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3 years ago

The display of the aurora and the reflection of radio waves back to earth result from the _____.

2 answers:

8 0

The layer of electrically charged molecules and atoms which spans 40-250 miles above ground called ionosphere causes the display of the aurora and the reflection of radio waves back to earth.

svetlana [45]3 years ago

8 0

Answer:

The display of dawn and the reflection of radio waves back to Earth result from the ionosphere.

Explanation:

Ionosphere is one of the layers of the earth's atmosphere, characterized by containing charges of ions and electrons and covering between 60 km and 500 km altitude. The ionosphere - also known as thermosphere - is located between the mesosphere and the exosphere. This layer of the atmosphere is constantly ionized due to the radiation it receives from cosmic and solar rays. This layer is responsible for the display of dawn and the reflection of radio waves back to Earth.

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A box weighing 52.4 N is sliding on a rough horizontal floor with a constant friction force of magnitude LaTeX: ff. The box's in

german

Answer:

The magnitude of the friction force exerted on the box is 2.614 newtons.

Explanation:

Since the box is sliding on a rough horizontal floor, then it is decelerated solely by friction force due to the contact of the box with floor. The free body diagram of the box is presented herein as attachment. The equation of equilbrium for the box is:

$\Sigma F = -f = m\cdot a$ (Eq. 1)

Where:

$f$ - Kinetic friction force, measured in newtons.

$m$ - Mass of the box, measured in kilograms.

$a$ - Acceleration experimented by the box, measured in meters per square second.

By applying definitions of weight ( $W = m\cdot g$ ) and uniform accelerated motion ( $v = v_{o}+a\cdot t$ ), we expand the previous expression:

$-f = \left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$

And the magnitude of the friction force exerted on the box is calculated by this formula:

$f = -\left(\frac{W}{g} \right)\cdot \left(\frac{v-v_{o}}{t}\right)$ (Eq. 1b)

Where:

$W$ - Weight, measured in newtons.

$g$ - Gravitational acceleration, measured in meters per square second.

$v_{o}$ - Initial speed, measured in meters per second.

$v$ - Final speed, measured in meters per second.

$t$ - Time, measured in seconds.

If we know that $W = 52.4\,N$ , $g = 9.807\,\frac{m}{s^{2}}$ , $v_{o} = 1.37\,\frac{m}{s}$ , $v = 0\,\frac{m}{s}$ and $t = 2.8\,s$ , the magnitud of the kinetic friction force exerted on the box is:

$f = -\left(\frac{52.4\,N}{9.807\,\frac{m}{s^{2}} } \right)\cdot \left(\frac{0\,\frac{m}{s}-1.37\,\frac{m}{s} }{2.8\,s} \right)$

$f = 2.614\,N$

The magnitude of the friction force exerted on the box is 2.614 newtons.

5 0

3 years ago

An 67-kg jogger is heading due east at a speed of 2.3 m/s. A 70-kg jogger is heading 61 ° north of east at a speed of 1.3 m/s. F

ololo11 [35]

Answer

given,

mass of jogger = 67 kg

speed in east direction = 2.3 m/s

mass of jogger 2 = 70 Kg

speed = 1.3 m/s in 61 ° north of east.

jogger one

$P_1 = m_1 v_1 \hat{i}$

$P_1 = 67 \times 2.3\hat{i}$

$P_1 = 154.1 \hat{i}$

$P_2 = m_2 v_2 \hat{i} +m_2 v_2 \hat{j}$

$P_2 = 70\times v cos \theta \hat{i} +70\times v sin \theta \hat{j}$

$P_2 = 70\times 1.3 cos 61^0 \hat{i} +70\times 1.3 sin 61^0\hat{j}$

$P_2 = 44.12\hat{i} +79.59\hat{j}$

now

P = P₁ + P₂

$P = 198.22 \hat{i} +79.59 \hat{j}$

magnitude

$P = \sqrt{198.22^2 + 79.59^2}$

$P =213.60 kg.m/s$

$\theta = tan^{-1}\dfrac{79.59}{198.22}$

$\theta = 21.87$

the angle is $\theta = 21.87$ north of east

7 0

3 years ago

Ch 31 HW Problem 31.63 10 of 15 Constants In an L-R-C series circuit, the source has a voltage amplitude of 116 V , R = 77.0 Ω ,

Degger [83]

Answer:

a. I = 0.76 A

b. Z = 150.74

c. RL₁ = 34.41 , RL₂ = 602.58

d. RL₂ = 602.58

Explanation:

V₁ = 116 V , R₁ = 77.0 Ω , Vc = 364 V , Rc = 473 Ω

Using law of Ohm

V = I * R

I = Vc / Rc = 364 V / 473 Ω

I = 0.76 A

The impedance of the circuit in this case the resistance, capacitance and inductor

V = I * Z

Z = V / I

Z = 116 v / 0.76 A

Z = 150.74

The reactance of the inductor can be find using

Z² = R² + (RL² - Rc²)

Solve to RL'

RL = Rc (+ / -) √ ( Z² - R²)

RL = 473 (+ / -) √ 150.74² 77.0²

RL = 473 (+ / -) (129.58)

RL₁ = 34.41 , RL₂ = 602.58

The higher value have the less angular frequency

RL₂ = 602.58

ω = 1 / √L*C

ω = 1 / √ 602.58 * 473

f = 285.02 Hz

6 0

3 years ago

The radius of Saturn is about 10 times the radius of Venus and the mass is about 100 times that of Venus. How much larger is the

lora16 [44]

let the mass of Venus is M then mass of Saturn is 100 M

similarly if the radius of Venus is R then the radius of Saturn is 10 R

now the force of gravity on a man of mass "m" at the surface of Venus is given by

$F_1 = \frac{GMm}{R^2}$

now similarly the gravitational force on the man if he is at the surface of Saturn

$F_2 = \frac{G*100M*m}{(10R)^2}$

$F_2 = \frac{GMm}{R^2}$

so here if we divide the two forces

$\frac{F_1}{F_2} = 1$

so here we can say

F1 = F2

so on both planets the gravitational force will be same

7 0

3 years ago

The moment of a force is calculated from the product of the ———— and the———— distance from the line of action of the force to th

bixtya [17]

Answer: It is the product of the (force)multiplied by the (perpendicular) distance from the line of action of the force to the pivot

Explanation:

7 0

3 years ago