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2 years ago

Which of the following statements about horizons is true?

Physics

2 answers:

MArishka [77]2 years ago

7 0

Answer:

The statement about horizons which is true is:

d. They are defined by their different physical features.

Explanation:

A soil horizon is the layers of the soil which are parallel to each other and they differ according to their characteristics. The difference in the physical features of the soil is termed to be a particular horizon. They have a distinct color, texture, and physical properties.

nalin [4]2 years ago

6 0

All soils have completely different horizon patterns.

You might be interested in

Q1: A runner is jogging in a straight line at a steady vr= 6.8 km/hr. When the runner is L= 2.4 km from the finish line, a bird

serious [3.7K]

Answer:

Q1: 3.2km

Q2: 4.8K

Explanation:

Q1:

So db is the distance of bird, and dr is the distance of runner

db = 2vr and the distance of bird is going to be 2 times greater than the runner.

formulas: db = 2vr & db = 2dr

db = 2dr
L + (L - x) = 2x
2L - x = 2x
2L = 3x
x = $\frac{2}{3}$ L

Insert it in x = $\frac{2}{3}$ L

$\frac{2}{3}$ (2.4km) = 1.6km

Now we use formula db = 2dr

db = 2L - x
db = 2(2.4km) - 1.6km
db = 3.2km

Q2:

Formulas: Vr = L /Δt & Vb = db/Δt

Vr = L/ Δt ⇒ Δt = $\frac{L}{Vr}$
$\frac{2.4km}{6.8km/hr}$
$\frac{6}{17}hr$

(Km cancel each other)

Vb = db/Δt ⇒ db = VbΔt
13.6km/hr $(\frac{6}{17}hr )$
4.8km

(hr cancel each other)

Hope it helps you :)

6 0

2 years ago

What is the threshold frequency for sodium metal if a photon with frequency 6.66 × 1014 s−1 ejects a photon with 7.74 × 10−20 J

FrozenT [24]

Answer:

5.5 × 10^14 Hz or s^-1

no orange light has less frequency so no photoelectric effect

Explanation:

hf = hf0 + K.E

HERE h is Planck 's constant having value 6.63 × 10 ^-34 J s

f is frequency of incident photon and f0 is threshold frequency

hf0 = hf- k.E

6.63 × 10 ^-34 × f0 = 6.63 × 10 ^-34× 6.66 × 10^14 - 7.74× 10^-20

6.63 × 10 ^-34 × f0 = 3.64158×10^-19

f0 = 3.64158×10^-19/ 6.63 × 10 ^-34

f0 = 5.4925 × 10^14

f0 =5.5 × 10^14 Hz or s^-1

frequency of orange light is 4.82 × 10^14 Hz which is less than threshold frequency hence photo electric effect will not be observed for orange light

8 0

2 years ago

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph from due north.

Afina-wow [57]

Answer:

a) 17.05 mph

b) 54.7° northeast direction

c) 10.71 mph

The direction is -22.58° relative to the east.

To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.

Explanation:

The question is a little confusing but, I guess the correct question should be;

You are flying a hang glider at 14 mph in the northeast direction (45°). The wind is blowing at 4 mph due north.

a) What is your airspeed?

b) What angle (direction) are you flying?

c) The wind increases to 14 mph from north. Now what is your airspeed and what direction are you flying? If your destination is to the northeast, how would you change your speed or direction so you might make it there?

NB: The difference in the question and my suggestion is highlighted boldly.

Your speed = 14 mph

direction is 45° northeast

Th wind speed = 4 mph

direction is north

We resolve the your speed and the wind speed into the horizontal and vertical components

For vertical the component component

$V_{y}$ = 14(sin 45) + 4 = 9.89 + 4 = 13.89 mph

For the horizontal speed component

$V_{x}$ = 14(cos 45) + 0 = 9.89 + 0 = 9.89 mph

Resultant speed = $\sqrt{V^{2} _{y}+V^{2} _{x} }$

==> $\sqrt{13.89^{2} +9.89^{2} }$ = 17.05 mph This is your airspeed

b) To get your direction, we use

tan ∅ = $V_{y}$ / $V_{x}$

tan ∅ = 13.89/9.89 = 1.413

∅ = $tan^{-1}$ (1.413) = 54.7° northeast direction

c) If the wind increases to 14 mph from the north, then it means the wind blows due south. As before, only the vertical component is affected .

In this case,

$V_{y}$ = 14(sin 45) - 14 = 9.89 - 14 = -4.11 mph

Resultant speed = $\sqrt{V^{2} _{y}+V^{2} _{x} }$

==> $\sqrt{4.11^{2} +9.89^{2} }$ = 10.71 mph This is your airspeed

Your direction will be,

tan ∅ = $V_{y}$ / $V_{x}$

tan ∅ = -4.11/9.89 = -0.416

∅ = $tan^{-1}$ (-0.416) = -22.58° this is the angle you'll travel relative to the east.

To head northeast, you must either increase your gliding speed or increase your angle relative to the x-axis greater than 45°.

5 0

2 years ago

A yo-yo can be thought of as a solid cylinder of mass m and radius r that has a light string wrapped around its circumference (s

lbvjy [14]

Answer:

6.5 m/s^2

Explanation:

The net force acting on the yo-yo is

F_net = mg-T

ma=mg-T

now T= mg-ma

net torque acting on the yo-yo is

τ_net = Iα

I= moment of inertia (= 0.5 mr^2 )

α = angular acceleration

τ_net = 0.5mr^2(a/r)

Tr= 0.5mr^2(a/r)

(mg-ma)r=0.5mr^2(a/r)

a(1/2+1)=g

a= 2g/3

a= 2×9.8/3 = 6.5 m/s^2

4 0

2 years ago

A 92-kg skier is sliding down a ski slope that makes an angle of 30 degrees above the horizontal direction. The coefficient of k

9966 [12]

Answer:

a = 4.05 m/s²

Explanation:

Known data

m= 92 kg : mass of the skier

θ =30° :angle θ of the ski slope with respect to the horizontal direction

μk= 0.10 : coefficient of kinetic friction

g = 9.8 m/s² : acceleration due to gravity

Newton's second law:

∑F = m*a Formula (1)

∑F : algebraic sum of the forces in Newton (N)

m : mass s (kg)

a : acceleration (m/s²)

We define the x-axis in the direction parallel to the movement of the block on the ramp and the y-axis in the direction perpendicular to it.

Forces acting on the skier

W: Weight of the skier : In vertical direction

N : Normal force : perpendicular to the ski slope

f : Friction force: parallel to the ski slope

Calculated of the W

W= m*g

W= 92kg* 9.8 m/s² = 901,6 N

x-y weight components

Wx= Wsin θ= 901,6 N *sin 30° = 450.8 N

Wy= Wcos θ = 901,6 N *cos 30° =780.8 N

Calculated of the N

We apply the formula (1)

∑Fy = m*ay ay = 0

N - Wy = 0

N = Wy

N = 780.8 N

Calculated of the f

f = μk* N= 0.10*780.8 N

f = 78.08 N

We apply the formula (1) to calculated acceleration of the skier:

∑Fx = m*ax , ax= a : acceleration of the block

Wx - f = m*a

450.8- 78.08 = ( 92)*a

372.72 = (92)*a

a = (372.72)/ (92)

a = 4.05 m/s²

6 0

2 years ago