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3 years ago

The image messed up but you get the point. Don't be mean and just answer to get the points, please and thank you.

Physics

1 answer:

Katarina [22]3 years ago

4 0

Answer:

Answer is A, it will pass through to focal point after reflecting.

Explanation:

I had the same question in a test, Sorry that you had to do this question in middle school.

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How is sound Important in different cultures

katrin2010 [14]

Well, some sounds come from instruments or it could be part of a certain practiced natural sound made by the human body. It can help define what the culture is by the certain sounds that are affiliated with. It shows that those sounds come from those specific people.

7 0

3 years ago

The froghopper, Philaenus spumarius, holds the world record for insect jumps. When leaping at an angle of 58.0 above the horizon

kirza4 [7]

Answer:

A) 3.79 m/s B) 1.33 m

Explanation:

Horizontal movement:
Once in the air, no forces act on the froghopper, so it keeps moving with the same initial horizontal speed.
This horizontal component, is the projection of the velocity vector on the horizontal direction (x-axis):

$v_{ox} = v_{o} *cos (58.0 deg)$

The horizontal displacement can be simply calculated as follows:

$x = v_{ox} *t$

Vertical movement:
As the vertical and horizontal are independent each other (due to they are perpendicular, so there is no projection of one movement on the other), in the vertical direction, all happens as if would be a body thrown upward with a given initial vertical velocity.
This velocity can be found as the projection of the velocity vector on the vertical direction (y-axis):

$v_{oy} = v_{o} *sin (58.0 deg) (1)$

Once in the air, the gravity will cause that the froghopper be slow down, till it reaches to the maximum height, where it will come momentarily to an stop.
In that moment, we can apply the following kinematic equation:

$v_{fy} ^{2} -v_{oy} ^{2} = 2*g*h_{max}$

where vfy = 0, g = -9.8m/s2, hmax = 52.7 cm= 0.527 m
Replacing by the givens, we can solve for voy:

$v_{oy} =\sqrt{2*g*h_{max}} = \sqrt{2*9.8m/s2*0.527m} =3.21 m/s$

From the equation (1), we can solve for the magnitude of the initial velocity, v₀:

$v_{o} = \frac{v_{oy}}{sin 58.0} =\frac{3.21m/s}{0.848} = 3.79 m/s$

With the value of the magnitude of the initial velocity, we can find the horizontal component vox, as follows:

$v_{ox} = v_{o} *cos (58.0 deg) =\\ \\ 3.79 m/s * cos (58.0deg) = 2.01 m/s$

In order to know the horizontal distance travelled, we need to find the time that the insect was in the air.
We can use the equation for the vertical displacement, replacing this value by 0, as follows:

$y = 0 = v_{oy} *t -\frac{1}{2} * g *t^{2}$

Replacing by the givens, and rearranging terms, we can solve for t:

$t_{air} =\frac{2*v_{oy} }{g} = \frac{2*3.21 m/s}{9.8 m/s} = 0.66 s$

Finally, we find the horizontal displacement, as follows:

$x_{max} = v_{ox} *t_{air} = 2.01 m/s * 0.66 s \\ \\ x_{max} = 1.33 m$

The horizontal distance covered by the froghopper was 1.33 m.

4 0

3 years ago

Consider a particle launched at a horizontal velocity v0 from a height h above the ground. 1. Derive an expression for the time

Rina8888 [55]

Answer: $t=\sqrt{\frac{2h}{g} }$

Explanation: The time taken by the object to reach the ground depends up on the vertical velocity of the object.

The object horizontal velocity component only moves the object in forward direction and does not have any effect vertically

as the object is launched horizontally its initial vertical velocity is zero

$u_y=0$

the object experiences downward fall due to gravitational acceleration only since there is no air resistance

therefore a=g

the distance covered =s=h

now the kinematic equation could be used

$h=u_yt+0.5at^2\\h=0+0.5gt^2\\h=\frac{gt^2}{2}\\t^2=\frac{2h}{g} \\t=\sqrt{\frac{2h}{g} }$

5 0

3 years ago

Draw a winding path that covers a distance of 70 miles and finishes with a displacement 20 miles soulthwest of the starting poin

EleoNora [17]

Distance is indeed a scalar amount that also refers to "how the ground an object has encased", and the Displacement is a vector thing that leads "to the extent to which an object is located", and the further calculation can be defined as follows:

Given:

distance= 70 miles

displacement = 20 miles