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2 years ago

8

What is the wavelength of a wave?

1 answer:

Sunny_sXe [5.5K]2 years ago

6 0

The wavelength is the spatial period of a periodic wave—the distance over which the wave's shape repeats.

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In developed nations, the water delivered to homes by pipe is generally potable. Please select the best answer from the choices

stiv31 [10]

True is ur answer sorry if i am too late

6 0

2 years ago

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According to Newton’s law of universal gravitation, which statement is true?

SIZIF [17.4K]

Option (a) is correct.

Falling objects accelerate as they approach the ground.This is because of the force of gravity acting on the falling objects. so the velocity of these objects increases continuously as they approach the ground. the acceleration acting on the falling objects is a constant ( close to the surface of earth) and is called as acceleration due to gravity denoted by g. value of g=9.8 m/s².

4 0

2 years ago

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Would someone just answer this?

Alecsey [184]

Answer:

I answered :)

Explanation:

i just dk the right answer

8 0

2 years ago

Assume that the Deschutes River has straight and parallel banks and that the current is 0.75 m/s. Drifting down the river, you f

DedPeter [7]

Answer:

d = 142.5 m

Explanation:

This is a vector exercise. Let's calculate how much the boat travels in the 40s

d₀ = $v_{b}$ t

d₀ = 0.75 40

d₀ = 30 m

Let's write the kinematic equations

Boat

x = d₀ + $v_{b}$ t

x = 0 + $v_{h}$ t

At the meeting point the coordinate is the same for both

d₀ + $v_{b}$ t = $v_{h}$ t

t ( $v_{h}$ - $v_{b}$ ) = d₀

t = d₀ / ( $v_{b}$ - $v_{h}$ )

The two go in the same direction therefore the speeds have the same sign

t = 30 / (0.95-0.775)

t = 150 s

The distance traveled by man is

d = $v_{h}$ t

d = 0.95 150

d = 142.5 m

3 0

3 years ago

A 21500 kg jet airplane is flying through some high winds. At some point in time, the airplane is pointing due north, while the

GalinKa [24]

Answer: $\theta =79.47 North\ of\ west$

Explanation:

Given

Mass of airplane =21500 kg

Force due to jet engines=35100 N

Force from wind is 14900 N at angle of 75 south of west

Resolving forces

$F_x=14900cos75=-3856.403 N$

$F_y=35100-14900sin75$

$F_y=35100-14392.29=20707.70 N$

Therefore acceleration in x direction

$a_x=\frac{F_x}{m}=\frac{-3856.403}{21500}=-0.179 m/s^2$

$a_y=\frac{F_y}{m}=\frac{20707.70}{21500}=0.963 m/s^2$

Direction of acceleration

$tan\theta =\frac{a_y}{a_x}=\frac{0.963}{0.179}=-5.38$

$\theta =79.47\ North\ of\ west$

5 0

3 years ago