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Fynjy0 [20]
3 years ago
5

Who might be suffering from dissociative fugue?

Physics
2 answers:
Oxana [17]3 years ago
7 0

Answer:

man you found his self in texas

Explanation:

ap.ex

svetlana [45]3 years ago
6 0

Answer:

More common in adults.

Explanation:

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For problems 2-3, a Ferrari accelerates from 0-60.0 miles per hour in 2.50 seconds.
Montano1993 [528]

Answer:

Explanation:

The first part of this question is simply asking us to convert the speed from miles per hour to meters per second:

60.0\frac{mi}{hr}*\frac{1hr}{3600sec}*\frac{1609.34m}{1mi}=26.8\frac{m}{s} choice C.

The next part wants us to use the equation for acceleration and find the acceleration:

a=\frac{v-v_0}{t} where v is final velocity, v0 is initial velocity, and t is time in seconds (which was one of the reasons we had to convert the initial velocity from 60.0 mph to m/s):

a=\frac{26.8-0}{2.5} and

a = 10.7 m/s/s, choice B.

6 0
3 years ago
The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This sp
Delvig [45]

Answer:

Explanation:

Here is the full question and answer,

The archerfish is a type of fish well known for its ability to catch resting insects by spitting a jet of water at them. This spitting ability is enabled by the presence of a groove in the roof of the mouth of the archerfish. The groove forms a long, narrow tube when the fish places its tongue against it and propels drops of water along the tube by compressing its gill covers.

When an archerfish is hunting, its body shape allows it to swim very close to the water surface and look upward without creating a disturbance. The fish can then bring the tip of its mouth close to the surface and shoot the drops of water at the insects resting on overhead vegetation or floating on the water surface.

Part A: At what speed v should an archerfish spit the water to shoot down a floating insect located at a distance 0.800 m from the fish? Assume that the fish is located very close to the surface of the pond and spits the water at an angle 60 degrees above the water surface.

Part B: Now assume that the insect, instead of floating on the surface, is resting on a leaf above the water surface at a horizontal distance 0.600 m away from the fish. The archerfish successfully shoots down the resting insect by spitting water drops at the same angle 60 degrees above the surface and with the same initial speed v as before. At what height h above the surface was the insect?

Answer

A.) The path of a projectile is horizontal and symmetrical ground. The time is taken to reach maximum height, the total time that the particle is in flight will be double that amount.

Calculate the speed of the archer fish.

The time of the flight of spitted water is,

t = \frac{{2v\sin \theta }}{g}

Substitute 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}} for g and 60^\circ  for \theta in above equation.

t = \frac{{2v\sin 60^\circ }}{{9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}}}\\\\ = \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\  

Spitted water will travel 0.80{\rm{ m}} horizontally.

Displacement of water in this time period is

x = vt\cos \theta

Substitute \left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2} for t\rm 60^\circ[tex] for [tex]\theta and 0.80{\rm{ m}} for x in above equation.

\\0.80{\rm{ m}} = v\left( {0.1767\;v} \right){{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\left( {\cos 60^\circ } \right)\\\\0.80{\rm{ m}} = {v^2}\left( {0.1767{\rm{ }}} \right)\frac{1}{2}{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}\\\\v = \sqrt {\frac{{2\left( {0.80{\rm{ m}}} \right)}}{{0.1767\;{{\rm{m}}^{ - 1}} \cdot {{\rm{s}}^2}}}} \\\\ = 3.01{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\

B.) There are two component of velocity vertical and horizontal. Calculate vertical velocity and horizontal velocity when the angle is given than calculate the time of flight when the horizontal distance is given. Value of the horizontal distance, angle and velocity are given. Use the kinematic equation to solve the height of insect above the surface.

Calculate the height of insect above the surface.

Vertical component of the velocity is,

{v_v} = v\sin \theta

Substitute 3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}} for v and 60^\circ  for \theta in above equation.

\\{v_v} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\sin 60^\circ \\\\ = 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\

Horizontal component of the velocity is,

{v_h} = v\cos \theta

Substitute 3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}} for v and 60^\circ  for \theta in above equation.

\\{v_h} = \left( {3.01\;{\rm{m}} \cdot {{\rm{s}}^{ - 1}}} \right)\cos 60^\circ \\\\ = 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}\\

When horizontal ({0.60\;{\rm{m}}} distance away from the fish.  

The time of flight for distance (d) is ,

t = \frac{d}{{{v_h}}}

Substitute 0.60\;{\rm{m}} for d and 1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}} for {v_h} in equation t = \frac{d}{{{v_h}}}

\\t = \frac{{0.60\;{\rm{m}}}}{{1.505{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}}}\\\\ = 0.3987{\rm{ s}}\\

Distance of the insect above the surface is,

s = {v_v}t + \frac{1}{2}g{t^2}

Substitute 2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}} for {v_v} and 0.3987{\rm{ s}} for t and - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}} for g in above equation.

\\s = \left( {2.6067{\rm{ m}} \cdot {{\rm{s}}^{ - 1}}} \right)\left( {0.3987{\rm{ s}}} \right) + \frac{1}{2}\left( { - 9.8{\rm{ m}} \cdot {{\rm{s}}^{ - 2}}} \right){\left( {0.3987{\rm{ s}}} \right)^2}\\\\ = 0.260{\rm{ m}}\\

7 0
3 years ago
Please please answer click on image to see full pic
NISA [10]
Hope this may help you

4 0
3 years ago
A box of weight w=2.0N accelerates down a rough plane that is inclined at an angle ϕ=30∘ above the horizontal, as shown (Figure
Nesterboy [21]

Answer:

<h2>The work done is 0.882 Joules.</h2>

Explanation:

To calculate the work, we need to find all forces that are involved in the movement.

As you can analyse, the body is the force that make the box to move, and the friction force is opposite to it, we cannot forget about the friction. So, we have to calculate the resultant force to this context.

F_{R}=W_{x}-F_{k}

So, to find W_{x} we use: W_{x}=W.sen\phi; where W=2N;\phi = 30\°

W_{x}=2N.sin30\°=1\frac{1}{2}N=1N

The friction force would be:

F_{k}=\mu_{k}N=(0.30)(1.7N)=0.51

Then, the resultant force is:

F_{R}=W_{x}-F_{k}

F_{R}=1N-0.51N=0.49N

Now, we calculate the work: W=F_{R} d

W=(0.49N)(1.8m)=0.882J

Therefore, the work done is 0.882 Joules.

7 0
4 years ago
Read 2 more answers
What causes earthquakes and volcanic erruptions
Scorpion4ik [409]
Tectonic plates under the ground ,either go on top of each other or crash.
6 0
4 years ago
Read 2 more answers
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