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What change of state occurs during vaporization? liquid to gas gas to liquid solid to liquid liquid to solid

nignag [31]

Answer:

A is correct!

Explanation:

e2020

8 0

2 years ago

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Study the four transverse waves shown. Compare the properties of waves B, C, & D to that of wave A.

erica [24]

Wave D has the same frequency

5 0

2 years ago

In the long jump, an athlete launches herself at an angle above the ground and lands at the same height, trying to travel the gr

NikAS [45]

A) 2.64t

B) 2.64h

C) 2.64D

Explanation:

A)

The motion of the athlete is equivalent to the motion of a projectile, which consists of two independent motions:

- A uniform motion (constant velocity) along the horizontal direction

- A uniformly accelerated motion (constant acceleration) along the vertical direction

The time of flight of a projectile can be found from the equations of motion, and it is found to be

$t=\frac{2u sin \theta}{g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the time of flight is $t$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the time of flight on Mars will be:

$t'=\frac{2usin \theta}{g'}=\frac{2u sin \theta}{0.379g}=\frac{1}{0.379}t=2.64t$

B)

The maximum height reached by a projectile can be also found using the equations of motion, and it is given by

$h=\frac{u^2 sin^2\theta}{2g}$

where

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the maximum height is $h$ .

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. So, the maximum height reached on Mars will be:

$h'=\frac{u^2 sin^2\theta}{2g'}=\frac{u^2 sin^2\theta}{(0.379)2g}=\frac{1}{0.379}h=2.64h$

C)

The horizontal distance covered by a projectile is also found from the equations of motion, and it is given by

$D=\frac{u^2 sin(2\theta)}{g}$

where:

u is the initial speed

$\theta$ is the angle of projection

g is the acceleration due to gravity

In this problem, when the athlete is on the Earth, the horizontal distance covered is D.

When she is on Mars, the acceleration due to gravity is:

$g'=0.379 g$

where g is the acceleration due to gravity on Earth. Therefore, the horizontal distance reached on Mars will be:

$D'=\frac{u^2 sin(2\theta)}{g'}=\frac{u^2 sin(2\theta)}{(0.379)g}=\frac{1}{0.379}D=2.64D$

7 0

3 years ago

-x times -x what is the answer

Brilliant_brown [7]

Answer:

Explanation:

x

5 0

2 years ago

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A 91-kg astronaut and a 1300-kg satellite are at rest relative to the space shuttle. The astronaut pushes on the satellite, givi

WARRIOR [948]

Answer:

18.2145 meters

Explanation:

Using the conservation of momentum, we have that:

$m1v1 + m2v2 = m1'v1' + m2'v2'$

m1 = m1' is the mass of the astronaut, m2=m2' is the mass of the satellite, v1 and v2 are the inicial speed of the astronaut and the satellite (v1 = v2 = 0), and v1' and v2' are the final speed of the astronaut and the satellite. Then we have that:

$0 + 0 = 91*v1' + 1300*0.17$

$v1' = -1300*0.17/91 = -2.4286\ m/s$

The negative sign of this speed just indicates the direction the astronaut goes, which is the opposite direction of the satellite.

If the astronaut takes 7.5 seconds to come into contact with the shuttle, their initial distance is:

$distance = 2.4286 * 7.5 = 18.2145\ meters$

8 0

3 years ago