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3 years ago

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You are in a hot air balloon, 650 m above the flat Texas plains. The Earth's radius is about 6400 km.You look out toward the hor

izon. How far out can you see−that is, how far is your horizon?

1 answer:

Molodets [167]3 years ago

4 0

Answer:

91 km

Explanation:

Draw a picture. The line of sight to the horizon forms a tangent line with the curvature of the Earth. The radius line to that point forms a right angle with the tangent line. So we can use Pythagorean theorem to find the distance.

c² = a² + b²

(6400 km + 0.650 km)² = (6400 km)² + d²

d ≈ 91 km

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A 10.0 N package of whole wheat flour is suddenly placed on the pan of a scale such as you find in grocery stores. The pan is su

inna [77]

Answer:

0.025 m

Explanation:

From the question,

Applying Hook's law

F = ke................... Equation 1

Where F = Force, k = spring constant of the scale, e = maximum distance at which the spring will compress.

make e the subject of the equation

e = F/k....................... Equation 2

Given: F = 10 N, e = 395 N/m

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e = 10/395

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3 years ago

Read 2 more answers

What is the slope of the line if the rise of a line on a distance versus-time graph is 900 meters and the run is 3 minutes?

Alexandra [31]

600/3 = 200
the slope is 200m/min

OR

600/ (3/60) =
600 x 60/3 =
600 x 20 = 12,000 meters per hour

6 0

3 years ago

When the play button is pressed, a CD accelerates uniformly from rest to 450 rev/min in 3.0 revolutions. If the CD has a radius

Marina CMI [18]

To solve this problem it is necessary to apply the kinematic equations of angular motion.

Torque from the rotational movement is defined as

$\tau = I\alpha$

where

I = Moment of inertia $\rightarrow \frac{1}{2}mr^2$ For a disk

$\alpha =$ Angular acceleration

The angular acceleration at the same time can be defined as function of angular velocity and angular displacement (Without considering time) through the expression:

$2 \alpha \theta = \omega_f^2-\omega_i^2$

Where

$\omega_{f,i} =$ Final and Initial Angular velocity

$\alpha =$ Angular acceleration

$\theta =$ Angular displacement

Our values are given as

$\omega_i = 0 rad/s$

$\omega_f = 450rev/min (\frac{1min}{60s})(\frac{2\pi rad}{1rev})$

$\omega_f = 47.12rad/s$

$\theta = 3 rev (\frac{2\pi rad}{1rev}) \rightarrow 6\pi rad$

$r = 7cm = 7*10^{-2}m$

$m = 17g = 17*10^{-3}kg$

Using the expression of angular acceleration we can find the to then find the torque, that is,

$2\alpha\theta=\omega_f^2-\omega_i^2$

$\alpha=\frac{\omega_f^2-\omega_i^2}{2\theta}$

$\alpha = \frac{47.12^2-0^2}{2*6\pi}$

$\alpha = 58.89rad/s^2$

With the expression of the acceleration found it is now necessary to replace it on the torque equation and the respective moment of inertia for the disk, so

$\tau = I\alpha$

$\tau = (\frac{1}{2}mr^2)\alpha$

$\tau = (\frac{1}{2}(17*10^{-3})(7*10^{-2})^2)(58.89)$

$\tau = 0.00245N\cdot m \approx 2.45*10^{-3}N\cdot m$

Therefore the torque exerted on it is $2.45*10^{-3}N\cdot m$

3 0

3 years ago

Question 1 of 10

Debora [2.8K]

Answer:

V = λ f (wavelength * frequency)

λ = V / f = 343 m/s / 262 / s = 1.3 m

7 0

2 years ago