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lesantik [10]
3 years ago
8

What is the lowest point of a transverse wave

Physics
2 answers:
Anna007 [38]3 years ago
8 0

Answer:

trough

Explanation:

The lowest point on a transverse wave is the <u>trough</u>.

Gwar [14]3 years ago
3 0
Sorry this is late...

The lowest point of a transverse wave is called the trough, while the highest is called the crest.
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What happens at<br> the Bernoulli effect
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3 years ago
The total distance from a house to a school to is 9.5 km. A student travels all the way from his house to the school and back to
marysya [2.9K]

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Particle A of charge 3.06 10-4 C is at the origin, particle B of charge -5.70 10-4 C is at (4.00 m, 0), and particle C of charge
Alex

Answer:

F_net = 26.512 N

Explanation:

Given:

Q_a = 3.06 * 10^(-4 ) C

Q_b = -5.7 * 10^(-4 ) C

Q_c = 1.08 * 10^(-4 ) C

R_ac = 3 m

R_bc = sqrt (3^2 + 4^2) = 5m

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Coulomb's Law:

F_i = k * Q_i * Q_j / R_ij^2

Compute F_ac and F_bc :

F_ac = k * Q_a * Q_c / R^2_ac

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F_ac = 33.01128 N

F_bc = k * Q_b * Q_c / R^2_bc

F_bc =  8.99 * 10^9* ( 5.7 * 10^(-4 ))* (1.08 * 10^(-4 )) / 5^2  

F_bc = - 22.137 N

Angle a is subtended between F_bc and y axis @ C

cos(a) = 3 / 5

sin (a) = 4 / 5

Compute F_net:

F_net = sqrt (F_x ^2 + F_y ^2)

F_x = sum of forces in x direction:

F_x = F_bc*sin(a) = 22.137*(4/5) = 17.71 N

F_y = sum of forces in y direction:

F_y = - F_bc*cos(a) + F_ac = - 22.137*(3/5) + 33.01128 = 19.72908 N

F_net = sqrt (17.71 ^2 + 19.72908 ^2) = 26.5119 N

Answer: F_net = 26.512 N

5 0
4 years ago
A bungee jumper of mass 75kg is attached to a bungee cord of length L=35m. She walks off a platform (with no initial speed), reac
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Answer:

1. 77.31 N/m

2. 26.2 m/s

3. increase

Explanation:

1. According to the law of energy conservation, when she jumps from the bridge to the point of maximum stretch, her potential energy would be converted to elastics energy. Her kinetic energy at both of those points are 0 as speed at those points are 0.

Let g = 9.8 m/s2. And the point where the bungee ropes are stretched to maximum be ground 0 for potential energy. We have the following energy conservation equation

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mgh = kx^2/2

where m = 75 kg is the mass of the jumper, h = 72 m is the vertical height from the jumping point to the lowest point, k (N/m) is the spring constant and x = 72 - 35 = 37 m is the length that the cord is stretched

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mv^2/2 = mgH

where H = 35m this time due to the height difference between the jumping point and the point 35m below the platform

v^2/2 = gH

v = \sqrt{2gH} = \sqrt{2*9.8*35} = 26.2 m/s

3. If she jumps from her platform with a velocity, then her starting kinetic energy is no longer 0. The energy conservation equation would then be

E_P + E_k = E_E

So the elastics energy would increase, which would lengthen the maximum displacement of the cord

5 0
4 years ago
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