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3 years ago

What mechanism is most responsible for generating the internal heat of Io that drives its volcanic activity?

Physics

1 answer:

Ghella [55]3 years ago

8 0

Answer:

Tidal heating

Explanation:

Tidal force is the ability of a massive body to produce tides on another body. The tidal force depends on the mass of the body that produces the tides and the distance between the two bodies.

Tidal forces can cause the destruction of a satellite that orbits a planet or a comet that is too close to the Sun or a planet. When the orbiting body crosses the "Roche boundary", the tidal forces along the body are more intense than the cohesion forces that hold the body together.

Tidal friction is the force between the Earth's oceans and ocean floors caused by the gravitational attraction of the Moon. The Earth tries to transport the waters of the oceans with it, while the Moon tries to keep them under it and on the opposite side of the Earth. In the long term, tidal friction causes the Earth's rotation speed to decrease, thus shortening the day. In turn, the Moon increases its angular momentum and gradually spirals away from Earth. Finally, when the day equals the orbital period of the Moon (which will be about 40 times the length of the current day), the process will cease. Subsequently, a new process will begin when the power to raise tides from the Sun takes angular momentum from the Earth-Moon system. The Moon will then spiral towards Earth until it is destroyed when it enters the "Roche boundary."

Tidal heating

It is the warming caused by the tidal action on a planet or satellite. The most important example of tidal heating in the Solar System is the effect of Jupiter on its Io satellite, in which the tidal effects produce such high temperatures that the interior of the satellite melts, producing volcanism.

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The horizontal beam in Fig. E11.14 weighs 190 N, and its center of gravity is at its center. Find (a) the tension in the cable a

grandymaker [24]

Answer:

(a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

Explanation:

Given that,

Weight of beam= 190 N

Here, The center of gravity is at its center

According to figure,

The angle is

$\sin\theta=\dfrac{3}{5}$

The horizontal component is

$T_{x}=T\cos\theta$

The vertical component is

$T_{y}=T\sin\theta$

(a). We need to calculate the tension in the cable

Using formula of net torque acting on the pivot

$\sum\tau=F_{b}\times r+F_{w}\times r'-T\sin \theta\times r'$

Put the value into the formula

$0=190\times2+300\times 4-T\sin\theta\times 4$

$T\sin\theta\times 4=380+1200$

$T=\dfrac{1580\times5}{3\times 4}$

$T=658.33\ N$

(b). We need to calculate the horizontal components of the force exerted on the beam at the wall

Using formula of horizontal component

$F_{x}=T\cos\theta$

Put the value into the formula

$F_{x}=658.33\times\dfrac{4}{5}$

$F_{x}=526.66\ N$

(c). We need to calculate the vertical components of the force exerted on the beam at the wall

Using formula of vertical component

$F_{y}=F_{b}+F_{w}-T\sin\theta$

Put the value into the formula

$F_{y}=190+300-658.33\times\dfrac{3}{5}$

$F_{y}=95.002\ N$

Hence, (a). The tension in the cable is 658.33 N.

(b). The horizontal components of the force exerted on the beam at the wall is 526.66 N.

(c). The vertical components of the force exerted on the beam at the wall is 95.002 N.

3 0

3 years ago

A uniform electric field of 2 kNC-1 is in the x-direction. A point charge of 3 μC initially at rest at the origin is released. W

ANEK [815]

The electric force acting on the charge is given by the charge multiplied by the electric field intensity:
$F=qE$
where in our problem $q=3 \mu C= 3 \cdot 10^{-6} C$ and $E=2 kN/C=2000 N/C$ , so the force is
$F=(3 \cdot 10^{-6} C)(2000 N/C)=0.006 N$

The initial kinetic energy of the particle is zero (because it is at rest), so its final kinetic energy corresponds to the work done by the electric force for a distance of x=4 m:
$K(4 m)=W=Fd=(0.006 N)(4 m)=0.024 J$

8 0

3 years ago

the density of a steel is 7.8.find the mass of steel cube of side 10 cm. find volume of steel if the mass is 8kg

lukranit [14]

Answer:

Mass of the steel cube = 7800 kg

Volume of the steel = 1.025 cubic centimetre

Explanation:

Given:

The density of the steel = 7.8

Side of the cube = 12 cm

(1)The mass of steel cube :

We know that,

$Density = \frac{Mass}{Volume}$

We are given with density and sides of the cube

then volume of the cube

= $(side)^3$

= $10^3$

= 1000 cubic centimetre

Now

$7.8 = \frac{mass}{1000}$

$mass = 7.8 \times 1000$

mass = 7800 kg

(2)volume of steel:

Given the mass = 8 kg

$Density = \frac{Mass}{Volume}$

Substituting the values

$7.8 = \frac{8}{volume}$

$volume = \frac{8}{7.8}$

volume = 1.025 cubic centimetre

3 0

3 years ago

A 20-Kg child is on a swing attached to 3.0 m-long chains. The child swings back and forth, swinging out to a 60-degree angle. (

kvv77 [185]

Answer:

v = 29.4 m / s

Explanation:

For this exercise we can use the conservation of mechanical energy

Lowest starting point.

Em₀ = K = ½ m v²

final point. Higher

$Em_{f}$ = U = m g h

Let's use trigonometry to lock her up

cos 60 = y / L

y = L cos 60

Height is the initial length minus the length at the maximum angle

h = L - L cos 60

h = L (1- cos 60)

energy is conserved

Em₀ = Em_{f}

½ m v² = mgL (1 - cos 60)

v = 2g L (1- cos 60)

let's calculate

v² = 2 9.8 3.0 (1- cos 60)

v = 29.4 m / s

6 0

3 years ago

How much displacement will a spring with a constant of 120N / m achieve if it is stretched by a force of 60N?

Andrew [12]

Answer:

Explanation:

There's a formula for this:

$F = k*displacement$

F being force, k being the spring constant, and displacement being the change in x

We are given the force and the spring constant, so this is essentially isolating the Δx term. Do 60N/120N per meter. The newtons cancel out and you get a final answer of Δx = 0.5 meters

3 0

3 years ago