Why are there two types of core?

The human ear is most sensitive to the sounds in what range?

jeyben [28]

Answer:

The human ear is most sensitive to sounds in the range of 2000 to 3000 Hz

3 0

3 years ago

What is relative density?

Artemon [7]

Relative density, or specific gravity, is the ratio of the density of a substance to the density of a given reference material. Specific gravity for liquids is nearly always measured with respect to water at its densest; for gases, the reference is air at room temperature.

3 0

2 years ago

A box slides down a ramp inclined at 24◦ to the horizontal with an acceleration of 1.7 m/s 2 . The acceleration of gravity is 9.

dsp73

Answer:

Explanation:

In friction, the coefficient of friction formula is expressed as;

$\mu = \frac{F_f}{R}$

Ff is the frictional force = Wsinθ

R is the reaction = Wcosθ

Substitute inti the equation;

$\mu = \frac{Wsin \theta}{W cos\theta} \\\mu = \frac{sin \theta}{cos\theta} \\\mu = tan \theta$

Given

θ = 24°

$\mu = tan 24^0\\\mu = 0.445\\$

Hence the coefficient of kinetic friction between the box and the ramp is 0.445

3 0

3 years ago

As a substance melts, heat energy is used to break the connections between molecules, and temperature __________ until all melti

mario62 [17]

Temperature stays the same

4 0

3 years ago

Scenario

Anvisha [2.4K]

Answer:

1) t = 23.26 s, x = 8527 m, 2) t = 97.145 s, v₀ = 6.4 m / s

Explanation:

1) First Scenario.

After reading your extensive problem, we are going to solve it, for this exercise we must use the parabolic motion relationships. Let's carry out an analysis of the situation, for deliveries the planes fly horizontally and we assume that the wind speed is zero or very small.

Before starting, let's reduce the magnitudes to the SI system

v₀ = 250 miles/h (5280 ft / 1 mile) (1h / 3600s) = 366.67 ft/s

y = 2650 m

Let's start by looking for the time it takes for the load to reach the ground.

y = y₀ + v_{oy} t - ½ g t²

in this case when it reaches the ground its height is zero and as the plane flies horizontally the vertical speed is zero

0 = y₀ + 0 - ½ g t2

t = $\sqrt{ \frac{2y_o}{g} }$

t = √(2 2650/9.8)

t = 23.26 s

this is the horizontal scrolling time

x = v₀ t

x = 366.67 23.26

x = 8527 m

the speed at the point of arrival is

v_y = v_{oy} - g t = 0 - gt

v_y = - 9.8 23.26

v_y = -227.95 m / s

Module and angle form

v = $\sqrt{v_x^2 + v_y^2}$

v = √(366.67² + 227.95²)

v = 431.75 m / s

θ = tan⁻¹ (v_y / vₓ)

θ = tan⁻¹ (227.95 / 366.67)

θ = - 31.97º

measured clockwise from x axis

We see that there must be a mechanism to reduce this speed and the merchandise is not damaged.

2) second scenario. A catapult located at the position x₀ = -400m y₀ = -50m with a launch angle of θ = 50º

we look for the components of speed

cos θ = v₀ₓ / v₀

sin θ = v_{oy} / v₀

v₀ₓ = v₀ cos θ

v_{oy} = v₀ sin θ

we look for the time for the arrival point that has coordinates x = 0, y = 0

y = y₀ + v_{oy} t - ½ g t²

0 = y₀ + vo sin θ t - ½ g t²

0 = -50 + vo sin 50 t - ½ 9.8 t²

x = x₀ + v₀ₓ t

0 = x₀ + vo cos θ t

0 = -400 + vo cos 50 t

podemos ver que tenemos un sistema de dos ecuación con dos incógnitas

50 = 0,766 vo t – 4,9 t²

400 = 0,643 vo t

resolved

50 = 0,766 ( $\frac{400}{0.643 \ t}$ ) t – 4,9 t²

50 = 476,52 t – 4,9 t²

t² – 97,25 t + 10,2 = 0

we solve the quadratic equation

t = [97.25 ± $\sqrt{97.25^2 - 4 \ 10.2}$ ] / 2

t = 97.25 ±97.04] 2

t₁ = 97.145 s

t₂ = 0.1 s≈0

the correct time is t1 the other time is the time to the launch point,

t = 97.145 s

let's find the initial velocity

x = x₀ + v₀ cos 50 t

0 = -400 + v₀ cos 50 97.145

v₀ = 400 / 62.44

v₀ = 6.4 m / s

5 0

3 years ago