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kirill [66]
3 years ago
12

In a certain chemical process, a lab technician supplies 292 J of heat to a system. At the same time, 68.0 J of work are done on

the system by its surroundings. What is the increase in the internal energy of the system?
Physics
1 answer:
vekshin13 years ago
3 0

Answer:

The increase in the internal energy of the system is 360 Joules.

Explanation:

Given that,

Heat supplied to a system, Q = 292 J

Work done on the system by its surroundings, W = 68 J

We need to find the increase in the internal energy of the system. It can be given by first law of thermodynamics. It is given by :

dE=dQ+dW\\\\dE=292\ J+68\ J\\\\dE=360\ J

So, the increase in the internal energy of the system is 360 Joules. Hence, this is the required solution.

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A baseball player hits a homerun, and the ball lands in the left field seats, which is 103m away from the point at which the bal
Sati [7]

(a) The ball has a final velocity vector

\mathbf v_f=v_{x,f}\,\mathbf i+v_{y,f}\,\mathbf j

with horizontal and vertical components, respectively,

v_{x,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\cos(-38^\circ)\approx16.2\dfrac{\rm m}{\rm s}

v_{y,f}=\left(20.5\dfrac{\rm m}{\rm s}\right)\sin(-38^\circ)\approx-12.6\dfrac{\rm m}{\rm s}

The horizontal component of the ball's velocity is constant throughout its trajectory, so v_{x,i}=v_{x,f}, and the horizontal distance <em>x</em> that it covers after time <em>t</em> is

x=v_{x,i}t=v_{x,f}t

It lands 103 m away from where it's hit, so we can determine the time it it spends in the air:

103\,\mathrm m=\left(16.2\dfrac{\rm m}{\rm s}\right)t\implies t\approx6.38\,\mathrm s

The vertical component of the ball's velocity at time <em>t</em> is

v_{y,f}=v_{y,i}-gt

where <em>g</em> = 9.80 m/s² is the magnitude of the acceleration due to gravity. Solve for the vertical component of the initial velocity:

-12.6\dfrac{\rm m}{\rm s}=v_{y,i}-\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)\implies v_{y,i}\approx49.9\dfrac{\rm m}{\rm s}

So, the initial velocity vector is

\mathbf v_i=v_{x,i}\,\mathbf i+v_{y,i}\,\mathbf j=\left(16.2\dfrac{\rm m}{\rm s}\right)\,\mathbf i+\left(49.9\dfrac{\rm m}{\rm s}\right)\,\mathbf j

which carries an initial speed of

\|\mathbf v_i\|=\sqrt{{v_{x,i}}^2+{v_{y,i}}^2}\approx\boxed{52.4\dfrac{\rm m}{\rm s}}

and direction <em>θ</em> such that

\tan\theta=\dfrac{v_{y,i}}{v_{x,i}}\implies\theta\approx\boxed{72.0^\circ}

(b) I assume you're supposed to find the height of the ball when it lands in the seats. The ball's height <em>y</em> at time <em>t</em> is

y=v_{y,i}t-\dfrac12gt^2

so that when it lands in the seats at <em>t</em> ≈ 6.38 s, it has a height of

y=\left(49.9\dfrac{\rm m}{\rm s}\right)(6.38\,\mathrm s)-\dfrac12\left(9.80\dfrac{\rm m}{\mathrm s^2}\right)(6.38\,\mathrm s)^2\approx\boxed{119\,\mathrm m}

6 0
4 years ago
A dog is walking back home and its motion is modeled by the graph above. What position was the dog at 3 seconds after it started
hammer [34]
6 meters because 6/3 = 10/5
4 0
3 years ago
An amusement park ride moves a rider at a constant speed of 14 meters per second in a horizontal circular path of radius 10. met
ASHA 777 [7]

Answer:

B) 2g

Explanation:

<u>Given the following data;</u>

Velocity, v = 14m/s

Radius, r = 10m

To find the centripetal acceleration;

Acceleration, a = \frac {v^{2}}{r}

Substituting into the equation, we have;

Acceleration, a = \frac {14^{2}}{10}

Acceleration, a = \frac {196}{10}

Acceleration, a = 19.6m/s²

In terms of acceleration due to gravity, g = 9.8m/s²

We would divide by g;

Acceleration, a = 19.6/9.8 = 2

Hence, centripetal acceleration = 2g

Therefore, the rider's centripetal acceleration in terms of g, the acceleration due to gravity is 2g.

4 0
3 years ago
Why do windows of distant houses appear to reflect the sun only when rising or setting
alisha [4.7K]
<span> most importantly, the sun is reflected back at the same angle as the angle of incidence, so with vertical panes of glass and the sun overhead, the reflection will just hit the ground before traveling any distance. </span>
3 0
3 years ago
Read 2 more answers
at which velocity would you expect an object to have the greatest amount of kinetic energy? a. 11-20 m/s b.0-10 m/s c. 21-30 m/s
UkoKoshka [18]

Answer:

31-40 m/s

Explanation:

The kinetic energy of an object is given by :

E=\dfrac{1}{2}mv^2

m is mass of the object

v is velocity

If mass of an object is constant, its kinetic energy depends only on velocity. It means if the velocity is more, it will have maximum kinetic energy. For the interval of 31-40 m/s, the kinetic energy of the object is maximum.

4 0
3 years ago
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