why does light refract when it encounters the glass in the lens
Kobotan [32]
Hey there!

<span>because it slows down, which causes it to bend is the reason why light refracts when it encounters the glass in the lens

Hope this helps!
</span>
5 0
2 years ago
Read 2 more answers
A car traveling on a flat (unbanked) circular track accelerates uniformly from rest with a tangential acceleration of 1.7 m/s^2.
Komok [63]

Answer:

The coefficient of static friction between the car and the track

u=0.572

Explanation:

We don't know the mass of the car or any other information so the acceleration is the reason to solve the friction coefficient

∑F=F_{f}=m*a_{t}

As we know

F_{f}=u*F_{N}=u*m*g

Also the center ward direction forces

F_{fc}=m*a_{c}

a_{c}=\frac{v_{t}^2}{r}

F_{fc}=m*\frac{v_{t}^2}{r}

But now vt relation with the tangential acceleration

v_{t}=2*a_{t}*\frac{\pi }{r}

replacing

F_{fc}=m*a_{t}*\frac{2\pi*r}{2r}

F_{fc}=m*a_{t}*\pi

So magnitude of the force can get by

F_{f}=\sqrt{(m*a_{t}*\pi)^{2}+(m*a_{t})^{2}}

Get the factor to simplify

F_{f}=a_{t}*m*\sqrt{(1+\pi^2)}

u*m*g=m*a_{t}*(\sqrt{1+\pi^2})

Solve to u'

u=\frac{a_{t}}{g}*(\sqrt{1+\pi^2})

u=\frac{17\frac{m}{s^2} }{9.8\frac{m}{s^2}}*(\sqrt{1+\pi^2})=0.572

5 0
3 years ago
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