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IrinaVladis [17]
2 years ago
15

Io, a satellite of Jupiter, is the most volcanically active moon or planet in the solar system. It has volcanoes that send plume

s of matter over 500 km high. Due to the satellite's small mass, the acceleration due to gravity on Io is only 1.81 m/s2, and Io has no appreciable atmosphere. Assume that there is no variation in gravity over the distance traveled. Part A What must be the speed of material just as it leaves the volcano to reach an altitude of 490 km
Physics
2 answers:
svetlana [45]2 years ago
4 0

Answer:

1.33 km/s

Explanation:

acceleration due to gravity, g = 1.81 m/s²

height, h = 490 km = 490000 m

Let u is the initial speed and the final speed is zero.

use third equation of motion

v² = u² - 2 g h

0 = u² - 2 x 1.81 x 490000

u² = 1773800

u = 1331.84 m/s

u = 1.33 km/s

Thus, the initial speed of the material is 1.33 km/s.

Mamont248 [21]2 years ago
3 0

Answer:

1331.84 m/s

Explanation:

t = Time taken

u = Initial velocity

v = Final velocity = 0

s = Displacement = 490 km

a = Acceleration

g = Acceleration due to gravity = 1.81 m/s² = a

From equation of linear motion

v^2-u^2=2as\\\Rightarrow -u^2=2as-v^2\\\Rightarrow u=\sqrt{v^2-2as}\\\Rightarrow u=\sqrt{0^2-2\times -1.81\times 490000}\\\Rightarrow u=1331.84\ m/s

The speed of the material must be 1331.84 m/s in order to reach the height of 490 km

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Answer:

acceleration, a = 9.8 m/s²

Explanation:

'A ball is dropped from the top of a building' indicates that the initial velocity of the ball is zero.

u = 0 m/s

After 2 seconds, velocity of the ball is 19.6 m/s.

t = 2s, v = 19.6 m/s

Using

v = u + at

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2 years ago
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10. A girl pulls a wagon along a level path for a distance of 44 m. The handle of
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The work done on the wagon is 3549 J

Explanation:

The work done by a force when moving an object is given by

W=Fd cos \theta

where :

F is the magnitude of the force

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In this problem we have the following data:

F = 87 N is the magnitude of the force

d = 44 m is the displacement of the wagon

\theta=22^{\circ} is the angle between the direction of the force and the displacement

Substituting, we find the work done

W=(87)(44)(cos 22^{\circ})=3549 J

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4 0
2 years ago
A golfer hits a golf ball at an angle of 25.0° to the ground. if the golf ball covers a horizontal distance of 301.5 m, what is
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<u>Answer:</u>

 Maximum height reached = 35.15 meter.

<u>Explanation:</u>

Projectile motion has two types of motion Horizontal and Vertical motion.

Vertical motion:

         We have equation of motion, v = u + at, where v is the final velocity, u is the initial velocity, a is the acceleration and t is the time taken.

         Considering upward vertical motion of projectile.

         In this case, Initial velocity = vertical component of velocity = u sin θ, acceleration = acceleration due to gravity = -g m/s^2 and final velocity = 0 m/s.

        0 = u sin θ - gt

         t = u sin θ/g

    Total time for vertical motion is two times time taken for upward vertical motion of projectile.

    So total travel time of projectile = 2u sin θ/g

Horizontal motion:

  We have equation of motion , s= ut+\frac{1}{2} at^2, s is the displacement, u is the initial velocity, a is the acceleration and t is the time.

  In this case Initial velocity = horizontal component of velocity = u cos θ, acceleration = 0 m/s^2 and time taken = 2u sin θ /g

 So range of projectile,  R=ucos\theta*\frac{2u sin\theta}{g} = \frac{u^2sin2\theta}{g}

 Vertical motion (Maximum height reached, H) :

     We have equation of motion, v^2=u^2+2as, where u is the initial velocity, v is the final velocity, s is the displacement and a is the acceleration.

   Initial velocity = vertical component of velocity = u sin θ, acceleration = -g, final velocity = 0 m/s at maximum height H

   0^2=(usin\theta) ^2-2gH\\ \\ H=\frac{u^2sin^2\theta}{2g}

In the give problem we have R = 301.5 m,  θ = 25° we need to find H.

So  \frac{u^2sin2\theta}{g}=301.5\\ \\ \frac{u^2sin(2*25)}{g}=301.5\\ \\ u^2=393.58g

Now we have H=\frac{u^2sin^2\theta}{2g}=\frac{393.58*g*sin^2 25}{2g}=35.15m

 So maximum height reached = 35.15 meter.

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1 year ago
An electric buzzer is activated, then sealed inside a glass chamber. When all of the air is pumped out of the chamber, how is th
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The sound is increased because sound waves are in fact mech. waves which means the  that they can't travel through empty space and thus need a medium to travel through
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2 years ago
To practice Problem-Solving Strategy 17.1 for wave interference problems. Two loudspeakers are placed side by side a distance d
Nimfa-mama [501]

Complete Question

The compete question is shown on the first uploaded question

Answer:

The speed is  v  =  350 \  m/s  

Explanation:

From the question we are told that

   The  distance of separation is  d =  4.00 m  

  The distance of the listener to the center between the speakers is  I =  5.00 m

  The change in the distance of the speaker is by k  =  60 cm  =  0.6 \  m

    The frequency of both speakers is f =  700 \  Hz

Generally the distance of the listener to the first speaker is mathematically represented as

       L_1  =  \sqrt{l^2 + [\frac{d}{2} ]^2}

       L_1  =  \sqrt{5^2 + [\frac{4}{2} ]^2}

        L_1  =   5.39 \  m

Generally the distance of the listener to second speaker at its new position is  

          L_2  =  \sqrt{l^2 + [\frac{d}{2} ]^2 + k}

       L_2  =  \sqrt{5^2 + [\frac{4}{2} ]^2 + 0.6}

        L_2  =   5.64  \  m  

Generally the path difference between the speakers is mathematically represented as

        pD  = L_2 - L_1  =  \frac{n  *  \lambda}{2}

Here \lambda is the wavelength which is mathematically represented as

         \lambda =  \frac{v}{f}

=>    L_2 - L_1  =  \frac{n  *  \frac{v}{f}}{2}

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

=>    L_2 - L_1  =  \frac{n  *  v}{2f}  

Here n is the order of the maxima with  value of  n =  1  this because we are considering two adjacent waves

=>    5.64 - 5.39   =  \frac{1  *  v}{2*700}      

=>    v  =  350 \  m/s  

7 0
2 years ago
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