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Makovka662 [10]
3 years ago
5

A 4.00-kg particle moves along the x axis. Its position varies with time according to x 5 t 1 2.0t 3, where x is in meters and t

is in seconds. Find (a) the kinetic energy of the particle at any time t, (b) the acceleration of the particle and the force acting on it at time t, (c) the power being delivered to the particle at time t, and (d) the work done on the particle in the interval t 5 0 to t 5 2.00 s
Physics
2 answers:
White raven [17]3 years ago
6 0

Answer:

Explanation:

Given equation is ,

x =t + 2 t³ ,

dx/dt = velocity ( v ) = 1 + 6 t²

a) kinetic energy = 1/2 m v² = .5 x 4 ( 1 + 6 t² )² = 2 ( 1 + 6 t²)²

b ) Acceleration = dv /dt = 12 t .

force( F ) = mass x acceleration = 4 x 12 t = 48 t

Power = force x velocity = 48 t x ( 1 + 6 t²). = 48 t + 288 t³ )

work done = ∫ F dx =∫ 48 t x( 1 + 6t² )dt ; =  [48t²/2 + 48 x 6 x t³ /3 = 24 t² + 96 t³ )]₀² =  864 J

Natali [406]3 years ago
4 0

Answer:

Givens:

  • x=t+2.0t^{3}
  • m=4.00 \ kg

We know that kinetic energy is:

K=\frac{1}{2}mv^{2}

So, we just need to calculate the speed. We have the equation of the movement, if we derivate that expression, we'll have the speed:

\frac{dx}{dt}=v\\ v=\frac{d}{dt}(t+2.0t^{3})\\v=1+2(3)t^{2} \\v=6t^{2}+1

Which is the speed at any time <em>t.</em>

Now, we replace the expression to find the kinetic energy at any time <em>t:</em>

K=\frac{1}{2}m(6t^{2}+1)^{2}

K=\frac{1}{2}(4)(36t^{4}+12t^{2}+1)\\K=2(36t^{4}+12t^{2}+1)\\K=72t^{4}+24t^{2}+2

So, this is the kinetic energy at energy at any time <em>t.</em>

Through derivation we can find the acceleration and force at any time <em>t:</em>

<em>a=\frac{dv}{dt}\\a=\frac{d}{dt}(6t^{2}+1)\\a=12t</em>

Also, we know that F=ma

Replacing values: F=4(12t)=48t

The power is define as the product of the force and the velocity:

P=Fv=48t(6t^{2}+1)=288t^{3}+48

At last, we now that the work is define as: W=\int Fdx

So, we just replace the force, and integrate it between <em>t=0 </em>and<em> t=2 sec.</em>

<em>W=\int _{0}^{2} 48t(6t^{2}+1)dt\\W=\int_{0}^{2} (288t^{3}+48t)dt\\W= \frac{288t^{4} }{4}+\frac{48t^{2}}{2} ]_{0}^{2} \\W= 72t^{4}+24t^{2}]_{0}^{2}\\W=72(2)^{4}+24(2)^{2}\\W= 1152 + 96=1248 \ J</em>

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d_{1x}=-(3.5 km)( cos 55^{\circ})=-2.01 km

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So, the total components on the two directions are

d_x = -2.01 km+0.74 km=-1.27 km

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d=\sqrt{(1.27 km)^2+(5.47 km)^2}=5.6 km

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3 years ago
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A cart moves along a track at a velocity of 3.5 cm/s. When a force is applied to the cart, its velocity increases to 8.2 cm/s. I
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Answer:

3.13cm/s²

Explanation:

Given

Initial velocity u = 3.5cm/s

Final velocity v = 8.2cm/s

Time t = 1.5secs

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Substitute the given parameters

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1.5a = 8.2-3.5

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At a distance of 41 ft , an ionizing radiation source delivers 5.0 rem of radiation. How close could you get to the source and s
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Answer:

at distance 18.33 ft no biological effects

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given data

distance = 41 ft

radiation = 5 rem

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how close get so no biological effects

solution

we know here that intensity of radiation R is inversely proportional to (radiation)²

so here equation will be

\frac{I1}{I2}  = (\frac{R2}{R1})^{2}   .............1

here I is intensity

so here I1 = 25

because for  0 to 25 there is  no detectable effects and for 25 to 100 it will . temporary decrease so

we take 25 because dose is less than 25 so we take that highest value

so

\frac{I1}{I2}  = (\frac{R2}{R1})^{2}

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R1 = 18.33 ft

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A cannonball is catapulted toward a castle. The cannonball's velocity when it leaves the catapult is 40 m/s at an angle of 37° w
sleet_krkn [62]

Answer:

a) Maximum height = 36.6 m

b) Horizontal distance at which the ball lands = 166.1 m

c) x-component = 32 m/s. y-component = - 27 m/s  

Explanation:

Please, see the attached figure for a description of the problem.

The velocity vector "v" of the cannonball has two components, a horizontal component, "vx", and a vertical component "vy". Notice that at the maximum height, the vertical component "vy" of the velocity vector is 0.

In the same way, the position vector "r" is composed by "rx", its horizontal component, and "ry", the vertical component.

The velocity vector "v" ad the position vector "r" at time "t" are given by the following equations:

v = (v0 * cos α, v0 * sin α + g * t)

r = (x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

Where

v0 = magnitude of the initial velocity vector

α = launching angle

g = gravity acceleration (-9.8 m/s², because the y-axis points up)

t = time

x0 = initial horizontal position

y0 = initial vertical position

If we consider the origin of the system of reference as the point at which the cannonball leaves tha catapult, then, x0 and y0 = 0

a) We know that at maximum height, the vertical component of the vector "v" is 0, because the ball does not move up nor down at that moment (see figure). Then:

0 = v0 * sin α + g * t

-v0 * sin α / g = t

-40 m/s * sin 37° / -9.8 m/s² = t

t = 2.5 s

We can now calculate the position of the cannonball at time t=2.5 s to obtain the maximum height:

r = (x0 + v0 * t cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

The max height is the magnitude of the vector ry max (see figure). The vector ry max is:

ry = (0, y0 + v0 t sin α + 1/2 g * t²)

magnitude of ry = |ry|= \sqrt{(0m)^{2} + (y0 + v0* t*sin \alpha+ 1/2*g*t^{2})^{2}}= y0 + v0*t*sin \alpha + 1/2*g*t^{2})

Then:

max height = y0 + v0 * t * sin α + 1/2 * g * t²

max height = 0 m + 40 m/s * 2.5 s * sin 37° - 1/2* 9.8 m/s² * (2.5 s)² = 29.6 m

Since the ball leaves the catapult 7 m above the ground, the max height above the ground will be 29.6 m + 7 m = 36.6m

<u>max height = 36.6 m</u>

b) When the ball hits the ground, the position is given by the vector "r final" (see figure). The magnitude of "rx", the horizontal component of "r final", is the horizontal distance between the catapult and the wall.

r final = ( x0 + v0 * t * cos α, y0 + v0 * t * sin α + 1/2 * g * t²)

We know that the vertical component of "r final" is -7 (see figure).

Then, we can obtain the time when the the ball hits the ground:

y0 + v0 * t * sin α + 1/2 * g * t² = -7 m

0 m + 40 m/s * t * sin 37° + 1/2 g * t² = -7 m

7 m + 40 m/s * t * sin 37° + 1/2 (-9.8 m/s²) * t² = 0

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solving the quadratic equation:

t = 5.2 s (The negative solution is discarded).

With this time, we can calculate the value of the horizontal component of "r final"

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v = (v0 * cos α, v0 * sin α + g * t)

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v =(32 m/s, -27 m)

x-component = 32 m/s

y-component = - 27 m/s

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