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nika2105 [10]
2 years ago
9

What’s your favorite stranger things character.

Physics
2 answers:
RoseWind [281]2 years ago
7 0

Answer:

<em>My favorite stranger things character is</em><em> Eleven</em><em> also known as</em><em> Millie Bobby Brown</em>

Explanation:

<u><em>I hope this helps and have a good day!</em></u>

VMariaS [17]2 years ago
3 0

Answer:

none

Explanation:

stranger mid mid mid mid

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When a skateboarder rides down a hill, the normal force exerted on the skater by<br> the hill is
Luda [366]

Newton's second law allows us to find that the correct answer is:

  • Less than weight of the skater

Newton's second law states that the net force and is propositional to the mass and acceleration of the body

For these problems it is essential to set a reference system, with respect to which to carry out the measurements, in this case we set a coordinate system with the x axis parallel to the plane and positive in the direction of movement and the y axis perpendicular to the plane.

In the attachment we can see a free-body diagram of the problem, let's work each axis separately

x-axis

         Wₓ -fr = m a

y-axis

         N - W_y = 0

         N = W_y

Where Wₓ and W_y are the components of the weight, fr the friction force that opposes the movement, m the mass and the acceleration of the body

let's use trigonometry to find the components of the weight

        cos θ = \frac{W_y}{W}

        sin θ = \frac{W_x}{W}

         W_y = W cos θ

         Wₓ = W sin θ

we substitute

          N = mg cos θ

From this equation we can see that the normal is less than the weight of the body.

In conclusion using Newton's second law we find that the correct answer is:

  • Less than th weight of the skater

 

Learn more about Newton's second law here:

brainly.com/question/13685393

3 0
2 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
A passenger at the rear of a train traveling at 15 m/s relative to earth throws a baseball with a speed of 15 m/s in the directi
matrenka [14]
Relative to the earth, the trains' speed is 15 m/s.
Relative to the train, the ball's speed is -15 m/s.

Therefore, relative to earth, the ball's speed is
15 - 15 = 0

Answer: 0
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Answer:

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Explanation:

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A bus travels North for 10 miles then turns around and
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The displacement of the bus is 6.25 miles.
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