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3 years ago

6

a water wave has a speed of 23.0 meters/second. if the waves frequency is 0.0680 hertz, what is the wavelength

1 answer:

USPshnik [31]3 years ago

7 0

Answer:

Using the Fundamental Equation of Wave, we have:

Explanation:

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Determine the correct name for the compound, C3H8. In organic chemistry, this is referred to as propane. A) Tricarbon hexahydrid

Leya [2.2K]

Well, since it's C3, we need Tricarbon, and H8 means Octa-, so B is your answer: Tricarbon Octahydride.

Hope this helps~!

8 0

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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4 years ago

is potential energy and kinetic energy real or is it theoretical and is used to explain how thing work? if it is real than where

vesna_86 [32]

It is is real because potential energy is stored inside gravitational energy.

5 0

3 years ago

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A 1-kg discus is thrown with a vertical velocity of 19 m/s at an angle of 35 degrees from a height of 1.94 m. Do not factor in a

Travka [436]

Answer:

Vertical velocity $V_y = 10.89 \ m/s$

Horizontal velocity $V_x = 15.57 \ m/s$

Explanation:

If a 1 -kg is thrown vertically with a velocity of 19 m/s at an angle of 35 °C from a height 1.94 m

The vertical and horizontal component can be resolved as:

$V_y = Vsin \theta \\ \\ \\\\V_x = Vcos \theta$

For Vertical component :

$V_y = V sin \theta \\\\V_y = 19 * sin 35 \\ \\ V_y = 19* 0..5736 \\ \\ V_y = 10.89 \ m/s$

For horizontal velocity

$V_x = V cos \theta \\\\V_x = 19 * cos 35 \\ \\ V_x = 19* 0.8192 \\ \\ V_x = 15.57 \ m/s$

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3 years ago

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Readme [11.4K]

Answer:

Explanation: Its your hairline

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3 years ago

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