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3 years ago

5

The temperature of a plastic cube is monitored while the cube is pushed 1.8 m across a floor at constant speed by a horizontal f

orce of 24 N. The monitoring reveals that the thermal energy of the cube increases by 18 J. What is the increase in the thermal energy of the floor along which the cube slides?

1 answer:

algol133 years ago

8 0

Answer: 25.2 Joules

Explanation:

Work = total heat energy = Fs = (24)(1.8) = 43.2 J

Heat = cube + floor = 43.2J

floor = 43.2 J - 18 J = 25.2 J

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A crate is pulled with a force of 165 N at an angle 30 ° northwest. What is the resultant horizontal force on the crate?

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Answer:

Resultant horizontal force = 143 N

Explanation:

Since the a gle is 30° northwest, then it means the resultant force will be horizontal and as such;

Resultant horizontal force = 165 * cos 30

Resultant horizontal force = 142.89

Approximating to a whole number gives;

Resultant horizontal force = 143 N

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The different between capactance wnd capacitor

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Answer:

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Explanation:

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3 years ago

Read 2 more answers

serg [7]

First compute the resultant force F:

$\mathbf F_1=(5.90\,\mathbf i-5.60\,\mathbf j)\,\mathrm N$

$\mathbf F_2=(4.65\,\mathbf i-5.55\,\mathbf j)\,\mathrm N$

$\implies\mathbf F=\mathbf F_1+\mathbf F_2=(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N$

Then use Newton's second law to determine the acceleration vector $\mathbf a$ for the particle:

$\mathbf F=m\mathbf a$

$(10.55\,\mathbf i-11.15\,\mathbf j)\,\mathrm N=(2.10\,\mathrm{kg})\mathbf a$

$\mathbf a\approx(5.02\,\mathbf i-5.31\,\mathbf j)\dfrac{\rm m}{\mathrm s^2}$

Let $\mathbf x(t)$ and $\mathbf v(t)$ denote the particle's position and velocity vectors, respectively.

(a) Use the fundamental theorem of calculus. The particle starts at rest, so $\mathbf v(0)=0$ . Then the particle's velocity vector at <em>t</em> = 10.4 s is

$\mathbf v(10.4\,\mathrm s)=\mathbf v(0)+\displaystyle\int_0^{10}\mathbf a(u)\,\mathrm du$

$\mathbf v(10.4\,\mathrm s)=\left((5.02\,\mathbf i-5.31\,\mathbf j)u\,\dfrac{\rm m}{\mathrm s^2}\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf v(10.4\,\mathrm s)\approx(52.2\,\mathbf i-55.2\,\mathbf j)\dfrac{\rm m}{\rm s}$

If you don't know calculus, then just use the formula,

$v_f=v_i+at$

So, for instance, the velocity vector at <em>t</em> = 10.4 s has <em>x</em>-component

$v_{f,x}=0+\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)(10.4\,\mathrm s)=52.2\dfrac{\rm m}{\mathrm s^2}$

(b) Compute the angle $\theta$ for $\mathbf v(10.4\,\mathrm s)$ :

$\tan\theta=\dfrac{-55.2}{52.2}\implies\theta\approx-46.6^\circ$

so that the particle is moving at an angle of about 313º counterclockwise from the positive <em>x</em> axis.

(c) We can find the velocity at any time <em>t</em> by generalizing the integral in part (a):

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

$\implies\mathbf v(t)=\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)t\,\mathbf j$

Then using the fundamental theorem of calculus again, we have

$\mathbf x(10.4\,\mathrm s)=\mathbf x(0)+\displaystyle\int_0^{10.4}\mathbf v(u)\,\mathrm du$

where $\mathbf x(0)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m$ is the particle's initial position. So we get

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\displaystyle\int_0^{10.4}\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u\,\mathbf j\right)\,\mathrm du$

$\mathbf x(10.4\,\mathrm s)=(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m+\dfrac12\left(\left(5.02\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf i+\left(-5.31\dfrac{\rm m}{\mathrm s^2}\right)u^2\,\mathbf j\right)\bigg|_{u=0}^{u=10.4}$

$\mathbf x(10.4\,\mathrm s)\approx(542\,\mathbf i-570\,\mathbf j)\,\mathrm m$

So over the first 10.4 s, the particle is displaced by the vector

$\mathbf x(10.4\,\mathrm s)-\mathbf x(0)\approx(270\,\mathbf i-283\,\mathbf j)\,\mathrm m-(-1.75\,\mathbf i+4.15\,\mathbf j)\,\mathrm m\approx(272\,\mathbf i-287\,\mathbf j)\,\mathrm m$

or a net distance of about 395 m away from its starting position, in the same direction as found in part (b).

(d) See part (c).

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3 years ago

A ball tossed vertically upward from the ground next to a building passes the bottom of a window 1.8 s after being tossed and pa

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The ball takes 0.2 sec to travel the height of the window, which is 2 m.

Then the speed of the ball at that time was

$v = \frac{d}{t}$

$v = \frac{2}{0.20}$

$v = 10\ \frac{m}{s}$

We know that:

$v = v_0 -gt$

Where $v_0$ is the initial velocity.

So:

$10 = v_0 -9.8(2)$

$10 + 9.8(2) = v_0$

$v_0 = 29.6\ m / s$

Then we have the equation for the position as a function of time.

$r = r_0 + v_0t - \frac{1}{2}gt^2$

Where

$r_0$ = initial position

r = position as a function of time

$v_0$ = initial velocity

g = acceleration of gravity

t = time.

If the ball is thrown from the ground then:

$r_0$ = 0 m

We want to find now the distance between the window and the ground.

When the ball reaches the bottom of the window t = 1.8s

So:

$r = 0 + 29.6(1.8) - 0.5(9.8)(1.8)^2$

$r = 37,404$ m

The window is 37,404 m high

Finally, to know how high the ball rises we must know at what moment the vertical velocity of the ball is zero.

$v = v_0 -gt\\\\0 = v_0 -gt\\\\gt = v_0\\\\t = \frac{29.6}{9.8}$

$t = 3.02\ s$

Now we replace t in the position equation

$r = 0 + 29.6(3.02) -0.5(9.8)(3.02)^2$

$r = 44.70\ m$

The ball reaches up to 44.70 m in height.

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What process is used in this example? You have learned that all living things use energy. Your dog is a living thing. She must u

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Option B.) Deductive Reasoning

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