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2 years ago

15

The forces in (Figure 1) are acting on a 4.3 kg object.

1 answer:

harina [27]2 years ago

5 0

Answer:

19computer deaseas

Explanation:

#having learn

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The coordinates of a bird flying in the xy-plane are given by x(t)=αt and y(t)=3.0m−βt2, where α=2.4m/s and β=1.2m/s2.part a:Cal

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Α $=2.4 \frac{m}{s}$

β $=1.2 \frac{m}{s^2}$

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$Vy(t)=-2$ β $t$

$vectorV=[$ α $;-2$ β $t]$

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$ay(t)=-2$ β $t$

$vector a [0;-2$ β $t]$

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3 years ago

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2 years ago

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: The maximum theoretical efficiency of a Carnot engine operating between reservoirs at the steam point and at room temperature

Hunter-Best [27]

Answer:

The value is $\eta = 0.2145$ or 21.45%

Explanation:

From the question we are told that

The first reservoir is at steam point $T_s = 100^o C = 100 + 273 = 373 \ K$

The second reservoir is at room temperature $T_r = 20^o C = 293 \ K$

Generally the maximum theoretical efficiency of a Carnot engine is mathematically evaluated as

$\eta = 1- \frac{T_r}{T_s}$

=> $1 - \frac{ 293}{373}$

=> $\eta = 0.2145$

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2 years ago

The earth's magnetic field is associated with the:

ra1l [238]

The earth's magnetic field is associated with the <span>core.</span>

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3 years ago

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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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3 years ago