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kumpel [21]
3 years ago
5

Io experiences tidal heating primarily because __________.

Physics
1 answer:
Inga [223]3 years ago
7 0

Answer:

the orbit of Io around Jupiter causes the tidal heating.

Explanation:

Io's elliptical orbit causes the tidal force to vary as it orbits Jupiter.

Because Jupiter is a very massive planet, it has a very large gravitational pull on its moons, so the side of Io facing the planet is experiencing a greater gravitational pull than the opposite side.  This causes a distortion in the shape of the Io, and a friction that causes the tidal heating.

This friction generates heat currents, which is what makes Io the place with the most volcanic activity in the solar system.

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Two identical waves are traveling toward each other in the same medium. One has a positive amplitude, meaning that its peaks onl
Olin [163]
The correct answer is Destructive Interference.

Consider the image attached below. Two waves are travelling towards each other. Blue wave always has a positive peak and the red wave always has a negative peak.

Now imagine these waves are moving through a rope. If blue waves will try to move the rope in positive direction, the red wave will pull it down, and thus the two waves will cancel the effect of each other. Thus resulting in a destructive interference. 

7 0
4 years ago
Read 2 more answers
If the airman had a mass of 80 kg, find the magnitude of the air drag acting on him when he reached terminal velocity of 54 m/s
Vlad1618 [11]

The magnitude of the air drag is 784 N

Explanation:

An object falling down reaches the terminal velocity when the magnitude of the air drag acting on it becomes equal to the weight of the object. Mathematically, this condition can be written as:

F_D = mg

where

F_D is the magnitude of the air drag

m is the mass of the object

g is the acceleration of gravity

In this problem, we have

m = 80 kg is the mass of the airman

g=9.8 m/s^2 is the acceleration of gravity

Substituting into the formula, we find:

F_D = (80)(9.8)=784 N

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4 0
3 years ago
How would the gravitational force between the earth and the moon change if the moon had half the mass?
Vikki [24]

The moon's gravity, combined with the waltz of Earth and the moon around their center of mass, forces the oceans into an oval shape, with two simultaneous high tides. ... If the moon were half its mass, then the ocean tides would have been correspondingly smaller and imparted less energy to it.

5 0
3 years ago
The route followed by a hiker consists of three displacement vectors, X, Y and Z. Vector X is along a measured trail and is 1430
poizon [28]

Answer:

  • magnitude : 1635.43 m
  • Angle: 130°28'20'' north of east

Explanation:

First, we will find the Cartesian Representation of the \vec{X} and \vec{Y} vectors. We can do this, using the formula

\vec{A}= | \vec{A} | \ ( \ cos(\theta) \ , \ sin (\theta) \ )

where | \vec{A} | its the magnitude of the vector and θ the angle. For  \vec{X} we have:

\vec{X}= 1430 m \ ( \ cos( 42 \°) \ , \ sin (42 \°) \ )

\vec{X}= ( \ 1062.70 m \ , \ 956.86 m \ )

where the unit vector \hat{i} points east, and \hat{j} points north. Now, the \vec{Y} will be:

\vec{Y}= - 2200 m \hat{j} = ( \ 0 \ , \ - 2200 m \ )

Now, taking the sum:

\vec{X} + \vec{Y} + \vec{Z} = 0

This is

\vec{Z} = - \vec{X} - \vec{Y}

(Z_x , Z_y) = - ( \ 1062.70 m \ , \ 956.86 m \ ) - ( \ 0 \ , \ - 2200 m \ )

(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 2200 m \ - \ 956.86 m \ )

(Z_x , Z_y) = ( \ - 1062.70 m \ ,  \ 1243.14 m\ )

Now, for the magnitude, we just have to take its length:

|\vec{Z}| = \sqrt{Z_x^2 + Z_y^2}

|\vec{Z}| = \sqrt{(- 1062.70 m)^2 + (1243.14 m)^2}

|\vec{Z}| = 1635.43 m

For its angle, as the vector lays in the second quadrant, we can use:

\theta = 180\° - arctan(\frac{1243.14 m}{ - 1062.70 m})

\theta = 180\° - arctan( -1.1720)

\theta = 180\° - 45\°31'40''

\theta = 130\°28'20''

5 0
3 years ago
You pull downward with a force of 27.3 N on a rope that passes over a disk-shaped pulley of mass 1.43 kg and radius 0.0792 m. Th
Salsk061 [2.6K]

Answer:

The tension in the left side string = 17.21 N

The tension in the right side string = F = 27.3 N

Explanation:

Given that

F= 27.3 N

M= 1.43 kg ,r= 0.0792 m

Moment of inertia of disk ,I = 0.5 m r²

I = 0.5 x 1.43 x 0.0792² = 0.0044 kg.m²

m= 0.7 kg

Lets take linear acceleration of system is a m/s²

Lets take tension in left side string = T

From Newtons law

T- mg = ma  

T- 0.7 x 10 = 0.7 a  ----------1

(F - T) r = I α      

α = Angular acceleration of disk

a = α  r

(F - T) r = I α  

(F - T) r² = I a  

( 27.3 - T) x 0.0792² = 0.0044 a        --------2

Form equation 1 and 2

a= 1.42 T - 10 m/s²

a = 1.42 ( 27.3 - T)  m/s²

1.42 T - 10 = 38.9 - 1.42 T

T=17.21 N

The tension in the right side string = F = 27.3 N

5 0
3 years ago
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