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4 years ago

If a system absorbs heat and also does work on its surroundings its energy

1 answer:

7 0

This is, I think, about heat engines and the first law of thermo ?Hot reservoir supplies heat to the system at a high temperature. System does work.System rejects some heat to a cold reservoir at a low temperature.That's a word description of a thermo block diagram. And I think it refers to the "steady state". Seems reasonable that the system would have to warm up (car engines, eg), during which process the (internal ?) energy would presumably rise. In the steady state, however, the energy will presumably be constant. If it continued to rise ... the thing would overheat ????

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A 96.0 kg hoop rolls along a horizontal floor so that its center of mass has a speed of 0.240 m/s. How much work must be done on

Ede4ka [16]

Answer:

The work done by the hoop is equal to 5.529 Joules.

Explanation:

Given that,

Mass of the hoop, m = 96 kg

The speed of the center of mass, v = 0.24 m/s

To find,

The work done by the hoop.

Solution,

The initial energy of the hoop is given by the sum of linear kinetic energy and the rotational kinetic energy. So,

$K_i=\dfrac{1}{2}mv^2+\dfrac{1}{2}I\omega^2$

I is the moment of inertia, $I=mr^2$

Since, $\omega=\dfrac{v}{r}$

$K_i=mv^2$

$K_i=96\times (0.24)^2=5.529\ J$

Finally it stops, so the final energy of the hoop will be, $K_f=0$

The work done by the hoop is equal to the change in kinetic energy as :

$W=K_f-K_i$

W = -5.529 Joules

So, the work done by the hoop is equal to 5.529 Joules. Therefore, this is the required solution.

4 0

3 years ago

Two forces,

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).

3 0

3 years ago

What is the measure of how much matter is in an object and that can be measured using a balance? a. height b. volume c. weight d

kirill [66]

the answer is D. mass

3 0

3 years ago

A. assuming a perpetual inventory system and using the weighted average method, determine the weighted average unit cost after t

Monica [59]

Assuming a perpetual inventory system and using the weighted average method, the weighted average unit is determined as $11.44 after the October 22 purchase.

<h3>What is Weighted Average Cost (WAC)?</h3>

The Weighted Average Cost (WAC) method of inventory valuation in accounting uses a weighted average to establish the COGS and inventory levels.

The price of the products up for grabs is divided by the quantity of them in the weighted average cost technique.

The WAC technique is appropriate under both GAAP and IFRS accounting. Weighted Average Cost (WAC) Method Formula

<h3>Weighted Average Cost</h3>

Weighted Average Unit Costs = [360 units×$12 + (320-180) ×$10] / [360+(320-180)]units}

Weighted Average Unit Costs = $5720 / 500 units

Weighted Average Unit Costs = $11.44

Costs of goods that are offered for sale are calculated using beginning inventory value plus acquisitions.

Units available for sale are the number of units that can be sold by a company or the total number of units that are in its inventory.

To know more about weighted average cost, Visit

brainly.com/question/13543092

#SPJ4

8 0

2 years ago

1. Earth releases about 44-46 Tw of heat, in fact heat can be converted into

Aleksandr [31]

Answer:

even if it all could be used, it wouldn't be enough

Explanation:

The land area of the US is about 5.45% of the world's area, so the amount of released heat over the area of the US is on the order of 2.4 Tw. Current technology for converting geothermal energy to electricity is about 12% efficient, so the available energy might amount to 0.29 Tw if it could all be captured.

Energy consumption in the US in 2019 was on the order of 0.46 Tw. This suggests that even if <em>all</em> of the thermal energy radiated by the Earth from the US could be turned to useful forms of energy, it would meet only about 60% of the US need for energy.

8 0

4 years ago