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3 years ago

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An electric current is flowing through a long cylindrical conductor with radius a. The current density J is uniform in the cylin

der. In this problem we consider an imaginary cylinder with radius r around the axis AB where r

1 answer:

RUDIKE [14]3 years ago

4 0

Answer:

Check the explanation

Explanation:

Kindly check the attached images below to see the step by step explanation to the question above.

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What happens to the mass and volume if density increases

nikitadnepr [17]

They all stay the same regardless

7 0

4 years ago

Two arrows are fired horizontally with the same speed of

Fed [463]

Answer:

Explanation:

Given

mass of each arrow=0.1 kg

velocity of arrow=30 m/s

One arrow is fired u=due to east and another towards south

Momentum of first arrow

$P_1=0.1\left ( 30\hat{i}\right )=3\hat{i}$

$P_2=0.1\left ( -30\hat{j}\right )=-3\hat{j}$

Total momentum P

$P=P_1+P_2$

$P=3\hat{i}-3\hat{j}$

magnitude $|P|=\sqrt{3^2+3^2}=3\sqrt{2}$

direction

$tan\theta =\frac{-3}{3}$

$\theta =45^{\circ}$ clockwise w.r.t to east

6 0

3 years ago

Deimos's Orbit. Deimos orbits Mars at a distance of 23,460 km from the center of the planet and has a period of 1.263 days. Assu

mihalych1998 [28]

Answer:

M = 5.882 10²³ kg

Explanation:

Let's use Newton's second law to analyze the satellite orbit around Mars.

F = m a

force is universal attraction and acceleration is centripetal

a = v²/ R

the modulus of velocity in a circular orbit is constant

v= d/T

the distance of the cicule is

d =2pi R

a = 2pi R/T

we substitute

- G m M / R² = m ( $- \frac{4\pi^2 R^2 }{T^2 R}$ )

G M = $\frac{ 4\pi ^2 R^3 }{T^2 }$

M = $\frac{4 \pi ^2 R^3 }{ G T^2 }$

the distance R is the distance from the center of the planet Mars to the center of the satellite Deimos

R = 23460 km = 2.3460 10⁷ m

the period of the orbit is

T = 1,263 days = 1,263 day (24 h / 1 day) (3600s / h)

T = 1.0912 10⁵ s

let's calculate

M = $\frac{4 \pi ^2 ( 2.3460 \ 10^7)^3 }{5.67 10^{-11} \ (1.0912 \ 10^5)^2 }$

M = 509.73418 10²¹ /8.66640 10⁻¹

M = 58.817 10²² kg

M = 5.882 10²³ kg

4 0

3 years ago

A typical ten-pound car wheel has a moment of inertia of about 0.35kg⋅m2. The wheel rotates about the axle at a constant angular

QveST [7]

Answer:

The rotational kinetic energy of the rotating wheel is 529.09 J

Explanation:

Given;

moment of inertia I = 0.35kg⋅m²

number of revolutions = 35.0

time of revolution, t = 4.00 s

Angular speed (in revolution per second), ω = 35/4 = 8.75 rev/s

Angular speed (in radian per second), ω = 8.75 rev/s x 2π = 54.985 rad/s

Rotational kinetic energy, K = ¹/₂Iω²

Rotational kinetic energy, K = ¹/₂ x 0.35 x (54.985)²

Rotational kinetic energy, K = 529.09 J

Therefore, the rotational kinetic energy of the rotating wheel is 529.09 J

7 0

3 years ago

1) What describes the unusually large release of plasma from the sun's corona?

sergey [27]

1) The correct answer is CME:
In fact, CME stands for "Coronal Mass Ejection", and they are huge release of plasma and magnetic field from the corona of the Sun.

2) The only statement that could be true is "<span>X and Y are both within the solar system."
In fact, the distance between X and Y is 30.2 AU (1 AU is the distance between Earth and Sun). Pluto, the farthest planet from the Sun, is located at approximately 40 AU from the Sun: the distance between X and Y is smaller than this value, so they could be both in the solar system.</span>

4 0

4 years ago