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

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A ball rolls up a slope. At the end of three seconds its velocity is 20 cm/s; at the end of eight seconds its velocity is 0. Wha

serg [3582]

Answer:

$a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

Explanation:

Data provided

initial velocity v₀=20 cm/s at time t=3s

final velocity vf=0 at time t=8 s

Required

Average Acceleration for the interval from 3s to 8s

Solution

Acceleration can be defined as the first derivative of velocity concerning time

$a_{acceleriation} =\frac{dv_{velocity}}{dt_{time}}\\a_{acceleriation} =\frac{v_{f}-v_{o} }{dt}\\ a_{acceleriation} =\frac{0-20cm/s }{8s-3s}\\ a_{acceleriation}=-4cm/s^{2}\\ or\\ a_{acceleriation}=-0.04m/s^{2}$

8 0

3 months ago

A locomotive is accelerating at 1.6 m/s2. it passes through a 20.0-m-wide crossing in a time of 2.4 s. after the locomotive leav

kicyunya [3294]

Response:

Once it has crossed, the locomotive requires 17.6 seconds to achieve a speed of 32 m/s.

Details:

The locomotive's acceleration is 1.6 $m/s^2$

The duration taken to pass the crossing is 2.4 seconds.

We can apply the motion equation, v = u + at, where v represents final velocity, u indicates initial velocity, a denotes acceleration, and t signifies time.

When the speed reaches 32 m/s, we have v = 32 m/s, u = 0 m/s, and a= 1.6 $m/s^2$ .

32 = 0 + 1.6 * t

t = 20 seconds.

Therefore, the locomotive attains a speed of 32 m/s after 20 seconds, and it passes the crossing in 2.4 seconds.

Thus, after clearing the crossing, it takes an additional 17.6 seconds to reach the speed of 32 m/s.

6 0

3 months ago

Read 2 more answers

For a given initial projectile speed Vo, calculate what launch angle A gives the longest range R. Show your work, don't just quo

ValentinkaMS [3465]

The ideal launch angle of 45° for achieving the greatest horizontal distance is only applicable when the starting height matches the final height.

In this scenario, you can demonstrate it as follows: 

the initial velocity is Vo 
the launch angle is α 

the initial vertical velocity is 
Vv = Vo×sin(α) 

horizontal velocity becomes 
Vh = Vo×cos(α) 

the total flight duration is the period required to return to a height of 0 m, thus 
d = v×t + a×t²/2 
where 
d = distance = 0 m 
v = initial vertical velocity = Vv = Vo×sin(α) 
t = time =? 
a = gravitational acceleration = g (= -9.8 m/s²) 
therefore 
0 = Vo×sin(α)×t + g×t²/2 
0 = (Vo×sin(α) + g×t/2)×t 
t = 0 (obviously, the projectile is at height 0 m at time = 0s) 
or 
Vo×sin(α) + g×t/2 = 0 
t = -2×Vo×sin(α)/g 

Now let's examine the horizontal distance. 
r = v × t 
where 
r = horizontal range =? 
v = horizontal velocity = Vh = Vo×cos(α) 
t = time = -2×Vo×sin(α)/g 
therefore 
r = (Vo×cos(α)) × (-2×Vo×sin(α)/g) 
r = -(Vo)²×sin(2α)/g 

To find the extreme points of r (max or min) with respect to α, the first derivative of r with regards to α must be determined and set to 0. 

dr/dα = d[-(Vo)²×sin(2α)/g] / dα 
dr/dα = -(Vo)²/g × d[sin(2α)] / dα 
dr/dα = -(Vo)²/g × cos(2α) × d(2α) / dα 
dr/dα = -2 × (Vo)² × cos(2α) / g 

As Vo and g are constants that are not equal to 0, the only solution for dr/dα to equal 0 is when 
cos(2α) = 0 
2α = 90° 
α = 45°

4 0

2 months ago