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nataly862011 [7]

3 years ago

15

A system dissipates 12 J of heat into the surroundings; meanwhile, 28 J of work is done on the system. What is the change of the

internal energy of the system?

1 answer:

Stels [109]3 years ago

3 0

Answer:

+16 J

Explanation:

We can solve the problem by using the 1st law of thermodynamics:

$\Delta U = Q-W$

where

$\Delta U$ is the change of the internal energy of the system

Q is the heat (positive if supplied to the system, negative if dissipated by the system)

W is the work done (positive if done by the system, negative if done by the surroundings on the system)

In this case we have:

Q = -12 J is the heat dissipated by the system

W = -28 J is the work done ON the system

Substituting into the equation, we find the change in internal energy of the system:

$\Delta U=-12 J-(-28 J)=+16 J$

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What is the direction of a vector with an x component of -12 m and a y component of -15 m?

dusya [7]

Answer:

$51.34^{\circ}$

Explanation:

Given that,

x component of a vector = -12 m

The y component of a vector = -15 m

We need to find the direction of a vector. The direction of a vector is given by :

$\tan\theta=\dfrac{y}{x}$

Put all the values,

$\tan\theta=\dfrac{-15}{-12}\\\\\theta=\tan^{-1}(\dfrac{-15}{-12})\\\\\theta=51.34^{\circ}$

So, the direction of vector is $51.34^{\circ}$ to x component.

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3 years ago

Which describes the best approach when conducting a scientific experiment

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I hope you found this helpful!

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3 years ago

determine the greates possile acceleration of the 975 kg race car so that its front wheels do not leace the gorund

wolverine [178]

<u>Answer</u>:

The greatest possible acceleration of the car is $a_G= 6.78 m/s^2$

<u>Explanation</u>:

$N_A+N_B-Mg = 0$

$-N_Aa +N_B(b-a)- \mu_s N_Bh - \mu_s N_Ah = 0$

$0.8N_B +0.8N_A = 975a_G$

$N_A+N_B = 9564.75$ -------------(1)

$-N_A(1.82) + N_B(2.20 -1.82) -0.9N_B(0.55)-0.8N_A(0.55)=0$

$-N_A(1.82) +0.38 N_B -0.44N_B -0.44N_A=0$

$-2.26N_A -0.06N_B= 0$ ----------------(2)

Solving the equation (1) and(2)

$N_A + N_B = 9564.75$

$-2.26N_A-0.06N_B=0$

$N_A = -260.85N$

$N_B = 9825.60N$

$\mu_s N_B + \mu_s N_A = 975a_G$

$0.8(9825.60)+0.8(-260.85) = 975a_G$ $a_G=\frac{7651.8}{975}$ $a_G_1=7.4848m/s^2$

Next lets assume that the front wheels contact with the ground N_A = 0

$F_B = Ma_G$

$N_B = M_g$

$N_B - M_g = 0$

$N_B(b-a) –F_Bh = 0$

$F_B = 975a_G$

$N_B-975(9.8) = 0$

$N_B=9564.75N$

$9564.75(2.20 -1.82) -F_B(0.55)=0$

$\frac{3634.605}{0.55}=F_B$

$F_B = 6608.3$

$F_B = Ma_G$

$6608.3 = 975a_G$

$a_G = 6.7778 m/s^2$

$a_G_2 = 6.78m/s^2$

Choosing the critical case

$a_G = min(a_G_1 ,a_G_2)$

$a_G = min(7.848, 6.78)$

$a_G= 6.78 m/s^2$

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