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3 years ago

What drives the movement of water through earths systems.

Physics

1 answer:

MrMuchimi3 years ago

8 0

Answer:

A. The Moon

Explanation:

I believe its the gravitational pool that effects the way out ocean tides moves

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Suppose an object in freefall is dropped from a building it’s starting velocity is 0m/s. Ignoring the facts of air resistance wh

Alja [10]

v = v₀ + at

v = final speed, v₀ = initial speed, a = acceleration, t = elapsed time

Given values:

v₀ = 0m/s (starts from rest), a = 9.81m/s², t = 3s

Plug in and solve for v:

v = 0 + 9.81(3)

v = 29.4m/s

3 0

3 years ago

A stretched string has a mass per unit length of 5.40 g/cm and a tension of 17.5 N. A sinusoidal wave on this string has an ampl

kondaur [170]

Answer:

Part a)

$y_m = 0.157 mm$

part b)

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

Explanation:

As we know that the speed of wave in string is given by

$v = \sqrt{\frac{T}{m/L}}$

so we have

$T = 17.5 N$

$m/L = 5.4 g/cm = 0.54 kg/m$

now we have

$v = \sqrt{\frac{17.5}{0.54}}$

$v = 5.69 m/s$

now we have

Part a)

$y_m$ = amplitude of wave

$y_m = 0.157 mm$

part b)

$k = \frac{\omega}{v}$

here we know that

$\omega = 2\pi f$

$\omega = 2\pi(92.2) = 579.3 rad/s$

so we have

$k = \frac{579.3}{5.69}$

$k = 101.8 rad/m$

Part c)

$\omega = 579.3 rad/s$

Part d)

here since wave is moving in negative direction so the sign of $\omega$ must be positive

4 0

3 years ago

Flura [38]

Answer:

about 4 km

Explanation:

15 minutes is a quarter of an hour, so you divide 16km by 4 to get your answer

8 0

3 years ago

In a tape recorder, the tape is pulled past the read-and-write heads at a constant speed by the drive mechanism. Consider the re

Irina-Kira [14]

Answer:

Torque decreases .

Explanation:

The tape is pulled at constant speed , speed v is constant , so there is

v = ω r where ω is angular speed and r is radius , As radius decreases , angular speed ω increases , So there is angular acceleration .

Let it be α . Let I be moment of inertia of reel .

Reel is in the form of disc

I = 1/2 m r²

I x α = torque

1/2 m r² x α = torque

As the reel is untapped , its mass decreases , r also decreases , so torques also decreases .

6 0

3 years ago

Three people pull simultaneously on a stubborn donkey. Jack pulls eastward with a force of 80.5 N, Jill pulls with 81.7 N in the

Gnesinka [82]

Answer:

F = 233.52 N, θ' = 351.41º

Explanation:

In this exercise we must find the net force applied on the donkey.

For this we use Newton's second law, where we create a reference frame with the horizontal x axis

let's decompose the forces

Jack

= 80.5 N

Jill

cos 45 = F_{2x} / F₂2

sin 45 = F_{2y} / F₂2

F_{2x} = F₂ cos 45

F_{2y} = F₂ sin 45

F_{2x} = 81.7 cos 45 = 57.77 N

F_{2y} = 81.7 sin 45 = 57.77 N

Jane

cos (270 + 45) = F_{3x} / F₃3

sin 315 = F_{3y} / F₃

F_{3x} = 131 cos 315 = 92.63 N

F_{3y} = 131 sin 315 = -92.63 N

the force can be found in each axis

X axis

F_{x} = F_{1x} + F_{2x} + F_{3x}

F_{x} = 80.5 +57.77 + 92.63

F_{x} = 230.9 N

Axis y

F_{y} = F_{1y} + F_{2y} + F_{3y}

F_{y} = 0 + 57.77 -92.63

F_{y} = -34.86 N

we can give the result in two ways

a) F = (230.9 i ^ - 34.86 j ^) N

b) in the form of module and angle

we use the Pythagorean theorem

F = √(Fₓ² + F_{y}²

F = √(230.9² + 34.86²)

F = 233.52 N

let's use trigonometry for the angle

tan θ = $\frac{F_y}{F_x} }$

θ = tan⁻¹ (\frac{F_y}{F_x} })

θ = tan⁻¹ (-34.86 / 230.9)

θ = -8.59º

if we measure this angle from the positive side of the x-axis counterclockwise

θ' = 360 -θ

θ‘= 360- 8.59

θ' = 351.41º

5 0

3 years ago