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1 month ago

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You have a remote-controlled car that has been programmed to have velocity v⃗ =(−3ti^+2t2j^)m/s, where t is in s. At t = 0 s, th

e car is at r⃗ 0=(3.0i^+2.0j^)mWhat is the x component of the car's position vector at 10 s?What is the y component of the car's position vector at 10 s?What is the x component of the car's acceleration vector at 10 s?What is the y component of the car's acceleration vector at 10 s?

1 answer:

Yuliya22 [2.4K]1 month ago

5 0

Answer:

The y-component of the position vector for the car is 670m/s.

The x-component of the acceleration vector is -3 and the y-component equals 40.

Explanation:

The car's displacement vector corresponding to the velocity

$\boldsymbol{v}= (-3t\boldsymbol{i}+2t^2\boldsymbol{j})m/s$

is derived from the integration of the velocity.

By integrating $\boldsymbol{v}$ , we obtain the displacement vector $\boldsymbol{d}$ :

$\boldsymbol{d}=(-\dfrac{3}{2}t^2\boldsymbol{i}+\dfrac{2}{3}t^3\boldsymbol{j} )$

Assuming the vehicle's starting position is

$\boldsymbol{r}= (3.0\boldsymbol{i}+2.0\boldsymbol{j})$

the displacement at time $t$ would then be

$\boldsymbol{d(t)}= \boldsymbol{r+d}$

$\boxed{\boldsymbol{d(t)}=(-\dfrac{3}{2}t^2+3.0\boldsymbol{i}+\dfrac{2}{3}t^3+2.0\boldsymbol{j} )}$

At the moment $t=10s$ , we find

$\boxed{\boldsymbol{d(t)}=(-147\boldsymbol{i}+670\boldsymbol{j} )}m$

The y-component of the position vector for the car is 670m/s.

The acceleration vector can be calculated as the derivative of the velocity vector:

$\boldsymbol{a(t)}=\dfrac{d\boldsymbol{v(t)}}{dt} =(-3\boldsymbol{i}+4t\boldsymbol{j})$

At $t=10s$ , it becomes

$\boldsymbol{a(t)}=(-3\boldsymbol{i}+40\boldsymbol{j})m/s^2$

The x-component of the acceleration vector is -3 and the y-component is 40.

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Answer:

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2)

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Explanation:

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When an object is projected near the surface of the Earth at an angle $\theta$ to the horizontal, it follows a trajectory known as a parabola. The only force acting on it (ignoring wind resistance) is gravity, affecting the vertical axis.

The height of a projectile can be calculated using

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

where $y_o$ represents the initial height from ground level, $v_{oy}$ is the vertical component of the initial velocity, and t is the elapsed time.

The vertical speed component is expressed as

$v_y=v_{oy}-gt$

1) To proceed, we will determine the initial vertical velocity component since we lack sufficient data to calculate the absolute value of $v_o$ .

The peak height is attained when $v_y=0$ , which allows us to compute the time to reach that height.

$v_{oy}-gt_m=0$

Solving for $t_m$

$\displaystyle t_m=\frac{v_{oy}}{g}$

Thus, the maximum height reached is

$\displaystyle y_m=y_o+\frac{v_{oy}^2}{2g}$

We know this value is equal to 8 meters

$\displaystyle y_o+\frac{v_{oy}^2}{2g}=8$

Continuing with the calculations for $v_{oy}$

$\displaystyle v_{oy}=\sqrt{2g(8-y_o)}$

Substituting known values yields

$\displaystyle v_{oy}=\sqrt{2(9.8)(8-1.5)}$

$\displaystyle v_{oy}=11.29\ m/s$

2) At t=1.505 seconds, the ball is positioned above Julie’s head; we can calculate

$\displaystyle y=y_o+V_{oy}t-\frac{gt^2}{2}$

$\displaystyle y=1.5+(11.29)(1.505)-\frac{9.8(1.505)^2}{2}$

$\displaystyle y=1.5\ m+16,991\ m-11.098\ m$

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Answer:

293.7 degrees

Explanation:

A = - 8 sin (37) i + 8 cos (37) j

A + B = -12 j

B = a i + b j, where a and b represent constants to solve for.

A + B = (a - 8 sin (37) ) i + ( 8cos(37) + b ) j

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By comparing the coefficients of i and j:

a = 8 sin (37) = 4.81452 m

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Thus,

B = 4.81452 i - 18.38908 j..... 4th quadrant

<pTherefore,

cos(Q) = 4.81452 / 12

Q = 66.346 degrees

360 - Q gives us 293.65 degrees from the + x-axis in a counterclockwise direction.

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For a particle with mass m in the nth energy level of an infinite square well potential of width L , the energy $E_{n}$ is given by:

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In the ground state (n=1) and in the first excited state (n=2), where the energy is noted as E₂= 6.0 eV. Substituting into the above equation yields:

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